Squaring the Circle Presentation
Copyright January 1, 2011 By Christopher Ricci
Email Address = Ricci4.4428828@yahoo.com
Introduction
I never cared much for academics. I graduated from High School by the skin of my teeth. I probably would've failed even at that had it not been for the perseverance of my dear mother. Undoubtedly, of all the subjects I took in school I despised mathematics the most. If someone would've told me back then that I would one day endeavor to solve a problem this complex, I certainly would not have believed them. Frankly, I couldn't care less whether or not the circle could be squared & would not have bothered with it at all had it not been for a single statement I heard about seven years ago. I was talking with an old friend of mine & he brought this subject to my attention. He asserted that even God Himself could not square the circle using only compass & straightedge. Something moved within me the very moment he spoke those words. It was like the Lord was reminding me that "with men this is impossible; but with God all things are possible." (Matt. 19:26). It was like He had taken that audacious statement as a challenge to His omniscience & was going to prove him wrong; "for with God nothing [including the squaring of the circle] shall be impossible." (Luke 1:37). Afterall, He created the compass and straightedge & the circle and square! I mean, can this trivial task be any more impossible than splitting the Red Sea or making the sun stand still in the sky? Of course not. There is & can be no doubt about it: God is certainly capable of accomplishing this. And if God can do it then Jesus Christ could do it too. For all power on earth was delivered unto him, (Luke 10:22; John 3:35). And if Christ could do it, then we can do it. (With Divine assistance of course). Jesus said that if we had faith as miniscule as a mustard seed we could move mountains, (Matt. 21:21). And through this faith we could do even greater things than he did, (John 14:12). Surely, squaring the circle cannot be any more impossible than walking on water, healing the sick, restoring sight to the blind, or raising the dead! So I embarked on this journey seven years ago in full faith that God would glorify His Holy Name. I prayed fervently along the way for a solution & the following demonstration is the answer I received.
Prelude
It never fails, every time I contemplate the circumference of a circle I am reminded of the mathematical numberline. In it we have positive & negative numbers traveling infinitely in opposite directions. They will never collide so long as they travel this way. But don't they also travel infinitely toward each other? The negative numbers can travel endlessly (or decimally) toward the positive, while the positive numbers can travel endlessly (or decimally) toward the negative. No matter how close they get they will never actually meet. Because to do so transforms the negative numbers into the positive & the positive numbers into the negative. And no matter how much the value of zero is crushed between them it will always be there; for it is intuitively obvious that something cannot become nothing. No matter how small something gets it can always be cut in half. The value of zero, therefore, becomes an infinity within itself even though it has circumscribed limitations.
This is precisely why the value of pi has never been ascertained. There is no beginning or ending to the geometrical shape of a circle. For as soon as we assign a beginning to it & draw a perfect arc back upon itself, it can never meet with the point of origin again. As the value of pi infinitely increases the gap between the starting point & the point of destination infinitely decreases. But they will never actually collide. Because to do so transforms the beginning of the arc into the ending & the ending of the arc into the beginning. What we really have then is a ray bent into an arc which has a beginning (only because we've assigned one to it), but no ending, for it can never meet with the point of origin again. For the point of destination to merge with the point of origin is to go too far. The ray overlaps itself; Alpha overtakes Omega; the two become one; either a point is lost or a single point has been counted twice. But for these two points not to merge at all doesn't go far enough & leaves the circle incomplete. This presents a mind-boggling prodigy: It shows that it is possible to have a ray travel INFINITELY toward a point that is directly in its path, and yet NEVER collide with it. It's like the Lord is even able to take eternity, an utterly boundless expanse, place it inside a bottle & simply pop a cork in the top. It retains its infinite, limitless nature despite the fact that it’s confined within a limited finite space. (This seems to support the assertions of a growing number of physicists regarding multiple dimensions & parallel universes. I think it also has a lot to do with predestination. But I won’t delve into that now). It’s like the Lord sees all of eternity all at once as if encapsulated in a child's snow globe even though there is no beginning or end to it. Friends, I'll tell ya, the Lord never ceases to amaze me!
The frustration man experiences over the value of pi lies within the square. Man can measure the square. But he becomes frustrated because he cannot do the same thing with a circle. He is frustrated for nothing because it stems from the presumption that he could measure the square in the first place. The fact is man cannot measure the perimeter of a square for the very same reason he cannot measure the circumference of a circle. In fact, pi will pose the same problem with all enclosed geometrical figures for they are just circles bent (not broken) into various arcs & angles. Surely, if we take a square swimming pool & fill it with water, the sides will bow out equidistant from the center point & form a perfect circle. Even though this is very easy to visualize & is most certainly true, it is not possible to duplicate the phenomenon geometrically. (So they say). That's because in order to accurately measure any enclosed figure we would need to establish a starting point and the moment we do, we "break" the line & it can never be "reconnected" because pi stands in the way. Inevitably, we run into the very same situation described in the above numberline analogy.
To demonstrate my point, try to think of it this way. You have one large square that is divided into 16 squares. There are four rows of four squares. Hence, there are four squares per side of the large square. The area of the large square is 4 rows x 4 squares per row = 16. The perimeter of the large square is 4 sides x 4 squares per side = 16. But if you consider these smaller squares as our unit of measurement, you will find that our calculations for the perimeter are incorrect. You will find that there are only 12 squares around the perimeter; not 16. In multiplying the four sides by the number of squares on each side we've counted the corner squares twice. Every child who has ever tried to build a square box has found this out. The child cuts out a square piece of plywood for the bottom. And then he cuts four boards for the sides; each equal to the side length of the plywood square. But as he desparately struggles to assemble the thing he begins to realize that there is something wrong. The square he has constructed from the four boards is larger than the square piece of plywood. Obviously, this is because he did not take into account the width of the boards. Now this is a very fundamental mistake. But it is perfectly analogous to the situation above. The way we measure squares is fundamentally flawed for in multiplying the measurement of one side by the number of sides, we’ve counted the corner points twice. Surely then, if we begin to build thereupon the finished structure is bound to collapse; for it is founded upon error.
I guess what I'm trying to say is that it is essential that we find a way to thrust the square into the same transcendental world as the circle; for there's no doubt it belongs there. In our quest to accomplish this feat it must be remembered that the square is a "slice" of a cube & is Euclidean in nature; while the circle is a "slice" of a sphere & is spherical in nature. When we take the two of them and lay them side by side within (not upon) a flat surface, we've unwittingly performed a mathematical function. Obviously, the square is perfectly at home there. But we've done injustice to the circle: We've deported it from its comfortable home in spherical geometry, stripped it of its double curvature, & thrust it into flatland; a world utterly foreign to it. If we hope to rectify and/or square the circle in a literal sense we need to balance the equation. We need to give to the square what we took from the circle. And we need to do it without the square being aware of it. In other words, the square must inherit an intimate relationship with pi, a kind of marriage with the circle. And this must be accomplished without making any visible changes to the diagram. That way it can still be drawn with the primitive instruments at our disposal; namely, a straightedge & unmarked compass. This is essentially what I've done. Consider now, if you will, the following proposal.
This presentation naturally divides itself into five parts. They are:
* Part One: Squaring the Circle precisely on Riemannian Manifolds.
* Part Two: [Plates 1-10] Compass & straightedge constructions.
* Part Three: [Plates 11-14] Squaring the Circle with the Golden Mean.
* Part Four: [Plates 15-28] Circle Rectification.
* Part Five: [Plates 29-33] Squaring the Circle in Area.
Part One
Squaring the Circle precisely on Riemannian Manifolds.
Here we have three circles on a flat sheet of paper: The top one is labeled Circle [A]; the middle one is Circle [B]; and the bottom one is Circle [C]. Inscribed within these circles are three of the most elementary shapes in all of geometry; a circle, a triangle, and a square respectively. All six of these figures are related geometrically & can be drawn with compass & straightedge. The technique for their construction will be outlined in forthcoming diagrams. This diagram is all I need to demonstrate that the circle can be squared precisely on Riemannian manifolds using only the primitive tools of plane geometry.
Squaring the Circle precisely on Riemannian Manifolds.
Here we have three circles on a flat sheet of paper: The top one is labeled Circle [A]; the middle one is Circle [B]; and the bottom one is Circle [C]. Inscribed within these circles are three of the most elementary shapes in all of geometry; a circle, a triangle, and a square respectively. All six of these figures are related geometrically & can be drawn with compass & straightedge. The technique for their construction will be outlined in forthcoming diagrams. This diagram is all I need to demonstrate that the circle can be squared precisely on Riemannian manifolds using only the primitive tools of plane geometry.
Part One: Circle [B] Construction
I will begin the demonstration by constructing the equilateral triangle inscribed within Circle [B], and make all subsequent constructions from there.
1). First, we draw a random line segment and label the endpoints [A] & [B] accordingly.
2). Anchor the compass on point [A], extend it out to point [B], and sweep a green arc counterclockwise.
3). Anchor the compass at point [B], extend it out to point [A], and sweep a yellow arc such that it intersects the green arc. Label the point of intersection with the letter [C].
4). Join points [AC] & [BC] as shown.
I will begin the demonstration by constructing the equilateral triangle inscribed within Circle [B], and make all subsequent constructions from there.
1). First, we draw a random line segment and label the endpoints [A] & [B] accordingly.
2). Anchor the compass on point [A], extend it out to point [B], and sweep a green arc counterclockwise.
3). Anchor the compass at point [B], extend it out to point [A], and sweep a yellow arc such that it intersects the green arc. Label the point of intersection with the letter [C].
4). Join points [AC] & [BC] as shown.
Now, we must determine the circumradius (i.e., the radius of Circle [B] around the triangle).
1). Anchor the compass on point [A], extend it out to point [C], and sweep the blue arc counterclockwise.
2). Anchor the compass on point [C], extend it out to point [A], and sweep the pink arc such that it intersects the blue arc. Label the point of intersection with letter [F].
3). Anchor the compass on point [A], extend it out to point [B], and sweep the orange arc counterclockwise.
4). Anchor the compass on point [B], extend it out to point [A], and sweep the yellow arc such that it intersects the orange arc. Label the point of intersection with letter [E].
5). Join points [FB] and points [CE] with a line segment as shown. Label the point of intersection with the letter [D].
1). Anchor the compass on point [A], extend it out to point [C], and sweep the blue arc counterclockwise.
2). Anchor the compass on point [C], extend it out to point [A], and sweep the pink arc such that it intersects the blue arc. Label the point of intersection with letter [F].
3). Anchor the compass on point [A], extend it out to point [B], and sweep the orange arc counterclockwise.
4). Anchor the compass on point [B], extend it out to point [A], and sweep the yellow arc such that it intersects the orange arc. Label the point of intersection with letter [E].
5). Join points [FB] and points [CE] with a line segment as shown. Label the point of intersection with the letter [D].
Complete Circle [B] around the equilateral triangle using line segment [DC] as the radius.
Part Two: Circle [A] Construction
To construct Circle [A] I will need to inscribe a square in Circle [B]. I prepare to do this by isolating Circle [B] & erasing the inscribed triangle for the time being to reduce clutter.
1). Begin with Circle [B] with diameter labeled [AC] & midpoint labeled with letter [B].
2). Anchor the compass on point [C], extend it out to point [B], and sweep a yellow arc such that it intersects the circumference of orange Circle [B]. Label the points of intersection with the letters [D] & [E] accordingly.
3). Draw the green line segment connecting points [D] & [E]. Label the point where the green line intersects the diameter with the letter [X] as shown.
4). Anchor the compass on point [X], extend it out to point [B], and sweep the pink arc such that it intersects the green line. Label the two points [F] & [G] respectively.
To construct Circle [A] I will need to inscribe a square in Circle [B]. I prepare to do this by isolating Circle [B] & erasing the inscribed triangle for the time being to reduce clutter.
1). Begin with Circle [B] with diameter labeled [AC] & midpoint labeled with letter [B].
2). Anchor the compass on point [C], extend it out to point [B], and sweep a yellow arc such that it intersects the circumference of orange Circle [B]. Label the points of intersection with the letters [D] & [E] accordingly.
3). Draw the green line segment connecting points [D] & [E]. Label the point where the green line intersects the diameter with the letter [X] as shown.
4). Anchor the compass on point [X], extend it out to point [B], and sweep the pink arc such that it intersects the green line. Label the two points [F] & [G] respectively.
Complete the inscribed square.
1). Draw a line through points [B] & [F] such that it intersects the circumference of orange Circle [B]. Label the points of intersection with the letters [I] & [J] as shown.
2). Draw a line through points [B] & [G] such that it intersects the circumference of orange Circle [B]. Label the points of intersection with the letters [H] & [K] as shown.
3). Complete the square by connecting points [H], [I], [J], & [K]; shown here as dotted black lines.
1). Draw a line through points [B] & [F] such that it intersects the circumference of orange Circle [B]. Label the points of intersection with the letters [I] & [J] as shown.
2). Draw a line through points [B] & [G] such that it intersects the circumference of orange Circle [B]. Label the points of intersection with the letters [H] & [K] as shown.
3). Complete the square by connecting points [H], [I], [J], & [K]; shown here as dotted black lines.
Complete Circle [A].
1). Anchor the compass on point [B], extend it out to the midpoint of one of the squares sides, and draw Circle [A]; colored red in the diagram. Circle [A] is now complete. Since it is formed within the inscribed square, Circle [A] is exactly half the size of Circle [B].
2). I now utilize the diagonals of the black square to inscribe a blue square. This square is not part of my original diagram, but it will be needed to construct the small circle inside Circle [A], as well as to construct Circle [C].
1). Anchor the compass on point [B], extend it out to the midpoint of one of the squares sides, and draw Circle [A]; colored red in the diagram. Circle [A] is now complete. Since it is formed within the inscribed square, Circle [A] is exactly half the size of Circle [B].
2). I now utilize the diagonals of the black square to inscribe a blue square. This square is not part of my original diagram, but it will be needed to construct the small circle inside Circle [A], as well as to construct Circle [C].
Now I will construct the smaller circle inside Circle [A]. Again, I will erase all unnecessary figures to reduce clutter.
1). Step #1: First, make one typical phi construction using a line segment from the midpoint of the blue square's baseline [X] to the square's vertex [Y] as a radius. Label the endpoint with the letter [Z] as shown.
1). Step #1: First, make one typical phi construction using a line segment from the midpoint of the blue square's baseline [X] to the square's vertex [Y] as a radius. Label the endpoint with the letter [Z] as shown.
Step #2: Anchor the compass on point [B], extend it out to point [Z], and draw the pink circle.
Step #3: Extend the diagonals of the blue square outwardly such that they intersect the circumference of the pink circle. Label the points of intersection with the letters [K], [L], [M], and [N] accordingly. Connect these four points to complete an inscribed square.
Steps 4-6: Complete the smaller circle within Circle [A]. Again, I've erased the blue square here in order to reduce clutter.
Step #4: Draw the green circle inside the black square as described earlier.
Step #5: Inscribe the yellow square using the diagonals of the black square.
Step #6: Draw the orange circle inside the yellow square. This completes the construction for the smaller circle inside Circle [A] in the base diagram.
Step #4: Draw the green circle inside the black square as described earlier.
Step #5: Inscribe the yellow square using the diagonals of the black square.
Step #6: Draw the orange circle inside the yellow square. This completes the construction for the smaller circle inside Circle [A] in the base diagram.
Part Three: Constructing Circle [C]. From henceforth I will consider Circle [A] as a Unit Circle (radius =1; area = pi), and make all subsequent calculations from that. Hence, the blue square would have side lengths of square root 2. Notice here, I have returned the blue square constructed previously & have rotated the diagram 45 degrees. This is not necessary, but it does make it much easier to describe & explain. From this diagram we will begin the process of constructing Circle [C]; located at the bottom of the base diagram. Draw a line through the vertices of the blue square & label all the points as shown; [BCA].
Step #1: Extend the diameter [BC] outwardly to point [X]. Next, extend a right angle line segment down to point [Y] as shown. This is a relatively simple task & can be accomplished in only four moves, so I won't bother with it's construction.
Step #1: Extend the diameter [BC] outwardly to point [X]. Next, extend a right angle line segment down to point [Y] as shown. This is a relatively simple task & can be accomplished in only four moves, so I won't bother with it's construction.
Step #2 in Circle [C] construction.
Anchor the compass on point [C], extend it out to point [D], and sweep a green arc counterclockwise such that it intersects the extended diameter line [BX], and label the point of intersection with the letter [E]. Line segment [AE] will have a total length equal to the Silver Ratio (1 + sq.rt.2 = 2.414213562).
Anchor the compass on point [C], extend it out to point [D], and sweep a green arc counterclockwise such that it intersects the extended diameter line [BX], and label the point of intersection with the letter [E]. Line segment [AE] will have a total length equal to the Silver Ratio (1 + sq.rt.2 = 2.414213562).
Step #3: Anchor the compass on point [A], extend it out to point [E], and sweep an orange arc counterclockwise until it intersects with vertical line extension [BY]. Label the point of intersection with the letter [F].
Step #4: The Silver Triangle: It's name obviously derived from the fact that it's hypotenuse is equal in length to the Silver Mean. Here we use the Pythagorean Theorem to calculate the length of segment [BF]. A squared - C squared = B squared. Hence, the length of this line segment is 2.197368227.
Step #5: Anchor the compass on point [B], extend it out to point [F], and draw the pink arc counterclockwise such that it intersects with the extended diameter [BX]. Label the point of intersection with the letter [G] accordingly. Line segment [BG] and line segment [BF] are now both equal in length.
Step #6: Now I will complete the square within which Circle [C] will be drawn. I have determined two sides of this square in previous steps. The remaining two sides are constructed as follows.
1). Anchor the compass on point [F], extend it out to point [B], and sweep a green arc clockwise.
2). Anchor the compass on point [G], extend it out to point [B], and sweep a purple arc counterclockwise such that the two arcs intersect at point [H].
3). Connect points [GH] & points [FH] with line segments as depicted. Now we have a completed square with side lengths of 2.197368227; which is the square root of the Silver Mean squared.
1). Anchor the compass on point [F], extend it out to point [B], and sweep a green arc clockwise.
2). Anchor the compass on point [G], extend it out to point [B], and sweep a purple arc counterclockwise such that the two arcs intersect at point [H].
3). Connect points [GH] & points [FH] with line segments as depicted. Now we have a completed square with side lengths of 2.197368227; which is the square root of the Silver Mean squared.
Step #7: Complete Circle [C]. Join the vertices of the square constructed in the previous frame to determine the center point; label it with the letter [I]. Anchor the compass there, extend it out to the midpoint of the square, and draw the orange circle. This completes the construction of Circle [C].
Step #8: Complete the inscribed square within Circle [C] as illustrated on the base diagram. Here, we simply repeat the procedure outlined in previous frames. Join the four points where the diagonals of the larger square intersect the circumference of the orange Circle [C] with line segments. The inscribed square is now complete.
Now, we have constructed all six figures represented in the base diagram. Considering Circle [A] as a unit circle, the following calculations will result.
Circle [A] Unit Circle: It has a diameter length of 2; Radius = 1; Area = pi. The smaller inscribed circle has a circumference of 5.441398093. This measurement is unaffected by the manifold; it remains the same whether it lies on a Euclidean or spherical surface.
Circle [B] has a diameter length of sq.rt. 8 = (2.828427125); Radius = The radius length = the square root of 2 (1.414213562); It's Area = 2 pi.
Circle [C] has a diameter length of the square root of those two values added together: The sq.rt. of (2 + sq.rt. 8) = 2.197368227; Radius = 1.0986841; It’s area = 1/2pi x Silver Mean = 3.7922376.
Circle [A] Unit Circle: It has a diameter length of 2; Radius = 1; Area = pi. The smaller inscribed circle has a circumference of 5.441398093. This measurement is unaffected by the manifold; it remains the same whether it lies on a Euclidean or spherical surface.
Circle [B] has a diameter length of sq.rt. 8 = (2.828427125); Radius = The radius length = the square root of 2 (1.414213562); It's Area = 2 pi.
Circle [C] has a diameter length of the square root of those two values added together: The sq.rt. of (2 + sq.rt. 8) = 2.197368227; Radius = 1.0986841; It’s area = 1/2pi x Silver Mean = 3.7922376.
Now, let's change the manifold. To demonstrate what I'm talking about imagine you walk outside in the middle of the night & gaze up into the pitch black sky and notice three full moons; each with it’s own unique shape inscribed within it’s equator circumference: In other words, the image you see in the night sky is exactly like the one appearing here on your screen.
Taking into consideration the aforementioned calculations, the circle is squared in two totally different worlds here; both Euclidean & Spherical Geometry. Under these conditions both the inscribed triangle and the inscribed square encompass the same amount of CURVED surface area as that which is bound by the small inscribed circle on Hemisphere [A]; with that being pi exactly. Also, the amount of curved surface area bound by the inscribed square & triangle is exactly equal to the amount of FLAT surface area bound by the flat equator Circle [A]; which is also pi exactly.
Several other neat things can be observed here. Since we are dealing with hemispheres rather than flat circles, their respective areas have now doubled in size. The area of Hemisphere [A] is now 2pi; the inscribed circle dividing the hemisphere’s surface into two separate sectors exactly equal to pi each. The area of hemisphere [B] is now 4pi; the inscribed triangle dividing it’s surface into four separate sectors with each one being exactly equal to pi. Hemisphere [C] is especially intriguing: The area bound by the square is pi exactly. The area of each of the four separate sectors lying outside the perimeter of that square are equal to the Form Factor Ratio (1.110720734). Hence, the two great geometries (Spherical & Euclidean) are brought together in the same diagram: One FLAT side length of the square (1.553773974) multiplied by the total amount of positively CURVED surface area lying outside the perimeter of that square (4.4428828) = The exact length of the equator circumference of the hemisphere!
Pretty neat huh?
I'm pretty sure this can be accomplished with other inscribed polygons as well (pentagons, hexagons, octagons, etc). For example: A pentagon embedded in the manifold of Sphere [A] will enclose a total amount of curved surface area equal to (2pi-3). But the math begins to get really complicated after that. Dealing with so many irrational numbers in combination with transcendental numbers makes the proof extremely difficult.
Taking into consideration the aforementioned calculations, the circle is squared in two totally different worlds here; both Euclidean & Spherical Geometry. Under these conditions both the inscribed triangle and the inscribed square encompass the same amount of CURVED surface area as that which is bound by the small inscribed circle on Hemisphere [A]; with that being pi exactly. Also, the amount of curved surface area bound by the inscribed square & triangle is exactly equal to the amount of FLAT surface area bound by the flat equator Circle [A]; which is also pi exactly.
Several other neat things can be observed here. Since we are dealing with hemispheres rather than flat circles, their respective areas have now doubled in size. The area of Hemisphere [A] is now 2pi; the inscribed circle dividing the hemisphere’s surface into two separate sectors exactly equal to pi each. The area of hemisphere [B] is now 4pi; the inscribed triangle dividing it’s surface into four separate sectors with each one being exactly equal to pi. Hemisphere [C] is especially intriguing: The area bound by the square is pi exactly. The area of each of the four separate sectors lying outside the perimeter of that square are equal to the Form Factor Ratio (1.110720734). Hence, the two great geometries (Spherical & Euclidean) are brought together in the same diagram: One FLAT side length of the square (1.553773974) multiplied by the total amount of positively CURVED surface area lying outside the perimeter of that square (4.4428828) = The exact length of the equator circumference of the hemisphere!
Pretty neat huh?
I'm pretty sure this can be accomplished with other inscribed polygons as well (pentagons, hexagons, octagons, etc). For example: A pentagon embedded in the manifold of Sphere [A] will enclose a total amount of curved surface area equal to (2pi-3). But the math begins to get really complicated after that. Dealing with so many irrational numbers in combination with transcendental numbers makes the proof extremely difficult.
Part Two: Compass & Straightedge Constructions
I'll need to employ a number of different geometrical shapes throughout the course of this presentation. All of them are easily constructed in quick succession using only compass & unmarked straightedge. Hence, I've decided to sweep through the drawing process first & produce all the required shapes in advance. That way they will be assembled collectively into a single base diagram [Plate #9]; and their respective dimensions will be delineated on [Plate #10]. This will help to move things along much more smoothly since I won't have to interrupt the demonstration each time I need to utilize a different shape. I can simply revert to the base diagram & extract the desired figures as I need them.
Plate #1: Open the compass to any random length & draw the Black Circle. Assign the letter (X) to the center point.
All subsequent shapes produced hereafter are intimately related to this initial circle since their creation is completely dependent upon it. It is also worth mentioning that this exhibit could not begin with the construction of a square instead. That's because a square cannot be created via compass & straightedge without the circle as the starting point. This is reminiscent of the Lord's relationship with man. In the world of Sacred Geometry it is universally accepted that the circle represents God & the spiritual world, since they are both infinite; having no beginning or ending. And the square represents man & the mortal world due to his obvious restrictions & inseparable connection to the temporal earth. Consequently, God is Absolute; the Origin & Creator of all things. Likewise, God (the Circle) is preeminent & without Him, man (the square) cannot exist. The problem of squaring the circle always has been, is now, and always will be a deeply spiritual one. So, there's really no way around it: The subject of Sacred Geometry is inevitable when discussing this particular topic; thus making it a recurring theme throughout the course of this demonstraion.
I'll need to employ a number of different geometrical shapes throughout the course of this presentation. All of them are easily constructed in quick succession using only compass & unmarked straightedge. Hence, I've decided to sweep through the drawing process first & produce all the required shapes in advance. That way they will be assembled collectively into a single base diagram [Plate #9]; and their respective dimensions will be delineated on [Plate #10]. This will help to move things along much more smoothly since I won't have to interrupt the demonstration each time I need to utilize a different shape. I can simply revert to the base diagram & extract the desired figures as I need them.
Plate #1: Open the compass to any random length & draw the Black Circle. Assign the letter (X) to the center point.
All subsequent shapes produced hereafter are intimately related to this initial circle since their creation is completely dependent upon it. It is also worth mentioning that this exhibit could not begin with the construction of a square instead. That's because a square cannot be created via compass & straightedge without the circle as the starting point. This is reminiscent of the Lord's relationship with man. In the world of Sacred Geometry it is universally accepted that the circle represents God & the spiritual world, since they are both infinite; having no beginning or ending. And the square represents man & the mortal world due to his obvious restrictions & inseparable connection to the temporal earth. Consequently, God is Absolute; the Origin & Creator of all things. Likewise, God (the Circle) is preeminent & without Him, man (the square) cannot exist. The problem of squaring the circle always has been, is now, and always will be a deeply spiritual one. So, there's really no way around it: The subject of Sacred Geometry is inevitable when discussing this particular topic; thus making it a recurring theme throughout the course of this demonstraion.
Plate #2: Inscribe the Blue Square.
I am not going to bother with the details of this construction. It can be accomplished with compass & straightedge in eleven moves. If you consider the Black Circle & it’s diameter as being already drawn (in Plate #1), then this can be done in only nine moves. This information is common knowledge & can be found on many math related websites. Once this first square has been completed, it’s diagonals can then be extended outwardly to provide the anchoring points for the vertices of all subsequent squares.
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* The next eight frames are basically self-explanatory. So rather than leave all this excess space blank, I've decided to utilize it to provide a few details of my journey along with other tidbits of pertinent information. These personal comments will be separated from the main body of the document by a row of x's (as above) and can be ignored since they are not of great importance to the overall demonstration.
I am not going to bother with the details of this construction. It can be accomplished with compass & straightedge in eleven moves. If you consider the Black Circle & it’s diameter as being already drawn (in Plate #1), then this can be done in only nine moves. This information is common knowledge & can be found on many math related websites. Once this first square has been completed, it’s diagonals can then be extended outwardly to provide the anchoring points for the vertices of all subsequent squares.
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* The next eight frames are basically self-explanatory. So rather than leave all this excess space blank, I've decided to utilize it to provide a few details of my journey along with other tidbits of pertinent information. These personal comments will be separated from the main body of the document by a row of x's (as above) and can be ignored since they are not of great importance to the overall demonstration.
PLATE #3: Make one standard phi construction off the right side of the Blue Square. Label the endpoint with the letter (Y).
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Here, we have the first of three Golden Mean constructions. This is a very special ratio & will be discussed several times throughout the course of this demonstration. It can be found throughout all of creation; from atoms to solar systems to galaxies. Gary Meisner is host to an excellent website that delineates many examples: www.goldennumber.net. Kepler compared this ratio to a “precious jewel” and said it was one of geometry’s greatest treasures. A friend of Leonardo da Vinci, Luca Pacioli, called it the Divine Proportion. Surely, he was Divinely inspired at the time he settled on that name: For there can be no doubt that it was considered the Divine Proportion long before he was even born.
As I indicated earlier, the problem of squaring the circle is as much theosophical as it is geometrical; if not more so. According to most major religious sects (including Jews, Muslims, Hindu, Catholics, and most Protestant denominations), the subject falls under a branch of sacred geometry which deals with the notion of an immortal God in heaven (i.e., the Circle) merging or reconciling Himself with mortal man on earth (i.e., the square). Of course in this particular case Christ would be the Golden Mean; for he is the quintessential element which makes this holy union possible. What a wonderful analogy: As Euclid so eloquently put it: "As the whole line is to the greater segment, so is the greater to the less.” Even so, As Almighty God is to Christ, so is Christ to man." And just as the Golden Ratio is generated by the combination of a circle & a square, even so, Christ was conceived by the Immaculate Conception of the Holy Spirit & the Virgin Mary. He represents the convergence of the eternal with the temporal; the sacred with the secular; the immortal with the mortal. The Divine and human nature are combined in a single person; Christ Jesus.
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Here, we have the first of three Golden Mean constructions. This is a very special ratio & will be discussed several times throughout the course of this demonstration. It can be found throughout all of creation; from atoms to solar systems to galaxies. Gary Meisner is host to an excellent website that delineates many examples: www.goldennumber.net. Kepler compared this ratio to a “precious jewel” and said it was one of geometry’s greatest treasures. A friend of Leonardo da Vinci, Luca Pacioli, called it the Divine Proportion. Surely, he was Divinely inspired at the time he settled on that name: For there can be no doubt that it was considered the Divine Proportion long before he was even born.
As I indicated earlier, the problem of squaring the circle is as much theosophical as it is geometrical; if not more so. According to most major religious sects (including Jews, Muslims, Hindu, Catholics, and most Protestant denominations), the subject falls under a branch of sacred geometry which deals with the notion of an immortal God in heaven (i.e., the Circle) merging or reconciling Himself with mortal man on earth (i.e., the square). Of course in this particular case Christ would be the Golden Mean; for he is the quintessential element which makes this holy union possible. What a wonderful analogy: As Euclid so eloquently put it: "As the whole line is to the greater segment, so is the greater to the less.” Even so, As Almighty God is to Christ, so is Christ to man." And just as the Golden Ratio is generated by the combination of a circle & a square, even so, Christ was conceived by the Immaculate Conception of the Holy Spirit & the Virgin Mary. He represents the convergence of the eternal with the temporal; the sacred with the secular; the immortal with the mortal. The Divine and human nature are combined in a single person; Christ Jesus.
PLATE #4: Using line segment XY as a radius, draw the Pink Circle.
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I began this journey with a simple set of nested squares & circles; four of each: Innermost square = (Radius squared); Circle around it = (1/2 pi x radius squared); Square around that = (2R squared); Circle around it = (pi x radius squared, or equator); Square around that = (4R squared); Circle around that = (2pi x radius squared, or hemisphere); Square around that = (8R Squared); Circle around that = (4pi x radius squared, or whole sphere area). I studied the daylights outta this. No kidding: I stared straight down on this diagram for hours on end every single day & every single night for three solid years. I honestly thought I had permanently injured my neck. I turned these shapes upside down & inside out. I began envisioning them as cubes & spheres. I started seeing certain numbers appearing on a regular basis. To make a very long story short, I eventually ended up with three sets of nested squares & circles. We'll call them sets A, B, & C. Each of them had the same eight elements; 4 squares & 4 circles. All 24 shapes were interconnected geometrically & could be drawn whenever I needed them. I could move upward from a smaller set to a larger one [A-B-C] by using an inscribed hexagon. And I could use an inscribed rhombus to move downward [C-B-A]. It wasn’t until just recently that I realized these diagrams were related through the Golden Mean. These three sets of shapes were actually three consecutive isometric drawings. In other words, the squares of Set A (if they were cubes) would perfectly fit in the corresponding circles of Set B (if they were spheres). The squares of Set B (if they were cubes) would perfectly fit in the corresponding circles of Set C (if they were spheres). The figures in Set C were exactly 1.5X the size of the corresponding shapes in Set B. The figures in Set B were exactly 1.5X the size of the corresponding shapes in Set A. This is where I obtained all the shapes for this demonstration. The Black Circle & the Blue Square belong to Set A. The Blue Square being the innermost square, (i.e., radius squared). The Red Square, Green Circle, Yellow Square, & Pink Circle all belong to Set B. The Red Square being the innermost square, (i.e., radius squared). The outermost Orange Circle was extracted from Set C (pi x radius squared).
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I began this journey with a simple set of nested squares & circles; four of each: Innermost square = (Radius squared); Circle around it = (1/2 pi x radius squared); Square around that = (2R squared); Circle around it = (pi x radius squared, or equator); Square around that = (4R squared); Circle around that = (2pi x radius squared, or hemisphere); Square around that = (8R Squared); Circle around that = (4pi x radius squared, or whole sphere area). I studied the daylights outta this. No kidding: I stared straight down on this diagram for hours on end every single day & every single night for three solid years. I honestly thought I had permanently injured my neck. I turned these shapes upside down & inside out. I began envisioning them as cubes & spheres. I started seeing certain numbers appearing on a regular basis. To make a very long story short, I eventually ended up with three sets of nested squares & circles. We'll call them sets A, B, & C. Each of them had the same eight elements; 4 squares & 4 circles. All 24 shapes were interconnected geometrically & could be drawn whenever I needed them. I could move upward from a smaller set to a larger one [A-B-C] by using an inscribed hexagon. And I could use an inscribed rhombus to move downward [C-B-A]. It wasn’t until just recently that I realized these diagrams were related through the Golden Mean. These three sets of shapes were actually three consecutive isometric drawings. In other words, the squares of Set A (if they were cubes) would perfectly fit in the corresponding circles of Set B (if they were spheres). The squares of Set B (if they were cubes) would perfectly fit in the corresponding circles of Set C (if they were spheres). The figures in Set C were exactly 1.5X the size of the corresponding shapes in Set B. The figures in Set B were exactly 1.5X the size of the corresponding shapes in Set A. This is where I obtained all the shapes for this demonstration. The Black Circle & the Blue Square belong to Set A. The Blue Square being the innermost square, (i.e., radius squared). The Red Square, Green Circle, Yellow Square, & Pink Circle all belong to Set B. The Red Square being the innermost square, (i.e., radius squared). The outermost Orange Circle was extracted from Set C (pi x radius squared).
PLATE #5: Inscribe the Yellow Square.
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While I studied those three sets of shapes, I also focused heavily on the Golden Mean & the hypocycloid. Surely, anything called the “Divine Proportion” just had to have something to do with this. I thoroughly researched the subject, but couldn't make a connection. It’s hard to believe that I could study both the Golden Mean & these three sets of shapes all at the same time & for that length of time (almost six years), and not know that they were all linked through the Golden Ratio. It still leaves me scratching my head. Just goes to show how easy it is to overlook the obvious. I guess it must not have been necessary for me to realize this during that time. Anyway, the good thing was I became very familiar with it & instantly recognized it when the time finally came.
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While I studied those three sets of shapes, I also focused heavily on the Golden Mean & the hypocycloid. Surely, anything called the “Divine Proportion” just had to have something to do with this. I thoroughly researched the subject, but couldn't make a connection. It’s hard to believe that I could study both the Golden Mean & these three sets of shapes all at the same time & for that length of time (almost six years), and not know that they were all linked through the Golden Ratio. It still leaves me scratching my head. Just goes to show how easy it is to overlook the obvious. I guess it must not have been necessary for me to realize this during that time. Anyway, the good thing was I became very familiar with it & instantly recognized it when the time finally came.
PLATE #6: Draw a Green Circle inside the Yellow Square.
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The hypocycloid (or astroid) totally fascinated me & opened many doors for me. This is what Pascal was working on when he made the dramatic decision to abandon mathematics in favor of religion. When a circle is inscribed within a square, the hypocycloid consists of all that area lying outside the circle but inside the square. It is the four curved shapes at the four corners. When folded inwardly to the center they come together to produce the single shape that is the hypocycloid. [See Plate #34 for diagram]. I knew it was important right away because:
(1). It had many properties of a square, yet it was comprised of nothing but curves.
(2). The length of it's perimeter was exactly equal to the length of the circumference of the inscribed circle.
(3). Although comprised solely of curves its area could be added to other shapes comprised solely of curves to attain an area exactly equal to a shape comprised solely of straight lines; namely a square. For example: One Hypocycloid + One Circle = One Square. Also, Two Hypocycloids + Four Vesica Pisces = One Square.
Due to its name (astroid; Gr. "Star"), I used to refer to it as the Star of David. But since it became the key to unlocking this mystery, I now call it the Key of David. For the first few years I couldn't figure out where the hypocycloid fit in. But now I know. It can be used to rectify the circle exactly. This has a lot to do with the Lune of Hippocrates. His drawings just had things a little "sideways" (sort of). I’ve also heard that the hypocycloid squares the circle hyperbolically in the Poincare Disk. That certainly wouldn’t surprise me. M.C. Escher offers a superb rendention of it in his artwork (Circle Limit III).
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The hypocycloid (or astroid) totally fascinated me & opened many doors for me. This is what Pascal was working on when he made the dramatic decision to abandon mathematics in favor of religion. When a circle is inscribed within a square, the hypocycloid consists of all that area lying outside the circle but inside the square. It is the four curved shapes at the four corners. When folded inwardly to the center they come together to produce the single shape that is the hypocycloid. [See Plate #34 for diagram]. I knew it was important right away because:
(1). It had many properties of a square, yet it was comprised of nothing but curves.
(2). The length of it's perimeter was exactly equal to the length of the circumference of the inscribed circle.
(3). Although comprised solely of curves its area could be added to other shapes comprised solely of curves to attain an area exactly equal to a shape comprised solely of straight lines; namely a square. For example: One Hypocycloid + One Circle = One Square. Also, Two Hypocycloids + Four Vesica Pisces = One Square.
Due to its name (astroid; Gr. "Star"), I used to refer to it as the Star of David. But since it became the key to unlocking this mystery, I now call it the Key of David. For the first few years I couldn't figure out where the hypocycloid fit in. But now I know. It can be used to rectify the circle exactly. This has a lot to do with the Lune of Hippocrates. His drawings just had things a little "sideways" (sort of). I’ve also heard that the hypocycloid squares the circle hyperbolically in the Poincare Disk. That certainly wouldn’t surprise me. M.C. Escher offers a superb rendention of it in his artwork (Circle Limit III).
PLATE #7: Draw the Red Square inside the Green Circle.
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It never fails: Every single time I think of the simplicity of this drawing I am reminded of the Biblical account where the hypocritical Pharisees brought the adulteress before Christ for judgment, (John 8:1-11). He began to draw in the sand as if he was ignoring them & said: "He that is without sin among you, let him first cast a stone at her." Now I have no Scriptural evidence to prove it, but I always envision him drawing out the solution to this problem right before their very eyes. And yet they could not see; for they suffered from self-inflicted blindness.
Of course it's not possible to determine whether or not these things are true. But there can be no doubt that Jesus was aware of this ancient mystery. It was very popular around his time. Well, actually, it's been popular since the dawn of recorded history. Men were trying to solve it for many centuries before the birth of Christ. Hippocrates drew his famous lune more than 300 years before then. Jesus was the son of a carpenter & a good understanding of geometry is absolutely essential to the trade. That's why, in religious iconography, his earthly father (Joseph) is often depicted with compass & straightedge in hand. Surely, Jesus contemplated this problem & it's quite probable he solved it (or at the very least, could've solved it).
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It never fails: Every single time I think of the simplicity of this drawing I am reminded of the Biblical account where the hypocritical Pharisees brought the adulteress before Christ for judgment, (John 8:1-11). He began to draw in the sand as if he was ignoring them & said: "He that is without sin among you, let him first cast a stone at her." Now I have no Scriptural evidence to prove it, but I always envision him drawing out the solution to this problem right before their very eyes. And yet they could not see; for they suffered from self-inflicted blindness.
Of course it's not possible to determine whether or not these things are true. But there can be no doubt that Jesus was aware of this ancient mystery. It was very popular around his time. Well, actually, it's been popular since the dawn of recorded history. Men were trying to solve it for many centuries before the birth of Christ. Hippocrates drew his famous lune more than 300 years before then. Jesus was the son of a carpenter & a good understanding of geometry is absolutely essential to the trade. That's why, in religious iconography, his earthly father (Joseph) is often depicted with compass & straightedge in hand. Surely, Jesus contemplated this problem & it's quite probable he solved it (or at the very least, could've solved it).
PLATE #8: Make on standard phi construction off the left side of the Red Square. Label the endpoint with the letter (B).
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The phrase "squaring the circle" can mean several different things, depending on the context within which it is employed. As mentioned earlier, it can be used in a theosophical sense; in which case it usually means the union or reconciliation of Almighty God (symbolized by the circle) with mortal man (symbolized by the square). The phrase can also be used in a general or allegorical way to describe the accomplishment of an extremely difficult task. Or, it can be used in a literal way to describe the accomplishment of an impossible task; in which case it’s often used in a derogatory manner. The reasoning being that since it is not possible to accomplish an impossible task, anyone claiming to have done so is either erroneous, or downright delusional. This would, of course, include those who claim to have squared the circle. Many individuals have made this claim over the course of history & every single one of them heretofore has been proven wrong. These people range from novice mathematicians who simply don’t understand the problem to egotistical megalomaniacs who only yearn for the respect & honor of men. The British mathematician Augustus de Morgan referred to these obsessed circle-squarers as cyclometers. Notwithstanding, it must be remembered that many things that were once “proven impossible” in the past were later proven to be very possible. One outstanding example of this is five-fold symmetry; which was considered to be impossible until Roger Penrose came along & proved it could be done. (Incidentally, he used the Golden Mean to accomplish the feat). A few reputable mathematicians have continued to work on this problem despite the fact that it was “proven impossible;” as was the case with Ramanujan. (I will discuss his contributions a little more later on). Einstein was cautious about the subject & simply called it “a frog & mouse game.” Who knows what he meant by that?! Anyway, in a strictly mathematical sense the phrase typically means one of two things:
(1). Constructing a square that is precisely equal in area to that of a circle. (Part Four of this demonstration).
(2). Constructing a square with a perimeter exactly equal in length to the circumference of a circle. (Part Three of this demonstration). To differentiate between the two we will call this "Circle Rectification."
Like I indicated earlier, both of these tasks are considered to be impossible due to the transcendental nature of pi.
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The phrase "squaring the circle" can mean several different things, depending on the context within which it is employed. As mentioned earlier, it can be used in a theosophical sense; in which case it usually means the union or reconciliation of Almighty God (symbolized by the circle) with mortal man (symbolized by the square). The phrase can also be used in a general or allegorical way to describe the accomplishment of an extremely difficult task. Or, it can be used in a literal way to describe the accomplishment of an impossible task; in which case it’s often used in a derogatory manner. The reasoning being that since it is not possible to accomplish an impossible task, anyone claiming to have done so is either erroneous, or downright delusional. This would, of course, include those who claim to have squared the circle. Many individuals have made this claim over the course of history & every single one of them heretofore has been proven wrong. These people range from novice mathematicians who simply don’t understand the problem to egotistical megalomaniacs who only yearn for the respect & honor of men. The British mathematician Augustus de Morgan referred to these obsessed circle-squarers as cyclometers. Notwithstanding, it must be remembered that many things that were once “proven impossible” in the past were later proven to be very possible. One outstanding example of this is five-fold symmetry; which was considered to be impossible until Roger Penrose came along & proved it could be done. (Incidentally, he used the Golden Mean to accomplish the feat). A few reputable mathematicians have continued to work on this problem despite the fact that it was “proven impossible;” as was the case with Ramanujan. (I will discuss his contributions a little more later on). Einstein was cautious about the subject & simply called it “a frog & mouse game.” Who knows what he meant by that?! Anyway, in a strictly mathematical sense the phrase typically means one of two things:
(1). Constructing a square that is precisely equal in area to that of a circle. (Part Four of this demonstration).
(2). Constructing a square with a perimeter exactly equal in length to the circumference of a circle. (Part Three of this demonstration). To differentiate between the two we will call this "Circle Rectification."
Like I indicated earlier, both of these tasks are considered to be impossible due to the transcendental nature of pi.
PLATE #9: Using line segment XB as a radius, draw the Orange Circle.
Now we all have all the shapes required for the presentation. The dimensions for each individual shape are delineated on the next page.
Now we all have all the shapes required for the presentation. The dimensions for each individual shape are delineated on the next page.
Plate #10: Here I treat the Red Square as a unit square: (Sides = 1; Area = 1). All other calculations are deduced from that. The category highlighted purple (Geodesic Circles) will come into play a little later. Their construction has not been provided because it's not really important at this time.
PART THREE: SQUARING THE CIRCLE WITH PHI
PLATE #11: The next four plates describe an elegant technique for squaring the circle by traditional methods. Here I employ the Divine Proportion, or Golden Mean. First, we need to isolate the following three shapes: The Blue Square, the Red Square, & the Orange Circle.
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This simple compass & straightedge method reflects the same ratio generated by the well-known equation, Phi Squared/5 = Pi/6. The value of the first divided by the value of the second = 1.000015319. When worked in reverse this yields a very tight approximation for pi: [Phi Squared/5 x 6 = 3.14164]. Obviously, the value of one will never perfectly match the value of the other due to the transcendental nature of Pi (not to mention the irrationality of Phi). There will always be a discrepency when one interjects Pi into the solution to a quadratic equation & mine is no exception. But it is minor. Algebraically, as opposed to geometrically, one can only hope to come close & this one comes about as close as it gets. (I will explain that statement a little later). I am not saying nobody has come closer, but as yet I don’t know of anyone. If, indeed, there are any, they are certainly very few & far between; especially in utilizing the Golden Mean to accomplish the task. Kepler's Golden Triangle can be used to obtain a value of 3.1446056 for pi, but that's still not as accurate as this.
The Divine Proportion is the limiting factor for the Fibonacci sequence & can be used to obtain very good approximations for pi. I've included a couple interesting results below:
Fibonacci Numbers = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987...
55/34 x 34/21 x 6/5 = 22/7 (3.1428).
89/55 x 55/34 x 6/5 = 267/85 (3.14117).
144/89 x 89/55 x 6/5 = 1398/445 (3.141573).
233/144 x 144/89 x 6/5 = 377/120 (3.14166).
610/377 x 377/233 x 6/5 = 732/233 (3.14163)…..
As we continue this exercise the ratios will ultimately converge on the very same ratio generated by the forthcoming diagrams, (3.14164). It might also be noteworthy to mention that if we take the square root of the 17th Fibonacci Number (987) and divide that by 10 = 3.14165; which is extremely close to our approximation. Here's another: (3 + 1/10 of 17/12 = 377/120); which yields 3.14166 for pi.
PLATE #11: The next four plates describe an elegant technique for squaring the circle by traditional methods. Here I employ the Divine Proportion, or Golden Mean. First, we need to isolate the following three shapes: The Blue Square, the Red Square, & the Orange Circle.
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This simple compass & straightedge method reflects the same ratio generated by the well-known equation, Phi Squared/5 = Pi/6. The value of the first divided by the value of the second = 1.000015319. When worked in reverse this yields a very tight approximation for pi: [Phi Squared/5 x 6 = 3.14164]. Obviously, the value of one will never perfectly match the value of the other due to the transcendental nature of Pi (not to mention the irrationality of Phi). There will always be a discrepency when one interjects Pi into the solution to a quadratic equation & mine is no exception. But it is minor. Algebraically, as opposed to geometrically, one can only hope to come close & this one comes about as close as it gets. (I will explain that statement a little later). I am not saying nobody has come closer, but as yet I don’t know of anyone. If, indeed, there are any, they are certainly very few & far between; especially in utilizing the Golden Mean to accomplish the task. Kepler's Golden Triangle can be used to obtain a value of 3.1446056 for pi, but that's still not as accurate as this.
The Divine Proportion is the limiting factor for the Fibonacci sequence & can be used to obtain very good approximations for pi. I've included a couple interesting results below:
Fibonacci Numbers = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987...
55/34 x 34/21 x 6/5 = 22/7 (3.1428).
89/55 x 55/34 x 6/5 = 267/85 (3.14117).
144/89 x 89/55 x 6/5 = 1398/445 (3.141573).
233/144 x 144/89 x 6/5 = 377/120 (3.14166).
610/377 x 377/233 x 6/5 = 732/233 (3.14163)…..
As we continue this exercise the ratios will ultimately converge on the very same ratio generated by the forthcoming diagrams, (3.14164). It might also be noteworthy to mention that if we take the square root of the 17th Fibonacci Number (987) and divide that by 10 = 3.14165; which is extremely close to our approximation. Here's another: (3 + 1/10 of 17/12 = 377/120); which yields 3.14166 for pi.
PLATE #12: Make one standard phi construction from the Blue Square. Continue the arc all the way around in one full rotation to complete the Golden Circle.
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Unbelievably, I cannot find anything resembling the diagrams provided here. The math is all over the place; especially in analogies dealing with the Great Pyramids of Egypt. But drawings are downright scarce. I suppose that's because many mathematicians have abandoned the drawing board in favor of the calculator. It seems that if a problem can be worked out mathematically, then they conclude it can be drawn geometrically. How to actually accomplish the task is rarely demonstrated. This is sad because many ancient problems were solved the other way around. They were solved by logic & then proven mathematically later on. This was often achieved using the age-old childlike method of paper-folding. This remains a viable & powerful tool in geometry today. For example, Hippocrates demonstrated that certain lunes could be squared even though the feat should be just as mathematically impossible as squaring the entire circle; for both shapes are inextricably connected to the same transcendental number (pi). Since then several other lunes have been squared in much the same way. These things were not accomplished mathematically & then proven logically, but vice versa. If these problems were approached with only a calculator in hand they may never have been solved.
Wikipedia, among several other math sites, mentions elaborate “CONSTRUCTIONS” by various mathematicians that have approximated pi to several digits deep. But that assertion can be a bit misleading. The “examples” delineated there are not actual constructions; they are ALGEBRAIC EQUATIONS. In fact, there is not a single example provided by the site. The same goes for Wolfram Mathworld, as well as many other sites. There's a big difference between algebraic equations & geometric constructions.
Ramanujan devised algebraic equations that have approximated pi to several decimal places. Incidentally, he also discovered a formula which produces the same close estimation created by the Golden Mean: 9/5 + sq.rt. 9/5 = 3.14164. Who knows how he came up with that? Anyway, he ultimately developed an "infinite series" for pi which, when plotted into a supercomputer, is capable of calculating pi to billions or even trillions of decimal places. However, I have serious doubts whether those alleged "constructions" could be converted into tangible diagrams. If it can be done I would absolutely love to see someone do it! Take a moment to research some of the complex equations proposed by him & try to imagine someone transforming one of them into an actual drawing! I wouldn't have the slightest idea where to begin & I doubt anyone else does either. It should also be noted that he continued to work on this problem even after it was proven that pi was transcendental; which shows that he apparently did not consider the matter closed as to whether or not the circle could be squared. He's not the only one either. There have been a few others.
This is precisely why we don't have many (actually, barely any) drawings that portray a detailed description of an actual procedure. Now, I have found a few diagrams here & there. But not one of them comes as close to solving the problem as this. Not even one. So if we set aside those complicated algebraic equations and focus solely on constructible illustrations, I'm convinced you won't find a better, or simpler, or more accurate method than the one presented here.
Some of the best diagrams I've encountered can be found in an article written by Rachel Fletcher, entitled, “Squaring the Circle: Marriage of Heaven and Earth.” Inside, Ms. Fletcher portrays a number of geometric constructions that come close to accomplishing the task of squaring the circle; which, as I indicated before, is all one can hope to accomplish by traditional methods. In the introduction to her book she observes,
"The problem cannot be solved with absolute precision, for circles are measured by the incommensurable value pi... which cannot be accurately expressed in finite whole numbers by which we measure squares. At the symbolic level, however, the quest to obtain circles and squares of equal measure is equivalent to seeking the union of transcendent and finite qualities, or the marriage of heaven and earth."
Exactly so. This is precisely why so many people continue to work on this mysterious question even though it’s supposed to be impossible. Some of the greatest discoveries in virtually every field of mathematics were made while trying to solve this riddle.
A friend of mine once challenged me to put these diagrams into algebraic form. I couldn't help but chuckle. I can’t even come close to imagining them in the form of a quadratic equation. I've always visualized them the old-fashioned way; much the same way the ancient geometers saw them: Simple compass & straightedge constructions that require no mathematics whatsoever to reproduce. You know, the kind of diagrams one might expect a child to draw on the sidewalk with a piece of chalk. That's the very thing that makes them so special. I don't know; maybe I see things backwards. But it just seems to me that converting these diagrams into algebraic equations would be more like complicating a simple task, rather than simplifying a complicated task. I must admit however, the challenge was quite ironic. I've always challenged mathematicians to convert their complex algebraic equations into tangible diagrams & now this fellow was challenging me to do the opposite!
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Unbelievably, I cannot find anything resembling the diagrams provided here. The math is all over the place; especially in analogies dealing with the Great Pyramids of Egypt. But drawings are downright scarce. I suppose that's because many mathematicians have abandoned the drawing board in favor of the calculator. It seems that if a problem can be worked out mathematically, then they conclude it can be drawn geometrically. How to actually accomplish the task is rarely demonstrated. This is sad because many ancient problems were solved the other way around. They were solved by logic & then proven mathematically later on. This was often achieved using the age-old childlike method of paper-folding. This remains a viable & powerful tool in geometry today. For example, Hippocrates demonstrated that certain lunes could be squared even though the feat should be just as mathematically impossible as squaring the entire circle; for both shapes are inextricably connected to the same transcendental number (pi). Since then several other lunes have been squared in much the same way. These things were not accomplished mathematically & then proven logically, but vice versa. If these problems were approached with only a calculator in hand they may never have been solved.
Wikipedia, among several other math sites, mentions elaborate “CONSTRUCTIONS” by various mathematicians that have approximated pi to several digits deep. But that assertion can be a bit misleading. The “examples” delineated there are not actual constructions; they are ALGEBRAIC EQUATIONS. In fact, there is not a single example provided by the site. The same goes for Wolfram Mathworld, as well as many other sites. There's a big difference between algebraic equations & geometric constructions.
Ramanujan devised algebraic equations that have approximated pi to several decimal places. Incidentally, he also discovered a formula which produces the same close estimation created by the Golden Mean: 9/5 + sq.rt. 9/5 = 3.14164. Who knows how he came up with that? Anyway, he ultimately developed an "infinite series" for pi which, when plotted into a supercomputer, is capable of calculating pi to billions or even trillions of decimal places. However, I have serious doubts whether those alleged "constructions" could be converted into tangible diagrams. If it can be done I would absolutely love to see someone do it! Take a moment to research some of the complex equations proposed by him & try to imagine someone transforming one of them into an actual drawing! I wouldn't have the slightest idea where to begin & I doubt anyone else does either. It should also be noted that he continued to work on this problem even after it was proven that pi was transcendental; which shows that he apparently did not consider the matter closed as to whether or not the circle could be squared. He's not the only one either. There have been a few others.
This is precisely why we don't have many (actually, barely any) drawings that portray a detailed description of an actual procedure. Now, I have found a few diagrams here & there. But not one of them comes as close to solving the problem as this. Not even one. So if we set aside those complicated algebraic equations and focus solely on constructible illustrations, I'm convinced you won't find a better, or simpler, or more accurate method than the one presented here.
Some of the best diagrams I've encountered can be found in an article written by Rachel Fletcher, entitled, “Squaring the Circle: Marriage of Heaven and Earth.” Inside, Ms. Fletcher portrays a number of geometric constructions that come close to accomplishing the task of squaring the circle; which, as I indicated before, is all one can hope to accomplish by traditional methods. In the introduction to her book she observes,
"The problem cannot be solved with absolute precision, for circles are measured by the incommensurable value pi... which cannot be accurately expressed in finite whole numbers by which we measure squares. At the symbolic level, however, the quest to obtain circles and squares of equal measure is equivalent to seeking the union of transcendent and finite qualities, or the marriage of heaven and earth."
Exactly so. This is precisely why so many people continue to work on this mysterious question even though it’s supposed to be impossible. Some of the greatest discoveries in virtually every field of mathematics were made while trying to solve this riddle.
A friend of mine once challenged me to put these diagrams into algebraic form. I couldn't help but chuckle. I can’t even come close to imagining them in the form of a quadratic equation. I've always visualized them the old-fashioned way; much the same way the ancient geometers saw them: Simple compass & straightedge constructions that require no mathematics whatsoever to reproduce. You know, the kind of diagrams one might expect a child to draw on the sidewalk with a piece of chalk. That's the very thing that makes them so special. I don't know; maybe I see things backwards. But it just seems to me that converting these diagrams into algebraic equations would be more like complicating a simple task, rather than simplifying a complicated task. I must admit however, the challenge was quite ironic. I've always challenged mathematicians to convert their complex algebraic equations into tangible diagrams & now this fellow was challenging me to do the opposite!
PLATE #13: Extend the selected lines off the Red Square as shown. Connect the dots to complete the Golden Square.
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The earliest efforts to square the circle often centered around the use of inscribed polygons: The idea being that the more sides one adds to the polygon the more area of the circle it would occupy. The hope was that eventually the entire circle would be filled in. For example: An inscribed pentagon would occupy more space within the circle than an inscribed square, and an inscribed hexagon would occupy more space that an inscribed pentagon, and an inscribed octagon would occupy more space within the circle than an inscribed hexagon, and so on. If proven it would also prove that the circle could be squared since polygons can be squared.
To create a square & a circle as close in area as the Golden Square is to the Golden Circle, one would require the use of inscribed polygons of multiple facets. Archimedes used a 96-gon to set upper & lower bounds for pi of 3.142857 & 3.140845 respectively. Ptolemy used a 360-gon to obtain a value of 3.141666. That’s a super close approximation. But the one generated by these diagrams is even better than that, (3.14164). Huygens defined this number as his upper limit for pi using a square & an octagon. Liu Hui obtained the exact same ratio by inscribing a polygon of 3,072 sides back in the 3rd Century. And Aryabhatta, some time around 500 AD, used a 384-gon to obtain the same close estimate. Imagine trying to draw any one of those with compass & straightedge! I doubt they ever drew any such thing though. Like I said before, mathematicians rarely drew these problems out on a sheet of paper in a literal sense.
Many respectable mathematicians, including Einstein, often rounded pi up to 3.1416. Many still do. Actually, this value sometimes naturally appears in algebraic equations involving electronics, physics, mechanics, & engineering. It was well-known & widely used in ancient architecture too. It is common to find this value connected to the pyramids. David Bowman has written a good article on this; (http://www.aiwaz.net). He makes a remarkable link between this number & the dimensions for the King's chamber. He likens it to what he calls a "double square." It consists of two squares of equal size set side by side; each with side lengths of .6. This arrangement forms a rectangle with a diagonal of .6 X sq.rt. 5 = 1.341640787. The sum of the sides of the two resulting triangles is equal to 3.14164 (.6 + 1.2 + 1.341640787). He further observes "this is an amazingly close approximation of the value of pi = 3.14159, with the error of 0.0015%." He adds that this is probably the main reason why this particular place was considered so sacred to the High Priest & why he often used this double square to communicate with the gods. This number is commonly associated with other ancient religious structures as well, such as Stonehenge. If you search the internet you'll find it is intimately connected to the Golden Ratio. Stephen Skinner, in his book “Sacred Geometry: Deciphering the Code,” links this ratio to the subject of squaring the circle. He suggests it has something to do with drawing a square with sidelengths of 3.14164. I have no idea where or how he got that. On twitter Dave Richeson says it is a good approximation for pi & explains that it is constructible via compass & straightedge. But, unfortunately, he gives no description on how to do it.
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The earliest efforts to square the circle often centered around the use of inscribed polygons: The idea being that the more sides one adds to the polygon the more area of the circle it would occupy. The hope was that eventually the entire circle would be filled in. For example: An inscribed pentagon would occupy more space within the circle than an inscribed square, and an inscribed hexagon would occupy more space that an inscribed pentagon, and an inscribed octagon would occupy more space within the circle than an inscribed hexagon, and so on. If proven it would also prove that the circle could be squared since polygons can be squared.
To create a square & a circle as close in area as the Golden Square is to the Golden Circle, one would require the use of inscribed polygons of multiple facets. Archimedes used a 96-gon to set upper & lower bounds for pi of 3.142857 & 3.140845 respectively. Ptolemy used a 360-gon to obtain a value of 3.141666. That’s a super close approximation. But the one generated by these diagrams is even better than that, (3.14164). Huygens defined this number as his upper limit for pi using a square & an octagon. Liu Hui obtained the exact same ratio by inscribing a polygon of 3,072 sides back in the 3rd Century. And Aryabhatta, some time around 500 AD, used a 384-gon to obtain the same close estimate. Imagine trying to draw any one of those with compass & straightedge! I doubt they ever drew any such thing though. Like I said before, mathematicians rarely drew these problems out on a sheet of paper in a literal sense.
Many respectable mathematicians, including Einstein, often rounded pi up to 3.1416. Many still do. Actually, this value sometimes naturally appears in algebraic equations involving electronics, physics, mechanics, & engineering. It was well-known & widely used in ancient architecture too. It is common to find this value connected to the pyramids. David Bowman has written a good article on this; (http://www.aiwaz.net). He makes a remarkable link between this number & the dimensions for the King's chamber. He likens it to what he calls a "double square." It consists of two squares of equal size set side by side; each with side lengths of .6. This arrangement forms a rectangle with a diagonal of .6 X sq.rt. 5 = 1.341640787. The sum of the sides of the two resulting triangles is equal to 3.14164 (.6 + 1.2 + 1.341640787). He further observes "this is an amazingly close approximation of the value of pi = 3.14159, with the error of 0.0015%." He adds that this is probably the main reason why this particular place was considered so sacred to the High Priest & why he often used this double square to communicate with the gods. This number is commonly associated with other ancient religious structures as well, such as Stonehenge. If you search the internet you'll find it is intimately connected to the Golden Ratio. Stephen Skinner, in his book “Sacred Geometry: Deciphering the Code,” links this ratio to the subject of squaring the circle. He suggests it has something to do with drawing a square with sidelengths of 3.14164. I have no idea where or how he got that. On twitter Dave Richeson says it is a good approximation for pi & explains that it is constructible via compass & straightedge. But, unfortunately, he gives no description on how to do it.
PLATE #14: Here we merge the Golden Circle & the Golden Square. This resembles most diagrams that one might typically expect to find when researching this topic.
If we consider the Red Square as a Unit Square [sides = 1; area = 1], the following calculations will result:
Golden Square = Phi (1.618033988). Area = Phi Squared (2.618033986).
Golden Circle = Radius (0.91287093); Radius Squared = (.833333334); Area = (2.61799388).
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With respect to area there is virtually no difference between these two shapes. Measured in inches the difference is literally microscopic. And even if we convert them into square feet the difference would not be detectable by the naked eye. (About the size of two red blood cells). This construction yields an extremely tight approximation for pi as well: (3.141640784). This is 99.85% accurate for true pi. To illustrate just how significant this is we would need to enlarge the two shapes astronomically. Imagine, for example, you have a planet with a diameter of a thousand miles. According to pi it would take a car racing along at 60 mph more than 52 hours & 21 minutes to circumnavigate the globe at its equator. If we were to extrapolate our travel time using phi instead, the difference between the two times would be less than three seconds! (Most of this material has been featured on Gary Meisner’s website, www.goldennumber.net. For more detailed information please visit his site).
The fact that we can attain such a high degree of precision without the aid of modern tools & in so few steps sets this construction apart from some of even the most ingenious techniques.
Although these diagrams reveal an extremely tight approximation for pi and this fact most definitely ought to be mentioned, I do not feel it is the most remarkable aspect to be highlighted. It must be remembered that pi was not even used to construct them. For that matter, I didn't use phi either. (By that I mean I didn't use their actual numerical values). The most remarkable thing about these drawings is that they were constructed without the use of any mathematics whatsoever. The instruments employed were primitive & utterly devoid of incremental markings. Actual measurement played no part in it; except, of course, in the proofs.
This procedure is so simple you'd expect to see these drawings surface whenever this subject appeared. Yet, diagrams that demonstrate this phenomenon are practically non-existant. For the life of me I cannot figure out why this is.
I think these diagrams are blueprints; literally. I think they are a kind of "signpost" pointing the way to the ultimate solution.
If we consider the Red Square as a Unit Square [sides = 1; area = 1], the following calculations will result:
Golden Square = Phi (1.618033988). Area = Phi Squared (2.618033986).
Golden Circle = Radius (0.91287093); Radius Squared = (.833333334); Area = (2.61799388).
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With respect to area there is virtually no difference between these two shapes. Measured in inches the difference is literally microscopic. And even if we convert them into square feet the difference would not be detectable by the naked eye. (About the size of two red blood cells). This construction yields an extremely tight approximation for pi as well: (3.141640784). This is 99.85% accurate for true pi. To illustrate just how significant this is we would need to enlarge the two shapes astronomically. Imagine, for example, you have a planet with a diameter of a thousand miles. According to pi it would take a car racing along at 60 mph more than 52 hours & 21 minutes to circumnavigate the globe at its equator. If we were to extrapolate our travel time using phi instead, the difference between the two times would be less than three seconds! (Most of this material has been featured on Gary Meisner’s website, www.goldennumber.net. For more detailed information please visit his site).
The fact that we can attain such a high degree of precision without the aid of modern tools & in so few steps sets this construction apart from some of even the most ingenious techniques.
Although these diagrams reveal an extremely tight approximation for pi and this fact most definitely ought to be mentioned, I do not feel it is the most remarkable aspect to be highlighted. It must be remembered that pi was not even used to construct them. For that matter, I didn't use phi either. (By that I mean I didn't use their actual numerical values). The most remarkable thing about these drawings is that they were constructed without the use of any mathematics whatsoever. The instruments employed were primitive & utterly devoid of incremental markings. Actual measurement played no part in it; except, of course, in the proofs.
This procedure is so simple you'd expect to see these drawings surface whenever this subject appeared. Yet, diagrams that demonstrate this phenomenon are practically non-existant. For the life of me I cannot figure out why this is.
I think these diagrams are blueprints; literally. I think they are a kind of "signpost" pointing the way to the ultimate solution.
PART FOUR: CIRCLE RECTIFICATION
Poposition #1: The perimeter of the Red Square is EXACTLY equal in length to the circumference of the Pink Circle. This probably sounds absurd. But please bear with me on this & follow along. You might find the journey worthwhile.
PLATE #15: Isolate the Pink Circle, the Green Circle, and the Red Square.
Poposition #1: The perimeter of the Red Square is EXACTLY equal in length to the circumference of the Pink Circle. This probably sounds absurd. But please bear with me on this & follow along. You might find the journey worthwhile.
PLATE #15: Isolate the Pink Circle, the Green Circle, and the Red Square.
PLATE #16: THE FOUR CORNERS OF THE EARTH = Here I would like to direct your attention to the two innermost shapes: The Red Square centered within the Green Circle.
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Normally, in flatland that is, the area of the Red Square would be equal to the area of the Green Circle divided by 1/2pi (1.5707963). The perimeter of the Red Square would be equal to the Green Circle's circumference divided by the Form Factor Ratio (1.1107207). The Form Factor Ratio is intimately connected to the subject of squaring the circle. Multiplied by sq.rt. 2 = 1/2pi & multiplied by sq.rt. 8 = pi exactly. It surfaces frequently in geometry; especially when dealing with Steinmetz Solids & octahedrons inscribed within spheres. It is also closely associated with the Silver Ratio via inscribed octagons.
I’m sure there are many people who think this subject is downright boring & that the squaring of the circle is totally unimportant in their lives. But they might want to reconsider that opinion. They might be surprised to find that the circle is being squared all around them on a daily basis & it plays a huge role in their lives. In electronics, for example, this ratio is used to convert sinusoidal (i.e., curved) waves into digital (i.e., square) waves. This new technology enables computers to process & transmit information more quickly & efficiently. It is also used in cameras, cell phones, televisions, clocks, radios, and a host of other electronic devices. I wonder how many people realize that when they hear music being played on the radio they are actually listening to the circle being squared. I wonder how many people realize that when they watch a movie on the television they are literally witnessing the circle being squared.
Anyway, things are not always what they initially seem to be. Consider now what happens when I change the manifold.
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Normally, in flatland that is, the area of the Red Square would be equal to the area of the Green Circle divided by 1/2pi (1.5707963). The perimeter of the Red Square would be equal to the Green Circle's circumference divided by the Form Factor Ratio (1.1107207). The Form Factor Ratio is intimately connected to the subject of squaring the circle. Multiplied by sq.rt. 2 = 1/2pi & multiplied by sq.rt. 8 = pi exactly. It surfaces frequently in geometry; especially when dealing with Steinmetz Solids & octahedrons inscribed within spheres. It is also closely associated with the Silver Ratio via inscribed octagons.
I’m sure there are many people who think this subject is downright boring & that the squaring of the circle is totally unimportant in their lives. But they might want to reconsider that opinion. They might be surprised to find that the circle is being squared all around them on a daily basis & it plays a huge role in their lives. In electronics, for example, this ratio is used to convert sinusoidal (i.e., curved) waves into digital (i.e., square) waves. This new technology enables computers to process & transmit information more quickly & efficiently. It is also used in cameras, cell phones, televisions, clocks, radios, and a host of other electronic devices. I wonder how many people realize that when they hear music being played on the radio they are actually listening to the circle being squared. I wonder how many people realize that when they watch a movie on the television they are literally witnessing the circle being squared.
Anyway, things are not always what they initially seem to be. Consider now what happens when I change the manifold.
PLATE #17: Here we have the same two shapes (the Red Square inside the Green Circle), but I've eliminated their colors. This will help to enhance the visual effect. I've also altered the manifold & have exemplified that change by darkening the background with a black magic marker. Otherwise, absolutely nothing has changed: The diagram remains intact and it would be totally impossible for a person to determine that any change had even taken place unless he was informed of it. All the dimensions are identical & it can still be drawn with compass & straightedge. Now, it is not the manifold within which the Green Circle is embedded that has been changed, but that within which the Red Square is embedded. In other words, the background is basically irrelevant. It's essentially empty space. I've only darkened it to help you visualize what I'm talking about. The surface area now bound by the Green Circle is actually a two-dimensional smooth Riemannian manifold of positive Ricci curvature; albeit, this curvature is not detectable from this vantage point. Imagine, if you will, the following scenario:
You walk outside on a dark winter night & gaze up into the pitch black sky. You notice there's a full moon tonight. (Of course, in reality, this is only half the moon; a hemisphere; which is the maximum amount of surface area one can see of any sphere at any one time). But what you see in the night sky is not what you'd typically expect to find. Instead, the image in the sky is exactly like the one appearing here on your screen. Please take a few moments to focus intently upon this drawing. It is a very important piece of the puzzle & your ability to see this from the proper perspective is paramount to understanding what I’m talking about.
Obviously, the sidelengths of the Red Square are curved; albeit, from this angle that cannot be established by simple observation. Consequently, the Red Square has developed an intimate relationship with the circle & gains many of its unique charateristics by inheritance; including the transcendental nature of pi.
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This always reminds me of DaVinci’s famous drawing, “The Vitruvian Man.” You know, the one with the man with outstretched arms & legs centered within the square, which is inside the circle. Coincidentally, that diagram is considered by many scholars to represent the squaring of the circle. Coincidentally, it appears on the patch awarded by NASA to all astronauts who complete a spacewalk. Speaking of NASA, my sister graduated from Notre Dame in 1985. One of her classmates had a friend (Philip T. Metzger) with a PhD in physics who works for NASA at the Kennedy Space Center. Coincidentally, his specialty is lunar exploration. He has written several scholarly articles on the subject. A few years ago she sent him Part Two of this demonstration (Squaring the Circle with the Golden Mean). He was impressed & said “this new proof’s main value lies in it’s being a vastly more elegant approximation, in that it achieves good numerical approximation in just a short series of steps without the need for a long series of tedious refinements.” And then he later adds, "I get the sense that this will be a welcome addition to the truly interesting proofs in geometry.” Whether or not this happens remains to be seen. The fact that he called it a “new proof” is what surprises me. The close approximation for pi generated by the Golden Mean (3.14164) is downright ubiquitous. I am constantly running into it. But like I said before, I have yet to find a single diagram that reflects this value.
Please turn to the next frame to see what happens when this diagram is turned forty-five degrees & viewed from another angle.
You walk outside on a dark winter night & gaze up into the pitch black sky. You notice there's a full moon tonight. (Of course, in reality, this is only half the moon; a hemisphere; which is the maximum amount of surface area one can see of any sphere at any one time). But what you see in the night sky is not what you'd typically expect to find. Instead, the image in the sky is exactly like the one appearing here on your screen. Please take a few moments to focus intently upon this drawing. It is a very important piece of the puzzle & your ability to see this from the proper perspective is paramount to understanding what I’m talking about.
Obviously, the sidelengths of the Red Square are curved; albeit, from this angle that cannot be established by simple observation. Consequently, the Red Square has developed an intimate relationship with the circle & gains many of its unique charateristics by inheritance; including the transcendental nature of pi.
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This always reminds me of DaVinci’s famous drawing, “The Vitruvian Man.” You know, the one with the man with outstretched arms & legs centered within the square, which is inside the circle. Coincidentally, that diagram is considered by many scholars to represent the squaring of the circle. Coincidentally, it appears on the patch awarded by NASA to all astronauts who complete a spacewalk. Speaking of NASA, my sister graduated from Notre Dame in 1985. One of her classmates had a friend (Philip T. Metzger) with a PhD in physics who works for NASA at the Kennedy Space Center. Coincidentally, his specialty is lunar exploration. He has written several scholarly articles on the subject. A few years ago she sent him Part Two of this demonstration (Squaring the Circle with the Golden Mean). He was impressed & said “this new proof’s main value lies in it’s being a vastly more elegant approximation, in that it achieves good numerical approximation in just a short series of steps without the need for a long series of tedious refinements.” And then he later adds, "I get the sense that this will be a welcome addition to the truly interesting proofs in geometry.” Whether or not this happens remains to be seen. The fact that he called it a “new proof” is what surprises me. The close approximation for pi generated by the Golden Mean (3.14164) is downright ubiquitous. I am constantly running into it. But like I said before, I have yet to find a single diagram that reflects this value.
Please turn to the next frame to see what happens when this diagram is turned forty-five degrees & viewed from another angle.
PLATE #18: The top drawing is the same one we've been viewing in the last three frames; the Red square centered within the Green Circle. The bottom drawing is what we would see if we rotate the diagram 45 degrees.
In essence then, we are looking at the same diagram; but from two different perspectives. If the square was shaded gray & the hemisphere's manifold was transparent, and you shined a light straight down on it from directly overhead, you would see that it would cast a perfect “shadow” on the flat Euclidean surface below which is identical to the top drawing. From blueprint perspective the two diagrams are the same; absolutely nothing has changed. In reality however, some things have changed. The manifold is not what we thought it was. Therefore, the geometry is not going to be what we thought it was. The circumference of the Green Circle hasn't changed, except now it represents an equator circle. The surface area bound by that Green Circle has now doubled because it represents a hemisphere (rather than a flat circle). Hence, it encompasses a positively curved area exactly equal to pi (which happens to be the same amount of area bound by the Pink Circle in flatland). The curved surface area now bound by the Red Square (i.e., pendentive dome) is equal to the hemispherical area bound by the Green Circle (pi) divided by the Silver Mean: pi/(1 + sq.rt.2) = 1.3012902. The sidelengths of the Red Square's perimeter have increased by exactly 1/2pi. Therefore, the total length of its perimeter is EXACTLY EQUAL to the circumference of the Pink Circle: (4 x 1/2pi = 2pi).
I occasionally refer to this diagram as the "Silver Dome" because the Silver Mean can also be employed here to square the circle in area; (albeit, I utilize a somewhat different approach in Part Four of this demonstration). For example, imagine the Red Square in flatland had a total area equal to the Silver Mean (1 + sq.rt. 2 = 2.414213562). If we project that Red Square onto a sphere such that it's vertices are anchored on the equator circle (as depicted in the top diagram), that Red Square (i.e., pendentive dome) would then encompass a total curved surface area EXACTLY EQUAL to pi. Interestingly, in this particular case, the total amount of hemispherical surface area lying outside the square (or pendentive dome) is precisely equal to 4 x the Form Factor Ratio = 4.4428828. If we take this number and multiply it by the side length of the Red Square it would be exactly equal in length to the circumference of the circumscribing Green Circle.
This clearly proves that Euclidean shapes projected onto spherical surfaces can indeed square the circle in both area & perimeter. Basically, a similar result can be achieved with an equilateral triangle. Please refer to Plates 37-38 & the 19-Step process that follows for additional details about this profoundly intriguing phenomenon.
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At the very beginning of this exhibit I mentioned that the square is a "slice" of a cube & is Euclidean in nature; while the circle is a "slice" of a sphere & is spherical in nature. When we take the two of them and lay them side by side within (not upon) a flat surface, we've unwittingly performed a mathematical function. Obviously, the square is perfectly at home there. But we've done injustice to the circle: We've deported it from its comfortable home in spherical geometry, stripped it of its double curvature, & thrust it into flatland; a world utterly foreign to it. If we hope to rectify and/or square the circle in a literal sense we need to balance the equation. We need to give to the square what we took from the circle. In other words, we need to find a way to thrust the square into the same transcendental world as the circle without making any visible changes to the square’s appearance.
Basically, this is what I’ve done. The two shapes have been reunited: Only this time the situation has been reversed: The circle is perfectly at home now, while it is the square that feels out of place. It has gained in this world what the circle relinquished in flatland; namely, double curvature.
Try to think of it this way: Imagine you & a friend go to a place where your vision is unobstructed for several miles in every direction; say, the Great Salt Flats of Utah. You draw a perfect 3’ x 3’ square in the sand. You step inside the square & center yourself in the middle. You look straight down & see the perfect square. Now you have your friend increase the size of the square around you concentrically. In other words, you have him utilize the same method required to produce a typical set of nested squares & circles. Consequently, each successive square would be exactly twice the size of the one before it. At first the squares are very easy to see. But before long you notice that although you can see your friend just fine, the lines he’s drawing in the sand are increasingly difficult to see. Finally, the lines vanish out of sight entirely. Obviously, this is because the lines have gone over the horizon. The medium (or manifold) within which the squares are embedded is curved (just like everything else in the known universe). Therefore, what appeared to be straight on a small scale was really curved afterall, but the curvature was so subtle that the miniscule amount could not be observed or even measured. If you continue this exercise long enough you will eventually find the vertices of your square anchored directly on the equator circle. Of course, from your original vantage point the square would no longer resemble a square. But imagine, if you will, that the initial 3’ x 3’ square within which you were standing was a kind of podium with a hydraulic lift attached to the bottom. Imagine you could use that lift to propel yourself upwardly to any height you desired. You then begin ascending at a constant rate. You would soon find that the higher you go the more of the square you would be able to see. Ultimately, you would be able to see the entire square. If you were to ascend to a great enough height, say, as far away as the moon, and you looked straight down on it (i.e., from God’s perspective), the square your friend has etched into the sand would look exactly like the inscribed square depicted at the top of this diagram. If you were to view the same image from the side it would look exactly like the bottom drawing. It would appear as an opened parachute; something like a square sheet draped over a sphere. This could pass for a snapshot of a sail vault, or pendentive dome.
Some will argue that what I've drawn is not actually a square. Some may cry “foul.” Others may call it a trick, or an illusion. Some will assert that I bent the rules. But that is not so. It is the lines that I bent, or more specifically, the surface upon which they appear; albeit, the curvature is impossible to detect from this angle. Hence, these diagrams are constructible via compass & straightedge. Both shapes (the curved pendentive dome & the flat Euclidean Square) share the same characteristics. They are both embedded in a two-dimensional manifold. All four sides are equal in length. The four sides intersect at right angles. And the diagonal obeys the Pythagorean Theorem. Anyway, I am not terribly concerned with what the mathematical community may think of this. It’s the solution the Lord gave me. It totally answers my prayer: A square has been drawn with compass & straightedge that has a perimeter EXACTLY equal in length to the circumference of a circle.
Expanding on our explanation of the bottom drawing: We see here that the four side lengths of the Red Square actually serve as four diameters of four geodesic circles upon the Green sphere's surface (highlighted red in the diagram). These four circles are arranged in a square shape; much like four paper plates set on edge. They intersect at right angles with each Red Circle being tangent to the adjacent one at a single point on the Green equator Circle. The diameter of each Red Circle is equal in length to one side of the Red Square (1). That means the circumference of each Red Circle would be exactly equal to pi. Since we are dealing with a hemisphere (as opposed to a whole sphere), the Red Circles are cut in half. Therefore, the length of each one is 1/2pi. Hence, all four sides would have a total length of 2pi; which is exactly equal to the Pink Circle's circumference. Perhaps this would be easier to understand if I folded the Red semi-circles to the side & down flat. Please turn to the next slide for additional details.
In essence then, we are looking at the same diagram; but from two different perspectives. If the square was shaded gray & the hemisphere's manifold was transparent, and you shined a light straight down on it from directly overhead, you would see that it would cast a perfect “shadow” on the flat Euclidean surface below which is identical to the top drawing. From blueprint perspective the two diagrams are the same; absolutely nothing has changed. In reality however, some things have changed. The manifold is not what we thought it was. Therefore, the geometry is not going to be what we thought it was. The circumference of the Green Circle hasn't changed, except now it represents an equator circle. The surface area bound by that Green Circle has now doubled because it represents a hemisphere (rather than a flat circle). Hence, it encompasses a positively curved area exactly equal to pi (which happens to be the same amount of area bound by the Pink Circle in flatland). The curved surface area now bound by the Red Square (i.e., pendentive dome) is equal to the hemispherical area bound by the Green Circle (pi) divided by the Silver Mean: pi/(1 + sq.rt.2) = 1.3012902. The sidelengths of the Red Square's perimeter have increased by exactly 1/2pi. Therefore, the total length of its perimeter is EXACTLY EQUAL to the circumference of the Pink Circle: (4 x 1/2pi = 2pi).
I occasionally refer to this diagram as the "Silver Dome" because the Silver Mean can also be employed here to square the circle in area; (albeit, I utilize a somewhat different approach in Part Four of this demonstration). For example, imagine the Red Square in flatland had a total area equal to the Silver Mean (1 + sq.rt. 2 = 2.414213562). If we project that Red Square onto a sphere such that it's vertices are anchored on the equator circle (as depicted in the top diagram), that Red Square (i.e., pendentive dome) would then encompass a total curved surface area EXACTLY EQUAL to pi. Interestingly, in this particular case, the total amount of hemispherical surface area lying outside the square (or pendentive dome) is precisely equal to 4 x the Form Factor Ratio = 4.4428828. If we take this number and multiply it by the side length of the Red Square it would be exactly equal in length to the circumference of the circumscribing Green Circle.
This clearly proves that Euclidean shapes projected onto spherical surfaces can indeed square the circle in both area & perimeter. Basically, a similar result can be achieved with an equilateral triangle. Please refer to Plates 37-38 & the 19-Step process that follows for additional details about this profoundly intriguing phenomenon.
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At the very beginning of this exhibit I mentioned that the square is a "slice" of a cube & is Euclidean in nature; while the circle is a "slice" of a sphere & is spherical in nature. When we take the two of them and lay them side by side within (not upon) a flat surface, we've unwittingly performed a mathematical function. Obviously, the square is perfectly at home there. But we've done injustice to the circle: We've deported it from its comfortable home in spherical geometry, stripped it of its double curvature, & thrust it into flatland; a world utterly foreign to it. If we hope to rectify and/or square the circle in a literal sense we need to balance the equation. We need to give to the square what we took from the circle. In other words, we need to find a way to thrust the square into the same transcendental world as the circle without making any visible changes to the square’s appearance.
Basically, this is what I’ve done. The two shapes have been reunited: Only this time the situation has been reversed: The circle is perfectly at home now, while it is the square that feels out of place. It has gained in this world what the circle relinquished in flatland; namely, double curvature.
Try to think of it this way: Imagine you & a friend go to a place where your vision is unobstructed for several miles in every direction; say, the Great Salt Flats of Utah. You draw a perfect 3’ x 3’ square in the sand. You step inside the square & center yourself in the middle. You look straight down & see the perfect square. Now you have your friend increase the size of the square around you concentrically. In other words, you have him utilize the same method required to produce a typical set of nested squares & circles. Consequently, each successive square would be exactly twice the size of the one before it. At first the squares are very easy to see. But before long you notice that although you can see your friend just fine, the lines he’s drawing in the sand are increasingly difficult to see. Finally, the lines vanish out of sight entirely. Obviously, this is because the lines have gone over the horizon. The medium (or manifold) within which the squares are embedded is curved (just like everything else in the known universe). Therefore, what appeared to be straight on a small scale was really curved afterall, but the curvature was so subtle that the miniscule amount could not be observed or even measured. If you continue this exercise long enough you will eventually find the vertices of your square anchored directly on the equator circle. Of course, from your original vantage point the square would no longer resemble a square. But imagine, if you will, that the initial 3’ x 3’ square within which you were standing was a kind of podium with a hydraulic lift attached to the bottom. Imagine you could use that lift to propel yourself upwardly to any height you desired. You then begin ascending at a constant rate. You would soon find that the higher you go the more of the square you would be able to see. Ultimately, you would be able to see the entire square. If you were to ascend to a great enough height, say, as far away as the moon, and you looked straight down on it (i.e., from God’s perspective), the square your friend has etched into the sand would look exactly like the inscribed square depicted at the top of this diagram. If you were to view the same image from the side it would look exactly like the bottom drawing. It would appear as an opened parachute; something like a square sheet draped over a sphere. This could pass for a snapshot of a sail vault, or pendentive dome.
Some will argue that what I've drawn is not actually a square. Some may cry “foul.” Others may call it a trick, or an illusion. Some will assert that I bent the rules. But that is not so. It is the lines that I bent, or more specifically, the surface upon which they appear; albeit, the curvature is impossible to detect from this angle. Hence, these diagrams are constructible via compass & straightedge. Both shapes (the curved pendentive dome & the flat Euclidean Square) share the same characteristics. They are both embedded in a two-dimensional manifold. All four sides are equal in length. The four sides intersect at right angles. And the diagonal obeys the Pythagorean Theorem. Anyway, I am not terribly concerned with what the mathematical community may think of this. It’s the solution the Lord gave me. It totally answers my prayer: A square has been drawn with compass & straightedge that has a perimeter EXACTLY equal in length to the circumference of a circle.
Expanding on our explanation of the bottom drawing: We see here that the four side lengths of the Red Square actually serve as four diameters of four geodesic circles upon the Green sphere's surface (highlighted red in the diagram). These four circles are arranged in a square shape; much like four paper plates set on edge. They intersect at right angles with each Red Circle being tangent to the adjacent one at a single point on the Green equator Circle. The diameter of each Red Circle is equal in length to one side of the Red Square (1). That means the circumference of each Red Circle would be exactly equal to pi. Since we are dealing with a hemisphere (as opposed to a whole sphere), the Red Circles are cut in half. Therefore, the length of each one is 1/2pi. Hence, all four sides would have a total length of 2pi; which is exactly equal to the Pink Circle's circumference. Perhaps this would be easier to understand if I folded the Red semi-circles to the side & down flat. Please turn to the next slide for additional details.
PLATE #19: Here, I've folded the Red Circles down flat so their dimensions could be seen more easily.
Now we state the obvious: Each Red half-circle is precisely equal in length to 1/4 the Pink Circle's circumference. All four would be equal to its entire circumference. Conclusion: The perimeter of the Red Square, when embedded in a two-dimensional smooth Riemannian manifold of positive curvature & bound by an equator circle the size of the Green Circle, is EXACTLY EQUAL in length to the circumference of the Pink Circle The circle is rectified via compass & straightedge.
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Incidentally, the famous "Lune of Hippocrates" makes an appearance here. Although unintentional I doubt it is sheer happenstance, since all these things are interrelated. The dimensions are very easy to see in this diagram: The Red semicircles are clearly equal in length to 1/2pi & all four added together = 2pi. If you were to eliminate the red arcs at the top & bottom, the resulting diagram would resemble the design plan for the well known Venetian Basilica, Santa Maria della Saluta. This kind of "sacred geometry" was (and still is) often incorporated into the very structure of religious buildings. This accentuates the close relationship between this particular problem & the Incarnation of the Son of God, Jesus Christ. See the next few slides for additional examples.
PLATE #20: Groin Vault: Also known as the “Cross Vault;” for obvious reasons. (In religiously oriented architecture it can be used to represent the Cross of Christ & I will exemplify that point as I proceed through the next few frames). This structure can be found most commonly in basement & ground floor levels of buildings due to its ability to support heavy payloads without the need for massive buttresses. It achieves this by directing the stresses almost vertically downward through the four corner piers. It was highly influential during the Middle Ages & was frequently employed in Gothic Cathedrals. Oftentime, it was used to construct long corridors by aligning several of them in a long row.
When compared to Plates 18 & 19 the relationship between this structure & the topic at hand is self-evident. The four red arcs & the Red Square show up quite nicely. The design of the Groin Vault was derived directly from a bicylinder which has been cut in half horizontally. (See Plate #21).
When compared to Plates 18 & 19 the relationship between this structure & the topic at hand is self-evident. The four red arcs & the Red Square show up quite nicely. The design of the Groin Vault was derived directly from a bicylinder which has been cut in half horizontally. (See Plate #21).
PLATE #21: Typical Bicylinder: It is formed by two cylinders of equal diameter intersecting at right angles. The interesting thing about cylinders is that Euclidean math is applicable. As you can see from the diagram on the left, the Red Square is very prominent. When the cylinder height (or length if you prefer) is reduced to its extreme minimum as portrayed in the bottom drawing, the four openings become tangent & meet at right angles. In other words, if the height of each cylinder was equal in length to its diameter (which, in this case is 1), then it would perfectly reflect the dimensions of the Red Square & the red arcs depicted in previous frames. If you were to cut it in half horizontally, you’d have the Groin Vault illustrated in the previous frame. The sphere within which this would be circumscribed would have a diameter equal to the cylinder’s diameter length (1) multiplied by sq.rt.2. In this case, the bottom bicylinder would perfectly fit inside the Green Circle; that is, of course, if it were actually a sphere. The four cylinder openings would come into contact with the surface of the sphere & trace out four perfect circles. Each Red Circle would be tangent to the adjacent one at a single point on the Green Sphere’s equator circle. They would meet at right angles & would look exactly like the diagrams of previous plates. From directly above curvature of the arcs would not be detectable & it would look exactly like the Red Square inscribed in the Green Circle [Plate #16].
If you were to fill the inside of this object with some sort of a sealant foam & then allowed it to harden, it would form a most interesting object known as the Steinmetz Solid. This particular object is downright fascinating & I've spent a great deal of time & energy studying its unique characteristics. Apparently, I’m not the only one who feels this way: Shakespeare also found it quite intriguing and specifically mentions it in his famous play, “Midsummer Night’s Dream.”
Turn to the next slide for more details.
If you were to fill the inside of this object with some sort of a sealant foam & then allowed it to harden, it would form a most interesting object known as the Steinmetz Solid. This particular object is downright fascinating & I've spent a great deal of time & energy studying its unique characteristics. Apparently, I’m not the only one who feels this way: Shakespeare also found it quite intriguing and specifically mentions it in his famous play, “Midsummer Night’s Dream.”
Turn to the next slide for more details.
PLATE #22: Steinmetz Solid: This is what it would look like if we filled the bicylinder with a substance & then allowed it to harden into a solid. The outer shell of this particular solid forms a four-sided polygonal dome; which has been used extensively in holy sites; especially the octagonal style. Probably the most famous example of this is the dome over the Florence Cathedral. It’s construction was shrouded in mystery for several centuries until a Florentine architect by the name of Massimo Ricci came along & finally solved the problem. He is pretty well-known for this accomplishment. I’ve thought about trying to contact him, but haven’t gotten around to it yet. Anyway, this Steinmetz Solid is quite remarkable in that it has perfectly SQUARE cross sections even though it was formed by the intersection of two perfectly ROUND cylinders. I knew I was on the right track when I saw that. The object is unusual; something like a queer deck of cards: It has the smallest square card on the bottom. From there the cards remain square in shape; but grow progressively larger as they approach the middle of the deck. And then they grow progressively smaller as they approach the top of the deck. Perfect circles the size of the Red Circle can be drawn around its midpoints; while perfect ellipses can be drawn around its vertices. Interestingly, those ellipses have a perimeter exactly equal in length to the circumference of a particular circle: 1/2pi x radius squared of Set C. (See my notes on Plate #4).
From these diagrams we can see what happens when we have two cylinders of equal diameter intersect horizontally at right angles. Now, let’s see what happens when we drive a third cylinder down vertically or perpendicularly through the center of those two cylinders.
From these diagrams we can see what happens when we have two cylinders of equal diameter intersect horizontally at right angles. Now, let’s see what happens when we drive a third cylinder down vertically or perpendicularly through the center of those two cylinders.
PLATE #23: Typical Tricylinder: It is formed by three cylinders of equal diameter intersecting at right angles. In this particular case the diameter of each cylinder would be the same length as the sidelength of the Red Square (1). If all three cylinders are reduced to their minimum height of 1 (as we did with the bicylinder; bottom drawing), and then inscribed this object within the Green sphere, it would trace out six circles upon its surface: Four around the equator circle & one each around the North & South Poles. It basically takes on the same shape as a cube, but instead of squares on each of it’s six faces, it has circles instead. Its Steinmetz Solid is the rhombic dodecahedron; which some say has a lot to do with the shape of the universe. In the next frame you will be able to see how this relates to the subject before us.
PLATE #24: Here, we've simply cut the Tricylinder in half on it’s horizontal axis. This reflects the floorplan for Hagia Sophia (Gr. Holy Wisdom). This is a Byzantine Church located in Istanbul Turkey, formerly Constantinople. It was constructed around 530 A.D. by Emperor Justinian. It was the first time the pendentive dome was used & it was immediately recognized as an architectural marvel. It still is. Upon it's completion Justinian is said to have exclaimed: "Solomon I have outdone thee!" Obviously, he was comparing it to Solomon’s Temple and referring to the sacred geometry & great wisdom utilized in its construction.
The fact that the "footprint" of this object is identical to the floorplan of the church is inescapable. Undoubtedly, the intention of the architects is both deliberate & unmistakable.
The fact that the "footprint" of this object is identical to the floorplan of the church is inescapable. Undoubtedly, the intention of the architects is both deliberate & unmistakable.
PLATE #25: Hagia Sophia: This was the seat of the Eastern Orthodox Church for nearly ten centuries. It's geometric design represents the marriage of heaven & earth; God & man through Christ Jesus. And "happy are those who are called to the Marriage Supper of the Lamb," (Rev. 19:7-9). As discussed earlier, the Circle represents God & the square represents man. The circular dome makes a graceful transition to the square base below via the pendentives. These pendentives are the strange looking green “triangles” at each of the four corners. [See Plate #27]. This evenly distributes the immense weight of the massive dome & channels it down through the four main columns at the four corners. Traditionally, the portraits of the four evangelists are depicted on these four pendentives: (Matthew, Mark, Luke & John). The theme is extremely symbolic. It represents God (the Circle) reconciling Himself with man (the square) by way of the pendentives (the Gospel of of our Lord Jesus Christ). As you can plainly see, the entire layout is shaped like a crucifix. Once in a while you might even encounter the phrase "square the circle" when researching this subject. It's usually employed figuratively or allegorically. But there's no doubt (at least in my mind) that it also solves this mystery in a literal sense.
PLATE #26: Here's another look at the floorplan of the Hagia Sophia. As you can see the whole thing is deliberately cruciform. Otherwise, the diagram is basically self-explanatory.
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This diagram could easily pass for a blueprint of Good Shepherd Catholic Church; a place of extraordinary beauty & very dear to me. I received all the Holy Sacrements there. Please bear with me for a moment while I tell you about something that happened to me there. I promise to be brief.
Now, the story of my life is pretty complicated. How I got to where I’m at right now is basically a long string of genuine miracles. My father was killed the day after my 7th birthday & I’ve been on one heck of a rollercoaster ride from that time forth. Since I grew up without a father I eventually became very rebellious. It goes without saying that I got into a lot of trouble. But I won’t bother you with all the details at this time. However, there is one incident that I’d like to mention which has a lot to do with the topic at hand. I attended classes at Good Shepherd Grade School. It was connected to Good Shepherd Church by way of a short corridor off the school gymnasium. One day I was down near the janitor’s office & noticed that he had left his keys lying on his workbench. Needless to say I grabbed them and slipped unnoticed down the hallway. Several days later the school principal announced the matter of the missing keys over the speaker system & asked that they be returned. Now, the janitor was a super nice guy & was old and somewhat crippled. I felt bad for him & returned them; annonymously of course. But I kept one key. That would be the key to the side door of the Church. That way, I could feel better about returning the stolen items & still gain entry to the school gymnasium late at night to play basketball with my buddies. There are a ton of stories to be told here! But like I said, I’ll try to be brief about this.
After graduating from grade school I moved on to High School.. After graduating from the third one I attended (which is a miracle in itself), I moved to Nevada. There are a ton of stories to be told there too. Anyway, I would often take very long walks deep into the desert wilderness. One day I encountered the Holy Spirit & although my life did not instantly turn around, it definitely was turned in the right direction. I immediately started reading the Bible on a daily basis. I was resolved to find out the truth about all this stuff for myself. I set my heart on the land of Israel & was determined to visit the place as soon as possible. I moved back home, got a job digging coal, applied for a passport, and began saving my money. I bought a plane ticket in 1982 & flew over there to see the Holy Land for myself.
Now the point I’ve been driving toward is this: I still had that key to the side door of the Church when I returned from Nevada. Just before & just after my trip to Israel I would slip into the Church very late at night; like two or three in the the morning. I would light a few of the candles up on the altar & pray for hours on end. And then leave; locking the door behind me. My God! What glorious memories! Just me & the Lord; all alone in that huge dark Church in the middle of the night! I won’t tell you about all the things I experienced there. But I will say that it was downright miraculous. Hmmm… I wonder what ever happened to that key.
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This diagram could easily pass for a blueprint of Good Shepherd Catholic Church; a place of extraordinary beauty & very dear to me. I received all the Holy Sacrements there. Please bear with me for a moment while I tell you about something that happened to me there. I promise to be brief.
Now, the story of my life is pretty complicated. How I got to where I’m at right now is basically a long string of genuine miracles. My father was killed the day after my 7th birthday & I’ve been on one heck of a rollercoaster ride from that time forth. Since I grew up without a father I eventually became very rebellious. It goes without saying that I got into a lot of trouble. But I won’t bother you with all the details at this time. However, there is one incident that I’d like to mention which has a lot to do with the topic at hand. I attended classes at Good Shepherd Grade School. It was connected to Good Shepherd Church by way of a short corridor off the school gymnasium. One day I was down near the janitor’s office & noticed that he had left his keys lying on his workbench. Needless to say I grabbed them and slipped unnoticed down the hallway. Several days later the school principal announced the matter of the missing keys over the speaker system & asked that they be returned. Now, the janitor was a super nice guy & was old and somewhat crippled. I felt bad for him & returned them; annonymously of course. But I kept one key. That would be the key to the side door of the Church. That way, I could feel better about returning the stolen items & still gain entry to the school gymnasium late at night to play basketball with my buddies. There are a ton of stories to be told here! But like I said, I’ll try to be brief about this.
After graduating from grade school I moved on to High School.. After graduating from the third one I attended (which is a miracle in itself), I moved to Nevada. There are a ton of stories to be told there too. Anyway, I would often take very long walks deep into the desert wilderness. One day I encountered the Holy Spirit & although my life did not instantly turn around, it definitely was turned in the right direction. I immediately started reading the Bible on a daily basis. I was resolved to find out the truth about all this stuff for myself. I set my heart on the land of Israel & was determined to visit the place as soon as possible. I moved back home, got a job digging coal, applied for a passport, and began saving my money. I bought a plane ticket in 1982 & flew over there to see the Holy Land for myself.
Now the point I’ve been driving toward is this: I still had that key to the side door of the Church when I returned from Nevada. Just before & just after my trip to Israel I would slip into the Church very late at night; like two or three in the the morning. I would light a few of the candles up on the altar & pray for hours on end. And then leave; locking the door behind me. My God! What glorious memories! Just me & the Lord; all alone in that huge dark Church in the middle of the night! I won’t tell you about all the things I experienced there. But I will say that it was downright miraculous. Hmmm… I wonder what ever happened to that key.
PLATE #27: Here is an excellent rendition of the pendentive dome. The relationship between the four Red semi-circles & the four sidelengths of the Red Square is very easy to see. I have figured out how to calculate it’s surface area: It is equal to the Green hemisphere’s surface area (pi) divided by the Silver Mean. (pi/1 + sq.rt 2) = 1.301290318. The pendentives are the strange looking “triangles” at each of the four corners. They can be concave or convex depending on what side of the structure you are on. They look like deltoids (that's a three-sided hypocycloid). But they're not. I’ve figured out how to calculate their surface area too: Area of the four pendentives is equal to the total surface area of the pendentive dome divided by (1 + the Silver Mean). In this particular case the four pendentives would have a total surface area of .381139; with each pendentive having a surface area of .095284779.
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I only have one more frame left in Part Three of this demonstration. But, before I wrap it up there’s a few more things I’d like to add to that story about Good Shepherd Church. Please bear with me for a few more moments as I finish.
As I indicated very early on, when I began this project I couldn’t remember even the most basic principles of geometry. I basically started from scratch. I spent the first three years focusing on the aforementioned three sets of nested squares & circles, the Divine Proportion, and the hypocycloid. At first I couldn’t figure out where the hypocycloid fit in. Some time around the beginning of my fourth year into this I met a guy who held a PhD in Mechanical Engineering. He spoke of lines on a sphere & Bingo! There’s where my hypocycloid belonged! It all seemed to make sense: Draw a hyperbolic shape of negative curvature (hypocycloid) on the surface of a parabolic shape of positive curvature (sphere) and the two cancel; leaving us with a Euclidean shape devoid of curvature (square) Well, sort of. Anyway, I was very excited about this development & for the next several months my focus centered upon what I called “the square on a sphere.” As I began to approach the beginning of my fifth year into this, progress had once again slowed to a snail’s pace.
So one day, kinda outta the blue, I decided to attend Easter mass at Good Shepherd Catholic Church. I hadn't been to mass in over 30 years. I arrived late so I stood in the very back. For whatever reason I glanced up at the massive dome & there it was; plain as day; my square on a sphere! I was utterly flabberghasted. All those long hours I spent praying in that Church in the middle of the night & it was right there all along! Needless to say, as soon as mass was over I rushed home to see what I could find out about this on the Internet. It wasn't long before I found that my square on a sphere had a name; it was called a pendentive dome. I examined hundreds of images of this unique structure & found that if I viewed it from a great enough height it would resemble a perfect inscribed square. And if I were to lay on my back in the middle of the Church & stare straight up at it, I’d see my hypocycloid staring right back down on me. Compare Plate #34 (the hypocycloid) with Plate #35. The image appearing on that frame (35) is from the dome of the Church of the Holy Sepulchre in Jerusalem. Incidentally, I stood under that dome as well; completely oblivious to the wisdom directly over my head. This event was a huge turning point in my research. My mission was back on track. Information began pouring in faster than I could process it. I began inscribing Bicylinders & Tricylinders inside spheres. I found that these objects carve out perfect circles on the sphere's surface identical to those we've been discussing. When viewing the inscribed Bicylinder from directly above, the curvature of each of its four openings could not be detected; much like looking at a paper plate sideways; and the four tangent circles appeared to be a perfect square (just like the Red Square in the previous diagrams). Later, I would study their respective Steinmetz Solids; which led me to investigate groin vaults & polygonal domes. I then began plucking different squares from my three sets of isometric drawings & projecting them onto different spheres. And the Golden Mean just fell outta the sky. Seriously. I wasn’t even looking for this or trying to square the circle from this angle. I found that certain squares were in Golden Ratio with the circumferences of certain circles and before I knew it, the method for squaring the circle depicted in Part Two of this demonstration was right there on the sheet of paper lying before me. I know it may sound ridiculous, but it’s almost as if I didn’t even draw it myself. It’s like it miraculously appeared on that sheet of paper like the handwriting on the wall in the Book of Daniel.
Ya know: It’s funny: It seems like I learned nearly everything I know about this problem in reverse. It's as if I slipped in through the back door. It was like the popular gameshow Jeopardy; where the contestants are given the answer & then challenged to frame the appropriate question to match it. You'd think I had started with the pendentive dome and then worked my way backwards; especially considering the fact that it represents the squaring of the circle and/or the Incarnation of the Son of God. (This is what surprises me the most. Why hasn't somebody already done this?) Anyway, as unbelieveable as it may sound, I had absolutely no idea that this math problem had such a deeply spiritual side to it. As far as I was concerned the only spiritual aspect of it was that a guy asserted God Himself couldn’t do it. It wasn’t until that Easter morning a couple years ago that I realized it had so many spiritual implications.
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I only have one more frame left in Part Three of this demonstration. But, before I wrap it up there’s a few more things I’d like to add to that story about Good Shepherd Church. Please bear with me for a few more moments as I finish.
As I indicated very early on, when I began this project I couldn’t remember even the most basic principles of geometry. I basically started from scratch. I spent the first three years focusing on the aforementioned three sets of nested squares & circles, the Divine Proportion, and the hypocycloid. At first I couldn’t figure out where the hypocycloid fit in. Some time around the beginning of my fourth year into this I met a guy who held a PhD in Mechanical Engineering. He spoke of lines on a sphere & Bingo! There’s where my hypocycloid belonged! It all seemed to make sense: Draw a hyperbolic shape of negative curvature (hypocycloid) on the surface of a parabolic shape of positive curvature (sphere) and the two cancel; leaving us with a Euclidean shape devoid of curvature (square) Well, sort of. Anyway, I was very excited about this development & for the next several months my focus centered upon what I called “the square on a sphere.” As I began to approach the beginning of my fifth year into this, progress had once again slowed to a snail’s pace.
So one day, kinda outta the blue, I decided to attend Easter mass at Good Shepherd Catholic Church. I hadn't been to mass in over 30 years. I arrived late so I stood in the very back. For whatever reason I glanced up at the massive dome & there it was; plain as day; my square on a sphere! I was utterly flabberghasted. All those long hours I spent praying in that Church in the middle of the night & it was right there all along! Needless to say, as soon as mass was over I rushed home to see what I could find out about this on the Internet. It wasn't long before I found that my square on a sphere had a name; it was called a pendentive dome. I examined hundreds of images of this unique structure & found that if I viewed it from a great enough height it would resemble a perfect inscribed square. And if I were to lay on my back in the middle of the Church & stare straight up at it, I’d see my hypocycloid staring right back down on me. Compare Plate #34 (the hypocycloid) with Plate #35. The image appearing on that frame (35) is from the dome of the Church of the Holy Sepulchre in Jerusalem. Incidentally, I stood under that dome as well; completely oblivious to the wisdom directly over my head. This event was a huge turning point in my research. My mission was back on track. Information began pouring in faster than I could process it. I began inscribing Bicylinders & Tricylinders inside spheres. I found that these objects carve out perfect circles on the sphere's surface identical to those we've been discussing. When viewing the inscribed Bicylinder from directly above, the curvature of each of its four openings could not be detected; much like looking at a paper plate sideways; and the four tangent circles appeared to be a perfect square (just like the Red Square in the previous diagrams). Later, I would study their respective Steinmetz Solids; which led me to investigate groin vaults & polygonal domes. I then began plucking different squares from my three sets of isometric drawings & projecting them onto different spheres. And the Golden Mean just fell outta the sky. Seriously. I wasn’t even looking for this or trying to square the circle from this angle. I found that certain squares were in Golden Ratio with the circumferences of certain circles and before I knew it, the method for squaring the circle depicted in Part Two of this demonstration was right there on the sheet of paper lying before me. I know it may sound ridiculous, but it’s almost as if I didn’t even draw it myself. It’s like it miraculously appeared on that sheet of paper like the handwriting on the wall in the Book of Daniel.
Ya know: It’s funny: It seems like I learned nearly everything I know about this problem in reverse. It's as if I slipped in through the back door. It was like the popular gameshow Jeopardy; where the contestants are given the answer & then challenged to frame the appropriate question to match it. You'd think I had started with the pendentive dome and then worked my way backwards; especially considering the fact that it represents the squaring of the circle and/or the Incarnation of the Son of God. (This is what surprises me the most. Why hasn't somebody already done this?) Anyway, as unbelieveable as it may sound, I had absolutely no idea that this math problem had such a deeply spiritual side to it. As far as I was concerned the only spiritual aspect of it was that a guy asserted God Himself couldn’t do it. It wasn’t until that Easter morning a couple years ago that I realized it had so many spiritual implications.
PLATE #28: Here I’ve simply “unfolded” the perimeter of the Red Square which converts it into a straight line. And then I did the same thing with the four red semi-circles. It is easy to see the relationship between the two figures. The perimeter of the Red Square increases by exactly 1/2pi when embedded in a two-dimensional smooth Riemannian manifold of positive Ricci curvature & bound by an equator circle the size of the Green Circle. Hence, the perimeter of the Red Square is exactly equal in length to the circumference of the Pink Circle (2pi). The circle is rectified via compass & straightedge.
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I realize my approach to this problem may seem somewhat "unorthodox." I don't know: Maybe that's not an accurate description; especially since I discovered this material dead center in the Eastern Orthodox Church! Maybe it's man's approach to this problem that is unorthodox & this has been the solution all along. But man, as he so often does, has spent his time "kicking against the pricks." There once was an Italian Jew by the name of Levi Cevita. He used Ricci's Absolute Differential Calculus to correct Einstein's field equations by calculating the curvature of spacetime. This eventually paved the way for his Theory of Relativity. Einstein would later prove that even light bends in space.
Question: If both time & space are curved, how can anything be straight? I mean, if the very medium within which all things exist is warped, how can anything embedded within it be flat?
And now, many decades later, we have another Jew, the reclusive genius Grigori Perelman, deforming Riemannian manifolds into cylinders via the Ricci Flow. By utilizing this extraordinary technique he eventually solved the Poincare Conjecture; considered to be the hardest question in all of mathematics. How much do you want to bet that this approach can somehow be applied to the solution presented above & eventually prove that this is & always has been the only viable solution to the problem?
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I realize my approach to this problem may seem somewhat "unorthodox." I don't know: Maybe that's not an accurate description; especially since I discovered this material dead center in the Eastern Orthodox Church! Maybe it's man's approach to this problem that is unorthodox & this has been the solution all along. But man, as he so often does, has spent his time "kicking against the pricks." There once was an Italian Jew by the name of Levi Cevita. He used Ricci's Absolute Differential Calculus to correct Einstein's field equations by calculating the curvature of spacetime. This eventually paved the way for his Theory of Relativity. Einstein would later prove that even light bends in space.
Question: If both time & space are curved, how can anything be straight? I mean, if the very medium within which all things exist is warped, how can anything embedded within it be flat?
And now, many decades later, we have another Jew, the reclusive genius Grigori Perelman, deforming Riemannian manifolds into cylinders via the Ricci Flow. By utilizing this extraordinary technique he eventually solved the Poincare Conjecture; considered to be the hardest question in all of mathematics. How much do you want to bet that this approach can somehow be applied to the solution presented above & eventually prove that this is & always has been the only viable solution to the problem?
PART FIVE: SQUARING THE CIRCLE IN AREA
PLATE #29: Isolate the Red Square & the Black Circle.
Proposition #2: The surface area bound by the Red Square is EXACTLY equal to the surface area bound by the Black Circle. Like Proposition #1 that probably sounds absurd; especially since I just indicated [in Plate #10] that the area of the Red Square was 1, and the area of the Black Circle was 1/3pi. But, again, things may not always be what they initially seem to be. So please bear with me on this. Hopefully, you & I both can gain something worthwhile.
PLATE #29: Isolate the Red Square & the Black Circle.
Proposition #2: The surface area bound by the Red Square is EXACTLY equal to the surface area bound by the Black Circle. Like Proposition #1 that probably sounds absurd; especially since I just indicated [in Plate #10] that the area of the Red Square was 1, and the area of the Black Circle was 1/3pi. But, again, things may not always be what they initially seem to be. So please bear with me on this. Hopefully, you & I both can gain something worthwhile.
PLATE #30: Isolate the Red Square & the Orange Circle. For the time being, eliminate the Black Circle.
Now I tread on uncharted territory. I’d like to say that this portion of the presentation is “hypothetical.” But since a hypothesis implies an educated guess & I am not educated, I cannot make that claim. Frankly, I am not qualified to solve this portion of the problem. I never took calculus or trigonometry. I dropped out of physics. For lack of a better word I guess you could call this a hunch. I’ve been drawn toward the Black Circle & the Red Square for a very long time. I am certain that the Red Square will gain area by exactly 1/3pi if embedded in the right sphere. Of course, the question is, Which sphere? Now, I know next to nothing about math, but it seems plausible that if we take a square with sides equal in length to the radius of a circle of area = pi, embed that square in a hemisphere with a total surface area 3x larger than that circle, that square will then encompass a total surface area 3x smaller than that initial circle. In other words, if we take the Red Square (which has sidelengths equal to the radius of a circle of area pi), embed that in a hemisphere with an equator circle the size of the Orange Circle (area = 3pi), then it will encompass an area equal to that of the Black Circle in flatland,(area = pi/3).
Now I tread on uncharted territory. I’d like to say that this portion of the presentation is “hypothetical.” But since a hypothesis implies an educated guess & I am not educated, I cannot make that claim. Frankly, I am not qualified to solve this portion of the problem. I never took calculus or trigonometry. I dropped out of physics. For lack of a better word I guess you could call this a hunch. I’ve been drawn toward the Black Circle & the Red Square for a very long time. I am certain that the Red Square will gain area by exactly 1/3pi if embedded in the right sphere. Of course, the question is, Which sphere? Now, I know next to nothing about math, but it seems plausible that if we take a square with sides equal in length to the radius of a circle of area = pi, embed that square in a hemisphere with a total surface area 3x larger than that circle, that square will then encompass a total surface area 3x smaller than that initial circle. In other words, if we take the Red Square (which has sidelengths equal to the radius of a circle of area pi), embed that in a hemisphere with an equator circle the size of the Orange Circle (area = 3pi), then it will encompass an area equal to that of the Black Circle in flatland,(area = pi/3).
PLATE #32: As noted earlier [Plate #13], under these circumstances the Red Square comes into Golden Ratio with the Orange Circle’s circumference.
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I had no idea the Golden Mean was going to make an appearance here. Had I not studied it early on I probably would’ve missed it altogether. Some may assert that this was a mere stroke of luck & that I stumbled upon this by chance. By definition I suppose one could say this was serendipitous. But I don’t believe that anything in this presentation is (or was) a matter of sheer happenstance. Although “time and chance happeneth to everything” (Ecc. 9:11), the Lord foreknows all things before they come to pass. And He is able to control these things if He sees fit. As Solomon once said, “The lot is cast into the lap; but the whole disposing thereof is of the Lord.” (Prov. 16:33). It is not possible for me to consider this to be a simple matter of chance. In fact, such an attitude on my part would be nothing less than a lack of faith. There have been far too many “tiny miracles” during this entire process for me to chalk it all up to mere chance. I’ve only listed a handful of them here. But there have been more; a lot more. I mean, working on this problem has drawn me back into the Church; both literally and geometrically. Now that alone is no small miracle!
Anyway, like I said before, I feel this reveals a kind of “signpost” pointing the way to the ultimate solution. But since my knowledge of mathematics is so limited, I’ve come to a kind of standstill. This is nothing new: I’ve experienced similar situations several times throughout the course of this endeavor. I’ve come a long way in the last seven years. Maybe it’s time to wait & see what the Lord has in mind.
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I had no idea the Golden Mean was going to make an appearance here. Had I not studied it early on I probably would’ve missed it altogether. Some may assert that this was a mere stroke of luck & that I stumbled upon this by chance. By definition I suppose one could say this was serendipitous. But I don’t believe that anything in this presentation is (or was) a matter of sheer happenstance. Although “time and chance happeneth to everything” (Ecc. 9:11), the Lord foreknows all things before they come to pass. And He is able to control these things if He sees fit. As Solomon once said, “The lot is cast into the lap; but the whole disposing thereof is of the Lord.” (Prov. 16:33). It is not possible for me to consider this to be a simple matter of chance. In fact, such an attitude on my part would be nothing less than a lack of faith. There have been far too many “tiny miracles” during this entire process for me to chalk it all up to mere chance. I’ve only listed a handful of them here. But there have been more; a lot more. I mean, working on this problem has drawn me back into the Church; both literally and geometrically. Now that alone is no small miracle!
Anyway, like I said before, I feel this reveals a kind of “signpost” pointing the way to the ultimate solution. But since my knowledge of mathematics is so limited, I’ve come to a kind of standstill. This is nothing new: I’ve experienced similar situations several times throughout the course of this endeavor. I’ve come a long way in the last seven years. Maybe it’s time to wait & see what the Lord has in mind.
PLATE #33: Four Ellipse Venn Diagram: Here we look at these things from a different perspective: Imagine the surface of the Orange Sphere was transparent. Now imagine the lines drawn on the previous diagram [Plate #32] as diameters of geodesic circles. In that diagram they appear as straight lines because they are being viewed from directly above. But if we turned the sphere 45 degrees & viewed them from a different angle, we would see that they are actually four geodesic circles & they would look much like those appearing here on the screen. Notice that each side of the Red Square falls on arcs of each of the four orthogonally intersecting geodesics. Of course this begs the question: How much of the sphere’s surface area is now encompassed by the Red Square?
PLATE #34: Here is a diagram of a typical hypocycloid; also known as an astroid (not asteroid). This, along with the Golden Mean, was the center of my attention for the first several years. It is a very special case in that it’s perimeter is exactly equal in length to the circumference of the circumscribing Red Circle & it’s area is exactly equal to the combined area of the four corner sections.
PLATE #35: Here is an image of a pendentive dome viewed from inside the Church. (In this case it is the Church of the Holy Sepulchre in Jerusalem). If you look up at it from directly below the hypocycloid comes into full view. If you view the same shape from directly above at great enough height, say, from God’s point of view, it resembles a perfect square (as depicted in Plate #16).
Here I advance two math problems. Thus far the solutions to these problems have eluded me. I would greatly appreciate any assistance my readers may have to offer.
PLATE #36: PROBLEM #1: COOKIE CUTTER: I have a spherical lump of cookie dough the size of the Orange Sphere below. I have a Red Square cookie cutter: Sides = 1; Area = 1. I depress that cookie cutter into the sphere such that its vertices sink deep into its surface (i.e., manifold). I carefully extract the cutter straight up & out of the sphere. I notice the cutter has left a “square” shape in the manifold of the sphere. Obviously, the sidelengths of the Red Square will gain curvature: They increase from a length of 1 to a length of 1.0367. The area also increases; but the amount of increase is unknown.
ORANGE SPHERE DIMENSIONS:
Diameter: (Sq.rt.6) = 2.4494897
Radius: (Sq.rt.1.5) = 1.2247448
Radius Squared = 1.5
Circumference = 7.6952987
Equator (flat) surface area: (pi x radius squared) = (1.5pi) 4.7123889
Hemisphere surface area: (2pi x radius squared) = (3pi) 9.4247778
Whole Sphere surface area: (4pi x radius squared) = (6pi) 18.849555
RED SQUARE (FLAT) DIMENSIONS
Unit Square: (Side = 1; Area = 1).
Question: What is the total surface area now encompassed by the red colored shape which has been etched into the sphere’s surface?
PROBLEM #2: OVERLAPPING SPHERICAL ZONES: Perhaps this would be easier to solve if it were viewed as two overlapping spherical zones? In actuality the size of the Red Square would be a bit larger than the one depicted here. But it still provides a good visual representation for my needs. As noted on [Plate #33], the sides of the Red Square, when depressed into the manifold of the sphere described above, will fall on arcs of four geodesic (small) circles of diameter sq.rt.5; (2.2360679). If these geodesics be thought of as boundaries for two overlapping spherical zones identical in size, the area encompassed by the Red Square would be the area shared by both. I have provided the dimensions for these zones below.
SPHERICAL ZONE DIMENSIONS
Spherical Zone Height = 1
Spherical Zone Area (each individual zone) = 7.6952987
Question: What is the total surface area of the overlapping portion of the two spherical zones (colored red in the diagram)?
PLATE #36: PROBLEM #1: COOKIE CUTTER: I have a spherical lump of cookie dough the size of the Orange Sphere below. I have a Red Square cookie cutter: Sides = 1; Area = 1. I depress that cookie cutter into the sphere such that its vertices sink deep into its surface (i.e., manifold). I carefully extract the cutter straight up & out of the sphere. I notice the cutter has left a “square” shape in the manifold of the sphere. Obviously, the sidelengths of the Red Square will gain curvature: They increase from a length of 1 to a length of 1.0367. The area also increases; but the amount of increase is unknown.
ORANGE SPHERE DIMENSIONS:
Diameter: (Sq.rt.6) = 2.4494897
Radius: (Sq.rt.1.5) = 1.2247448
Radius Squared = 1.5
Circumference = 7.6952987
Equator (flat) surface area: (pi x radius squared) = (1.5pi) 4.7123889
Hemisphere surface area: (2pi x radius squared) = (3pi) 9.4247778
Whole Sphere surface area: (4pi x radius squared) = (6pi) 18.849555
RED SQUARE (FLAT) DIMENSIONS
Unit Square: (Side = 1; Area = 1).
Question: What is the total surface area now encompassed by the red colored shape which has been etched into the sphere’s surface?
PROBLEM #2: OVERLAPPING SPHERICAL ZONES: Perhaps this would be easier to solve if it were viewed as two overlapping spherical zones? In actuality the size of the Red Square would be a bit larger than the one depicted here. But it still provides a good visual representation for my needs. As noted on [Plate #33], the sides of the Red Square, when depressed into the manifold of the sphere described above, will fall on arcs of four geodesic (small) circles of diameter sq.rt.5; (2.2360679). If these geodesics be thought of as boundaries for two overlapping spherical zones identical in size, the area encompassed by the Red Square would be the area shared by both. I have provided the dimensions for these zones below.
SPHERICAL ZONE DIMENSIONS
Spherical Zone Height = 1
Spherical Zone Area (each individual zone) = 7.6952987
Question: What is the total surface area of the overlapping portion of the two spherical zones (colored red in the diagram)?
Part Five:
Squaring the Circle precisely on Riemannian Manifolds
Creating a Triangle that is equal in area to that of a Circle
Heretofore I've focused on the four-cusped version of the hypocycloid (i.e. astroid; Gr. "Star"). Now, I would like to direct your attention to the three-cusped version (i.e. the deltoid).
Here, we repeat the same exercise outlined on Plates #17 & #31. Imagine you walk outside at midnight & gaze up at the full moon. But the image appearing in the night sky is not what you'd typically expect to find. Instead, it appears to have an equilateral triangle inscribed within it's equator circumference. If we consider the moon's diameter length to be the square root of 8, the surface area bound by this "spherical triangle" would be EXACTLY equal to pi: (3.14159265358979323846264338327950288...).
This finding is significant. It clearly demonstrates the fact that when Euclidean shapes are projected onto Riemannian manifolds they are, indeed, capable of "squaring the circle" exactly in area; right on down to the transcendental nature of pi.
Only 10 easy moves are required to construct the actual circle that is equal in area to that of the "triangle." First, inscribe a square inside the equator circle (9 moves). Then inscribe a circle inside that square (1 move). That innermost circle when embedded in a Ricci-flat two-dimensional manifold with zero curvature is EXACTLY equal in area to that which is bound by the "triangle" when embedded in a two-dimensional smooth Riemannian manifold of positive Ricci curvature.
Calculations & Dimensions:
Sphere Diameter = (Square Root of 8) = 2.828427125
Sphere Circumference (at equator) = 8.885765876
Radius = (Square Root of 2) = 1.414213562
Radius Squared = 2
Area of Equator Circle (flat) = (2 X pi) = 6.283185307
Hemisphere Surface Area = (4 x pi) = 12.56637061
Calculating the Area of each of the Three Caps cut by the Triangle:
The Spherical Triangle cuts three identical spherical caps & divides the hemisphere into four zones; each equal to pi exactly: (3.1415926535...).
Chord (side length of the triangle) = (Square Root of 6) = 2.449489743
Chord Height = (One fourth X diameter length) .707106781
Area of Each of the Three Caps: Chord Height (.707106781) X Circumference (8.885765876) = (2pi) 6.283185307.
Since we're dealing with a hemisphere here, as opposed to the whole sphere, each cap should be divided by 2 = (pi) 3.1415926535.
Hence, all three Hemispherical Caps would be = (3pi).
Conclusion: Total Hemisphere Area (4pi) - Three Hemispherical Caps (3pi) = Area of Triangle (pi).
Interestingly, the triangle divides the hemisphere into four equal sectors; each with a total surface area exactly equal to pi.
Squaring the Circle precisely on Riemannian Manifolds
Creating a Triangle that is equal in area to that of a Circle
Heretofore I've focused on the four-cusped version of the hypocycloid (i.e. astroid; Gr. "Star"). Now, I would like to direct your attention to the three-cusped version (i.e. the deltoid).
Here, we repeat the same exercise outlined on Plates #17 & #31. Imagine you walk outside at midnight & gaze up at the full moon. But the image appearing in the night sky is not what you'd typically expect to find. Instead, it appears to have an equilateral triangle inscribed within it's equator circumference. If we consider the moon's diameter length to be the square root of 8, the surface area bound by this "spherical triangle" would be EXACTLY equal to pi: (3.14159265358979323846264338327950288...).
This finding is significant. It clearly demonstrates the fact that when Euclidean shapes are projected onto Riemannian manifolds they are, indeed, capable of "squaring the circle" exactly in area; right on down to the transcendental nature of pi.
Only 10 easy moves are required to construct the actual circle that is equal in area to that of the "triangle." First, inscribe a square inside the equator circle (9 moves). Then inscribe a circle inside that square (1 move). That innermost circle when embedded in a Ricci-flat two-dimensional manifold with zero curvature is EXACTLY equal in area to that which is bound by the "triangle" when embedded in a two-dimensional smooth Riemannian manifold of positive Ricci curvature.
Calculations & Dimensions:
Sphere Diameter = (Square Root of 8) = 2.828427125
Sphere Circumference (at equator) = 8.885765876
Radius = (Square Root of 2) = 1.414213562
Radius Squared = 2
Area of Equator Circle (flat) = (2 X pi) = 6.283185307
Hemisphere Surface Area = (4 x pi) = 12.56637061
Calculating the Area of each of the Three Caps cut by the Triangle:
The Spherical Triangle cuts three identical spherical caps & divides the hemisphere into four zones; each equal to pi exactly: (3.1415926535...).
Chord (side length of the triangle) = (Square Root of 6) = 2.449489743
Chord Height = (One fourth X diameter length) .707106781
Area of Each of the Three Caps: Chord Height (.707106781) X Circumference (8.885765876) = (2pi) 6.283185307.
Since we're dealing with a hemisphere here, as opposed to the whole sphere, each cap should be divided by 2 = (pi) 3.1415926535.
Hence, all three Hemispherical Caps would be = (3pi).
Conclusion: Total Hemisphere Area (4pi) - Three Hemispherical Caps (3pi) = Area of Triangle (pi).
Interestingly, the triangle divides the hemisphere into four equal sectors; each with a total surface area exactly equal to pi.
The Silver Dome
Creating a Square that is equal in area to the of a Circle
According to the Jewish Encyclopedia the Dutch philosopher Baruch Spinoza “expressly declared that to him the notion that God took upon Himself the nature of man (in the person of Jesus Christ) seemed as self-contradictory as would be the statement that the circle has taken on the nature of the square.”
That statement was made in ignorance because squares & circles are just different ways of looking at the same things. It's not possible to "square the circle" only in that it's not possible to do that which is already done. It's not that it can never be done: It's that it can never be undone. To demonstrate my point, imagine the following scenario:
You take a compass & straightedge & construct a simple inscribed square. This is a relatively simple task. Now, imagine the square's area to be equal to the Silver Mean (1 + sq.rt. 2 = 2.414213562). Consequently, each of it's four sides would then be equal to 1.553773974 & it's diagonal would be equal to 2.197368226. Obviously, the length of the square's diagonal would also be equal in length to the circumscribing circle's diameter. Hence, the circle's radius length would be equal to 1.098684113; it's radius squared would be equal to 1.207106781; it's circumference would be equal to 6.903235877; and it's area would be equal to 3.792237794. (Please see the next 12 frames for details on the actual procedure required to produce such a square).
Now let's change the surface from Euclidean to Riemannian. To do this we simply repeat the exercises outlined on Plates 17, 31, & 37. But let me remind you: Although we're altering the surface upon which the square is embedded, the diagram itself remains intact; from this angle it's appearance does not change in any way. To visualize what I'm talking about, imagine you walk outside in the middle of the night & gaze up at the full moon (which is not really a full moon at all, but rather, only half the moon; a hemisphere). But the image appearing in the night sky is not what you'd typically expect to find. Instead, the image you see is identical to the diagram depicted here. Taking into consideration the aforementioned calculations, the area bound by the inscribed square would be exactly equal to pi (3.141592653589793238462643383279...).
So then, under these conditions the square inherits an intimate relationship with the circle; a marriage if you will; including even it's transcendental nature. So maybe the incarnation of God in the person of Jesus Christ is not so “self-contradictory” after all. Perhaps then, Spinoza would've displayed far more wisdom had he adopted the philosophy of that great Christian apologist Tertullian instead, when he said, “Credo quia absurdum;" (I believe BECAUSE it is absurd).
Creating a Square that is equal in area to the of a Circle
According to the Jewish Encyclopedia the Dutch philosopher Baruch Spinoza “expressly declared that to him the notion that God took upon Himself the nature of man (in the person of Jesus Christ) seemed as self-contradictory as would be the statement that the circle has taken on the nature of the square.”
That statement was made in ignorance because squares & circles are just different ways of looking at the same things. It's not possible to "square the circle" only in that it's not possible to do that which is already done. It's not that it can never be done: It's that it can never be undone. To demonstrate my point, imagine the following scenario:
You take a compass & straightedge & construct a simple inscribed square. This is a relatively simple task. Now, imagine the square's area to be equal to the Silver Mean (1 + sq.rt. 2 = 2.414213562). Consequently, each of it's four sides would then be equal to 1.553773974 & it's diagonal would be equal to 2.197368226. Obviously, the length of the square's diagonal would also be equal in length to the circumscribing circle's diameter. Hence, the circle's radius length would be equal to 1.098684113; it's radius squared would be equal to 1.207106781; it's circumference would be equal to 6.903235877; and it's area would be equal to 3.792237794. (Please see the next 12 frames for details on the actual procedure required to produce such a square).
Now let's change the surface from Euclidean to Riemannian. To do this we simply repeat the exercises outlined on Plates 17, 31, & 37. But let me remind you: Although we're altering the surface upon which the square is embedded, the diagram itself remains intact; from this angle it's appearance does not change in any way. To visualize what I'm talking about, imagine you walk outside in the middle of the night & gaze up at the full moon (which is not really a full moon at all, but rather, only half the moon; a hemisphere). But the image appearing in the night sky is not what you'd typically expect to find. Instead, the image you see is identical to the diagram depicted here. Taking into consideration the aforementioned calculations, the area bound by the inscribed square would be exactly equal to pi (3.141592653589793238462643383279...).
So then, under these conditions the square inherits an intimate relationship with the circle; a marriage if you will; including even it's transcendental nature. So maybe the incarnation of God in the person of Jesus Christ is not so “self-contradictory” after all. Perhaps then, Spinoza would've displayed far more wisdom had he adopted the philosophy of that great Christian apologist Tertullian instead, when he said, “Credo quia absurdum;" (I believe BECAUSE it is absurd).
Step #11: You may recall that at the very beginning of this demonstration I made the following statement:
"It is essential that we find a way to thrust the square into the same transcendental world as the circle; for there's no doubt it belongs there. In our quest to accomplish this feat it must be remembered that the square is a "slice" of a cube & is Euclidean in nature; while the circle is a "slice" of a sphere & is spherical in nature. When we take the two of them and lay them side by side within (not upon) a flat surface, we've unwittingly performed a mathematical function. Obviously, the square is perfectly at home there. But we've done injustice to the circle: We've deported it from its comfortable home in spherical geometry, stripped it of its double curvature, & thrust it into flatland; a world utterly foreign to it. If we hope to rectify and/or square the circle in a literal sense we need to balance the equation. We need to give to the square what we took from the circle. And we need to do it without the square being aware of it. In other words, the square must inherit an intimate relationship with pi, a kind of marriage with the circle. And this must be accomplished without making any visible changes to the diagram. That way it can still be drawn with the primitive instruments at our disposal; namely, a straightedge & unmarked compass. This is essentially what I've done."
Consider now, the diagram illustrated on the screen: Here, I've separated the two figures & placed them on two different manifolds. The top drawing depicts the Red Circle on a Euclidean Ricci-flat manifold. Nothing whatsoever has changed; including even the surface upon which it rests.
The bottom diagram still depicts the "Silver Square" centered within the Orange Circle, except now the Orange Circle represents a smooth Riemannian manifold of positive Ricci curvature; (i.e., a hemisphere). Nevertheless, let it be known: Although we're altering the surface upon which the square is embedded, the diagram itself remains intact; from this angle it's appearance does not change in any way: It looks exactly like it did before the change took place & can still be drawn with compass & straightedge. In fact, even under these unique circumstances it still retains most of it's Euclidean characteristics: The manifold is still two-dimensional, the four sides are equal in length, they intersect orthogonally (that is, at right angles), sides of the square opposite one another remain equidistant throughout their entire length, and the diagonal still obeys the Pythagorean Theorem.
The only obvious difference in appearance is that I've eliminated the colors. But I've done this merely to enhance the visual effect. As has been customary throughout this entire demonstration I have found it much easier to explain what I'm talking about, and for others to understand these complex principles, if I can persuade them to view this diagram as a snapshot of a full moon; which by comparison is smoother than a billiard ball. As described on Plate #38, under these conditions the Red Circle & the inscribed Silver Square are exactly equal in area; despite the transcendental nature of pi.
"It is essential that we find a way to thrust the square into the same transcendental world as the circle; for there's no doubt it belongs there. In our quest to accomplish this feat it must be remembered that the square is a "slice" of a cube & is Euclidean in nature; while the circle is a "slice" of a sphere & is spherical in nature. When we take the two of them and lay them side by side within (not upon) a flat surface, we've unwittingly performed a mathematical function. Obviously, the square is perfectly at home there. But we've done injustice to the circle: We've deported it from its comfortable home in spherical geometry, stripped it of its double curvature, & thrust it into flatland; a world utterly foreign to it. If we hope to rectify and/or square the circle in a literal sense we need to balance the equation. We need to give to the square what we took from the circle. And we need to do it without the square being aware of it. In other words, the square must inherit an intimate relationship with pi, a kind of marriage with the circle. And this must be accomplished without making any visible changes to the diagram. That way it can still be drawn with the primitive instruments at our disposal; namely, a straightedge & unmarked compass. This is essentially what I've done."
Consider now, the diagram illustrated on the screen: Here, I've separated the two figures & placed them on two different manifolds. The top drawing depicts the Red Circle on a Euclidean Ricci-flat manifold. Nothing whatsoever has changed; including even the surface upon which it rests.
The bottom diagram still depicts the "Silver Square" centered within the Orange Circle, except now the Orange Circle represents a smooth Riemannian manifold of positive Ricci curvature; (i.e., a hemisphere). Nevertheless, let it be known: Although we're altering the surface upon which the square is embedded, the diagram itself remains intact; from this angle it's appearance does not change in any way: It looks exactly like it did before the change took place & can still be drawn with compass & straightedge. In fact, even under these unique circumstances it still retains most of it's Euclidean characteristics: The manifold is still two-dimensional, the four sides are equal in length, they intersect orthogonally (that is, at right angles), sides of the square opposite one another remain equidistant throughout their entire length, and the diagonal still obeys the Pythagorean Theorem.
The only obvious difference in appearance is that I've eliminated the colors. But I've done this merely to enhance the visual effect. As has been customary throughout this entire demonstration I have found it much easier to explain what I'm talking about, and for others to understand these complex principles, if I can persuade them to view this diagram as a snapshot of a full moon; which by comparison is smoother than a billiard ball. As described on Plate #38, under these conditions the Red Circle & the inscribed Silver Square are exactly equal in area; despite the transcendental nature of pi.
Step #12: If we anchor the compass on point [I], extend it out to point [H], and draw a circle around the square [B-G-H-F], this diagram can also be used to "rectify the circle." In other words, under these conditions the perimeter of the innermost blue square would be exactly equal in length to the circumference of the outermost blue circle. (Note: Most of this material has already been covered in Part Three of the main presentation; plates 15-28. Please consult that section for additional details & diagrams).
As explained earlier, this diagram is a perfect rendition of the pendentive dome from blueprint perspective. (See Plates 17-27). Under these conditions the innermost square's dimensions would be transformed as follows:
SILVER SQUARE SUPERIMPOSED ON RIEMANNIAN MANIFOLD
Side: Increases by 1/2 pi = 2.440662451
Diagonal: Increases by 1/2 pi = 3.45161794
Perimeter: Increases by 1/2 pi = 9.762649804
Area = Increases by 1.3012902 = Pi exactly: 3.1415926535897932384626433832795028841971...
There are a couple of things that immediately capture the attention here: First of all we can plainly see that the area of the Red Circle & the innermost Silver Square are equal to pi exactly. Hence, the circle has been squared in area. But we also see that under these conditions the sides, diagonal, and perimeter of the Silver Square all increase by exactly 1/2 pi. Consequently, the perimeter of the Silver Square increases to 9.762649805; which is the exact same size of the outermost blue circle's circumference. Therefore, the circle has also been rectified exactly here. Interestingly, this iis how big the hemisphere would appear if we took a giant spatula & smashed it's curved surface down flat into a Euclidean surface.
It is also worthy to note that in Euclidean Geometry the ratio of a square's perimeter length to the length of the circumscribing circle's circumference is the Form Factor Ratio (1.1107207345). In other words, the perimeter length of a standard inscribed square multiplied by this ratio will provide the circumference length of the circle around it. Certainly, it is by some grand design, rather than sheer coincidence, that this very same ratio makes an appearance right here in this diagram. Each of the four hemispherical sections lying outside the boundaries of the innermost square (i.e., silver square) have a surface area exactly equivalent to this unique ratio (1.1107207345). Therefore, the two great geometries (Euclidean & Spherical) are brought together in the same diagram. If we take one (flat) side length of the innermost square (1.553773974) & multiply that by the total amount of hemispherical (curved) area lying outside that square (4.4428828), the result would be precisely equal to the circumference length of the equator circle!
As explained earlier, this diagram is a perfect rendition of the pendentive dome from blueprint perspective. (See Plates 17-27). Under these conditions the innermost square's dimensions would be transformed as follows:
SILVER SQUARE SUPERIMPOSED ON RIEMANNIAN MANIFOLD
Side: Increases by 1/2 pi = 2.440662451
Diagonal: Increases by 1/2 pi = 3.45161794
Perimeter: Increases by 1/2 pi = 9.762649804
Area = Increases by 1.3012902 = Pi exactly: 3.1415926535897932384626433832795028841971...
There are a couple of things that immediately capture the attention here: First of all we can plainly see that the area of the Red Circle & the innermost Silver Square are equal to pi exactly. Hence, the circle has been squared in area. But we also see that under these conditions the sides, diagonal, and perimeter of the Silver Square all increase by exactly 1/2 pi. Consequently, the perimeter of the Silver Square increases to 9.762649805; which is the exact same size of the outermost blue circle's circumference. Therefore, the circle has also been rectified exactly here. Interestingly, this iis how big the hemisphere would appear if we took a giant spatula & smashed it's curved surface down flat into a Euclidean surface.
It is also worthy to note that in Euclidean Geometry the ratio of a square's perimeter length to the length of the circumscribing circle's circumference is the Form Factor Ratio (1.1107207345). In other words, the perimeter length of a standard inscribed square multiplied by this ratio will provide the circumference length of the circle around it. Certainly, it is by some grand design, rather than sheer coincidence, that this very same ratio makes an appearance right here in this diagram. Each of the four hemispherical sections lying outside the boundaries of the innermost square (i.e., silver square) have a surface area exactly equivalent to this unique ratio (1.1107207345). Therefore, the two great geometries (Euclidean & Spherical) are brought together in the same diagram. If we take one (flat) side length of the innermost square (1.553773974) & multiply that by the total amount of hemispherical (curved) area lying outside that square (4.4428828), the result would be precisely equal to the circumference length of the equator circle!
Step #19: Changed Manifold: Here we have two shapes embedded in similar manifolds: The smaller circle we produced above is now embedded in a sphere with a diameter of 2. The circumference of this smaller circle is 5.441398093. The inscribed square below is our Silver Square & is embedded in a sphere with a diameter of the square root of (2 + sq.rt. 8) = 2.197368227. The circle is now squared on a Riemannian manifold; the inscribed circle & the inscribed square are both exactly equal to pi.
So now we see that the Silver Square (under these conditions) is capable of squaring the circle in both ways: It is exactly equal in positively curved area to the amount of FLAT area encompassed by the white circle above (3.1415926535...). It is also exactly equal in size to the amount of positively CURVED area bound by the smaller black circle when embedded in the surface of a sphere of this size; which is also equal to pi exactly (3.1415926535...).
So now we see that the Silver Square (under these conditions) is capable of squaring the circle in both ways: It is exactly equal in positively curved area to the amount of FLAT area encompassed by the white circle above (3.1415926535...). It is also exactly equal in size to the amount of positively CURVED area bound by the smaller black circle when embedded in the surface of a sphere of this size; which is also equal to pi exactly (3.1415926535...).
Three Moons
Okay... today is pi day; 3/14. So to commemorate the occasion I've decided to post something thought-provoking in honor of this mysterious number:
Take a look at the diagram before you: Here we have three plain white circles of differing sizes lying on a flat sheet of paper. We will label the top one Circle [A], the middle one Circle [B], and the bottom one Circle [C]. They have three of the most fundamental shapes in all of geometry inscribed within them; a circle, a triangle, and a square respectively. Please note: All the shapes shown here are interrelated geometrically & can be drawn on a single diagram with only a compass & unmarked straightedge in under thirty minutes.
Circle [A] is considered a Unit Circle: It has a diameter length of 2; Area = pi. The smaller inscribed circle has a circumference of 5.441398093.
Circle [B] has a diameter length of sq.rt. 8 = (2.828427125); Area = 2 pi.
Circle [C] has a diameter length of the square root of those two values added together: The sq.rt. of (2 + sq.rt. 8) = 2.197368227. It’s area = 1/2pi x Silver Mean.
Now, let's change the manifold. To demonstrate what I'm talking about imagine you walk outside in the middle of the night & gaze up into the pitch black sky and notice three full moons; each with it’s own unique shape inscribed within it’s equator circumference: In other words, the image you see in the night sky is exactly like the one appearing here on your screen.
Taking into consideration the aforementioned calculations, the circle is squared in two totally different worlds here; both Euclidean & Spherical Geometry. Under these conditions both the inscribed triangle and the inscribed square encompass the same amount of CURVED surface area as that which is bound by the small inscribed circle on Hemisphere [A]; with that being pi exactly. Also, the amount of curved surface area bound by the inscribed square & triangle is exactly equal to the amount of FLAT surface area bound by the flat equator Circle [A]; which is also pi exactly.
Several other neat things can be observed here. Since we are dealing with hemispheres rather than flat circles, their respective areas have now doubled in size. The area of Hemisphere [A] is now 2pi; the inscribed circle dividing the hemisphere’s surface into two separate sectors exactly equal to pi each. The area of hemisphere [B] is now 4pi; the inscribed triangle dividing it’s surface into four separate sectors with each one being exactly equal to pi. Hemisphere [C] is especially intriguing: The area bound by the square is pi exactly. The area of each of the four separate sectors lying outside the perimeter of that square are equal to the Form Factor Ratio (1.110720734). Hence, the two great geometries (Spherical & Euclidean) are brought together in the same diagram: One FLAT side length of the square (1.553773974) multiplied by the total amount of positively CURVED surface area lying outside the perimeter of that square (4.4428828) = The exact length of the equator circumference of the hemisphere!
Pretty neat huh?
I'm pretty sure this can be accomplished with other inscribed polygons as well (pentagons, hexagons, octagons, etc). For example: A pentagon embedded in the manifold of Sphere [A] will enclose a total amount of curved surface area equal to (2pi-3). But the math begins to get really complicated after that. Dealing with so many irrational numbers in combination with transcendental numbers makes the proof extremely difficult.
Okay... today is pi day; 3/14. So to commemorate the occasion I've decided to post something thought-provoking in honor of this mysterious number:
Take a look at the diagram before you: Here we have three plain white circles of differing sizes lying on a flat sheet of paper. We will label the top one Circle [A], the middle one Circle [B], and the bottom one Circle [C]. They have three of the most fundamental shapes in all of geometry inscribed within them; a circle, a triangle, and a square respectively. Please note: All the shapes shown here are interrelated geometrically & can be drawn on a single diagram with only a compass & unmarked straightedge in under thirty minutes.
Circle [A] is considered a Unit Circle: It has a diameter length of 2; Area = pi. The smaller inscribed circle has a circumference of 5.441398093.
Circle [B] has a diameter length of sq.rt. 8 = (2.828427125); Area = 2 pi.
Circle [C] has a diameter length of the square root of those two values added together: The sq.rt. of (2 + sq.rt. 8) = 2.197368227. It’s area = 1/2pi x Silver Mean.
Now, let's change the manifold. To demonstrate what I'm talking about imagine you walk outside in the middle of the night & gaze up into the pitch black sky and notice three full moons; each with it’s own unique shape inscribed within it’s equator circumference: In other words, the image you see in the night sky is exactly like the one appearing here on your screen.
Taking into consideration the aforementioned calculations, the circle is squared in two totally different worlds here; both Euclidean & Spherical Geometry. Under these conditions both the inscribed triangle and the inscribed square encompass the same amount of CURVED surface area as that which is bound by the small inscribed circle on Hemisphere [A]; with that being pi exactly. Also, the amount of curved surface area bound by the inscribed square & triangle is exactly equal to the amount of FLAT surface area bound by the flat equator Circle [A]; which is also pi exactly.
Several other neat things can be observed here. Since we are dealing with hemispheres rather than flat circles, their respective areas have now doubled in size. The area of Hemisphere [A] is now 2pi; the inscribed circle dividing the hemisphere’s surface into two separate sectors exactly equal to pi each. The area of hemisphere [B] is now 4pi; the inscribed triangle dividing it’s surface into four separate sectors with each one being exactly equal to pi. Hemisphere [C] is especially intriguing: The area bound by the square is pi exactly. The area of each of the four separate sectors lying outside the perimeter of that square are equal to the Form Factor Ratio (1.110720734). Hence, the two great geometries (Spherical & Euclidean) are brought together in the same diagram: One FLAT side length of the square (1.553773974) multiplied by the total amount of positively CURVED surface area lying outside the perimeter of that square (4.4428828) = The exact length of the equator circumference of the hemisphere!
Pretty neat huh?
I'm pretty sure this can be accomplished with other inscribed polygons as well (pentagons, hexagons, octagons, etc). For example: A pentagon embedded in the manifold of Sphere [A] will enclose a total amount of curved surface area equal to (2pi-3). But the math begins to get really complicated after that. Dealing with so many irrational numbers in combination with transcendental numbers makes the proof extremely difficult.
