TETRAGONOS The Four Corners of the Earth

...In the beginning YHWH 'Elohiym created the heaven and the earth. And the earth was without form, and void; and darkness was upon the face of the deep. And the Spirit of 'Elohiym moved upon the face of the waters. And 'Elohiym said, Let there be light: and there was light... What more is light than the stretching of a line? What more is the beam of light than a measuring rod in the hand of the Almighty? Oh YHWH, holy and true are thy ways! Just and upright are all thy thoughts! I want to spend the remainder of days with you, and when all of my days are gone, let me spend eternity with you. You alone are Holy and there is none other: No not one... Any perfect circle has a particular equilateral triangle that fits perfectly within its circumference. Likewise; any equilateral triangle also has a corresponding perfect circle. This got me thinking one day: Does an equilateral triangle, placed inside of its corresponding perfect circle, also represent an equilateral pyramid inside of a corresponding perfect sphere? Does an equilateral pyramid even have a corresponding sphere? What Yeshua has shown me is extraordinary, at least I thought so. Any perfect circle may be drawn, and its corresponding equilateral triangle may then be found by "squaring" the circle as in this first diagram, (Fig. A). Once a square is drawn outside of the circle as shown, its diagonals will betray the baseline of the triangle as they intersect the inner square marking the two lower "X's" shown in Fig. A. The two upper "X's" may be used to draw an inverted equilateral triangle, (a second upside-down triangle,) thus forming a six-pointed “Star of David” inside the circle; and yes, the triangle is always equilateral if the drawing is correct. Copy this diagram and try it yourself with any size circle.

So the question is: Does this equilateral triangle represent the base of an equilateral pyramid inside of a corresponding sphere? Here is what I found, glory be to YHWH: The objective was to find out if this equilateral triangle is also representative of an equilateral pyramid inside of a corresponding sphere. The triangle has 2 ¼” sides and the measurement of each side of the outer square is 2 9/16” thus revealing that the circle has a pole/diameter of the same measurement, (2 9/16”) see Fig. D below. I will also use some of the original drawings which were measured by hand, on paper, to help validate this article.

Technically speaking; a pyramid, by definition, cannot be equilateral since by its very own definition it is a four, (or more,) sided object with a polygonal base, and with triangular sides meeting at an apex at its top. Though a triangle is of the polygon family, it is not the base of any pyramidal shape by the very definition of a pyramid. The base of any pyramid has four sides or more. The object in question here is a tetrahedron. A tetrahedron is a four sided object with triangular sides including its base, (three sides plus the base). In the case of an equilateral tetrahedron; there is no top, or apex, since all four vertices are equal; all six linear sides are equal; all four surfaces are equal in area and dimension; and all of its angles are equal. Therefore the equilateral tetrahedron does not have a highest point; it may be tossed like dice, (if it is small enough,) and it will always land upright. No matter which side it lands on, that side becomes its base. In Fig.  B, the triangle ABC is currently the base, and D is the fourth vertex, but ABC, ABD, ACD, and BCD are all four equal sides with equal angles. Since all triangles have three angles which must add up to 180°; all four equilateral ,triangular, sides of this tetrahedron have three angles, of 60° each.

Also of note is that an equilateral tetrahedron, when held at the proper angle from a distance, has the optical illusion of a square and actually forms, and defines, the overall dimensions of a perfect cube. Its four vertices also point in four opposing directions, i.e. North, South, East, and West. However; I do not mean the four cardinal points on a flat-Earth map. The equilateral tetrahedron points out into space in four opposing directions.

Figure C appears to be a square, or even the top view of a cube with diagonals running through it, but it is also a view of the equilateral tetrahedron. Side AC is at the backside of the tetrahedron; side BD is at the front, (or visa-versa).

The 2 ¼” equilateral tetrahedron does not fit into a corresponding 2 9/16” sphere but it does fit into a knowable sphere. Herein is the math:

If we imagine that the 2 9/16” circle is a sphere, in this “from above view,” (Fig. D) then the four vertices of the proposed tetrahedron are A, B, C, and D. The fourth vertex (D) is at the top, and center, of the tetrahedron as well as at the top of the 2 9/16” poled sphere, (the sphere is 2 9/16” at any given pole). A, B,  & C, are the three vertices that comprise the base of the equilateral tetrahedron, thus the equilateral triangle, ABC, is the base of the equilateral tetrahedron ABCD.

This being the case it becomes obvious that the tetrahedron cannot fit into the 2 9/16” sphere. For if the 2 9/16” circle is now a sphere; and vertex D is at the top of the sphere in this “above view” then the base of the tetrahedron, (ABC) is positioned planally congruent with, and also coexists with, the equatorial plane of the 2 9/16” sphere. This means that the entire tetrahedron must fit into one hemisphere, or exactly half of the sphere in question. Since the height of one hemisphere is a radius of the sphere, (half of 2 9/16”), the tetrahedron cannot fit. To find the true size of the sphere we must “cube” the tetrahedron; for then we may easily discover the true nature of these things. Believe it, or not, the “cubing” of the tetrahedron has already been done for us here in Figure C above. Since the tetrahedron has six equal, linear, sides that are all 2 ¼” then it fits within a cube that has the same diagonals, (as shown in Fig. C). If we now picture Fig. D as a “side view” of the tetrahedron, inside of a sphere, it becomes evident that the chord BC rests on a 2 ¼” latitudenal circular plane around the 2 9/16” imagined sphere. This means the base triangle ABC also exists on a latitudenal plane of 2 9/16” as pictured in the same diagram, (Fig. D,) but the sphere in question is larger than 2 9/16” To find the cube of the 2 ¼” tetrahedron then, we must draw a new circle with an inner square that has diagonals of 2 ¼” and it will contain the tetrahedron if it becomes a cube. This is why the 2 ¼” latitudenal plane is important. I will provide all of the necessary mathematical proofs at the end of the article.

So then; if the diagonals of the square in Fig. E are 2 ¼” and it becomes a three dimensional, perfect, cube; then it contains the 2 ¼” tetrahedron inside of it. If we take a side view of this same cube, with 1 19/32” sides and 2 ¼” top and bottom diagonals, we have the next diagram; Fig. F, which is a horizontal side view of the cube:

The eight vertices of the cube are A, B, C, D, E, F, G, and H, (E and F hide behind C and D in this view). The vertical side in the center (CD) is the closest to the point of observation. The top and bottom, (AG and BH,) are the 2 ¼” diagonals of the cube.

The dotted lines are the longest diagonals of any cube; they run through the center of the cube from one vertex to its extreme opposite vertex, (i.e. A to H). Very simply now all we must do is place this diagram of the cube into its corresponding sphere and the matter will be resolved; for this diagram also contains the tetrahedron. This becomes even more simple if we know one more math factor: Every right triangle which has opposing angles of 35° and 55° (degrees) represents a potential cube; and its hypotenuse is the longest diagonal of that same cube, (in Fig. F these are the dotted lines). Its hypotenuse is also the diameter of a specified circle around the cube, and that circle is also its potential sphere. A case in point: All of the circles in Fig. G are defined by various points on the hypotenuse and the vertex point at the 35° angle.

Therefore now, the size of the sphere that fits the prescribed tetrahedron and its corresponding cube is ascertainable by dividing the hypotenuse in figure G. The circle drawn using the center point of the divided hypotenuse will denote the same sphere that encloses the cube and tetrahedron in question. All eight vertices of the cube will touch the sphere and therefore all four vertices of the tetrahedron will also touch the sphere’s “crust” since the four corners of the tetrahedron share points with four vertices of the cube. Simply divide the hypotenuse in half and that point becomes the center of two circles which encompass all eight vertices of the equilateral cube. The two circles are perpendicular to each other and form the four main longitudes of a perfect sphere.

As will be seen; the hypotenuse of the highlighted right triangle, (which is half the parallelogram ABCD, which is also the cube in question); is also the diameter of a specified circle and a pole of the same size sphere. The equilateral four-sided tetrahedron is perfectly placed within the lines of the cube; but the diameter of the circle is 2 ¾”, as the hypotenuse proves. Thus, a pole of the same sphere is then 2 ¾” and therefore the tetrahedron and cube fit into a 2 ¾” sphere; not the 2 9/16” sphere. Though it is true that all rectangles may be circled in this way; all rectangles do not represent the possibility of a perfect cube. Divide a planal drawing of the cube into four equal right triangles, such as in fig. H, and the hypotenuse angles will always be 35° and 55° if it is the side section of a true and perfect cube. The hypotenuse sides are the chords AD: (which is the shared hypotenuse of the right triangles ABD and ACD,) and the other chord BC: (which is the shared hypotenuse of the right triangles ABC and BCD). If we rotate the cube and sphere on the N-S axis pole 90°, (¼ turn,) the dimensions are the same for vertices E, F, G, and H, and circles 1 and 2 switch positions. The top and bottom diagonals remain 2 ¼”, the vertical sides all remain 1 19/32”, and therefore the hypotenuse chords, (now EH and FG,) are also the same. All of the cube’s vertices are 90°+90°+90° and arrive at apexes from three dehydral angles each, (all eight vertices do this). Therefore all dehydral angles inside the cube are also 90°, as this is what makes it a cube in the first place. A dehydral angle is simply two planal surfaces that meet and form an angle. Therefore if the top and bottom measurements remain 2 ¼” after a 90° (quarter) turn of the sphere on its N-S axis pole, and the vertical side measurements remain 1 19/32” as they all do; then it is an equilateral cube inside of a sphere. All six of its planal surfaces are equally 1 19/32” squared planes, meeting at 90° dehydral angles, which dehydral angles are also all equal in length, (1 19/32”). So then; all twelve dehydral angles intersect at the eight common vertices of the cube; at three dehydral angles per vertex; at 90° each. Three dehydral angles, multiplied by the eight vertices, equals the twenty-four separate angles of the perimeter of the cube. All twenty-four angles are 90°; therefore we can be sure: If all twenty-four angles are 90°, and all twelve linear measurements which define the cube are equal at 1 19/32”; then it is a perfectly symmetrical, equilateral, cube.

The 2 ¾” Sphere With Horizontal Placement of 2 ¼” Tetrahedron inside:

Thus we have a proper view of the equilateral tetrahedron and its horizontal placement inside the 2 ¾” poled sphere. The two sides shown traversing through the sphere, up to the apex, appear to be shorter than the base but this is an illusion due to the fact that the sides tilt away from the 2 ¼” base-line in this particular point of viewing. The height of the tetrahedron is the center line and is also an illusion of one of the 2 ¼” sides rising up to the apex. The height is 1 27/32”. The rest of the article will now focus on the math involved in proving all of the mathematical statements previously asserted herein.

The Tetrahedron and its Four Vertices:

A, B, C, and D are the four vertices of the tetrahedron and D is at the top in this “top view.” A, B, and C form the base. AD is an illusion of one 2 ¼” side, but also marks the center of the base (E) below; thus forming a right triangle, including the height, inside the tetrahedron.

The right triangle is ADE, (as well as BDE and CDE). AE is 1 9/32” as are BE and CE. AD is one 2 ¼” side, as are BD and CD. The height of the tetrahedron is center of base to vertex D, or DE.  If AE is 1 9/32” and AD is 2 ¼” and becomes the hypotenuse of the right triangle ADE, then the height (DE) is ascertainable likewise. (I will prove the measurement of 1 9/32” shortly).

The Height of the Tetrahedron:

The good ole Pythagoras Theorem: a (squared) + b (squared) = c (squared), where c is the hypotenuse. Henceforth I write “squared” as * Thus: a*+b*=c* 1 9/32*+b*=2 ¼* 1.28125*+b*=2.25* 1.64160+b*=5.0625 5.0625 - 1.64160=3.4209 b*=3.4209 Square root of 3.4209=1.8495” b=1.8495” =1 27/32” =Height of Tetrahedron

The 1 9/32" Interior Leg:

In an earlier section it is simply stated that AE is 1 9/32” as well as BE and CE. The measurement may be found because the three exterior base angles A, B, and C are all known to be 60° each, (because it is an equilateral triangle). The three exterior sides are also all known to be 2 ¼” or 2.25”. One may use the three congruent isosceles triangles shown in this diagram (Fig. L) created by the three interior legs. They split the three 60° angles of the base because they are a product of the opposing base side’s median.

Therefore the three isosceles triangles are all 30° 30° 120° triangles. Simply continue one of the legs through to its opposing 2 ¼” base side median and one isosceles triangle is dissected into two congruent right triangles which happen to be 30° 60° 90° right triangles; (the longest side is 2x the shortest side). However; none of this is necessary yet. One of the diagrams above, (Fig. D,) details AE, BE, and CE in dotted lines, as well as AD, BD, and CD, inside the 2 9/16” circle. It is quickly evident that AE, BE, and CE are all three radii of the 2 9/16” diameter circle: see Fig. D above. 2 9/16” x ½ = 1 9/32”

The All Important Cube Side of 1 19/32":

There are four main isosceles triangles inside the square in Fig. M. One is highlighted with dotted lines. The angles of each, at the center point, are all 90°. Therefore they are all 45° 45° 90° right triangles. The two equal legs of each large isosceles triangle are also radii, (of the 2 ¼” circle). Therefore they are 1 1/8” and the Pythagoras Theorem will, once again, give the answer.

a*+b*=c* (where c = Hypotenuse). 1/8” =.12500” 1.125*+1.125*=c* 1.26562+1.26562=c* 2.53124=c* c= square root of 2.53124 c=1.5909 1.5909” = 1 19/32”

Proving the Size of the Original Equilateral Triangle and the Measurements of its Three Equal Sides:

Back to the three congruent isosceles triangles; The angles of all equilateral triangles are 60° 60° 60°. Therefore, as shown earlier, the three isosceles triangles are 30° 30° 120°. If interior leg D is continued down, (dotted line,) to its median point on side C; it dissects the lower isosceles triangle, EFC, into two right triangles. Angle G is a dissection from 120° to 60°. Angle I is 30°, and the new angle H is 90°. Therefore the new triangle, (sides GH, F, and C2,) is a 30° 60° 90° right triangle. As with all 30° 60° 90° right triangles: the longest side is 2x the shortest side. The longest side is F; which we know to be a radius of the 2 9/16” circle. Therefore F=1.28125” and side GH is half of F. So then 1.28125 x .5 = .640625” = side GH. Since we are presently dealing with inches, the Pythagorean Theorem will not work with .640625 because it is less than one inch. Though it is a positive number, it is less than the minimal unit of current measure (1 inch), and therefore works as a negative number. Any negative number multiplied by another negative number increases the quantity negatively, (i.e. .1 x .1 = .01). We will convert to sixteenths of an inch to use the Pythagorean Theorem.

For this equation: F =  1.28125 x 16 = 20.5 GH = .640625 x 16 = 10.25 C2 = C2 x 16 GH* + C2* = F* 10.25* + C2* = 20.5* 105.0625 + C2* = 420.25 420.25 - 105.0625 = 315.1875 C2* = 315.1875 Square root of 315.1875 = 17.75352 = C2 Now simply convert back to inches:

17.7535 divided by 16 = 1.1096 C2 = 1 1/11” (and C2 is equal to C1 in fig. O). 1.1096 x 2 = 2.2192 2.2192 x 16 = 35.51/16ths,  35.51 sixteenths is equal to 2 3.51/16ths. Slightly more than 2 7/32” which is less than a 32nd of an inch from 2 ¼”. When I originally did all of this on paper it was not possible to draw within a 32nd of an inch. The finding is well within the realm of tolerance.

Proving The 2 ¾” Diameter Sphere:

A* + B* = C* 2.25* + 1.5909* = C* 5.0625 + 2.5309 = 7.5934 = C* Hypotenuse/Diameter/C = 2.7556 = 2 ¾” (+ .0056)

Why the 2 ¼” Latitudenal Circle Betrays the Identity of the Cube:

Every straight line in figure Q is 2 ¼”. The inner circle, (which contains the square cubed,) is also a 2 ¼” diameter circle. The outer circle is the 2 ¾” circle; the known sphere of the tetrahedron. The chords AG and BH are horizontal 2 ¼” latitudes around the 2 ¾” sphere. The chords CD and EF are vertical 2 ¼” latitudes, (not longitudes,) around the 2 ¾” sphere. The diagonals of the square are the 2 ¼” equilateral tetrahedron if the square is cubed. The intersection of AG and CD is one vertex of the square, as well as BH and CD, AG and EF, and BH and EF.

If this diagram were three dimensional; one could turn it any one of the four compass point directions North, South, East, or West a ¼ turn, and the diagram would still appear the same though the vertex points and chords would change whatever names we happen to allot to them. (A 1/8 turn horizontally would produce, essentially, the diagram in Fig. H as will be seen here by drawing a chord from A to B and from G to H). Also if the square is cubed, and the outer circle is the 2 ¾” sphere, then the inner 2 ¼” circle becomes one of two more latitudenal circles around the outer sphere. The other lies directly behind it on the other side of the outer sphere, and all eight vertices of the cube touch the outer 2 ¾” sphere at equidistant points marked by the intersections of each of the 2 ¼” chords drawn here. Each vertex of the cube is locked in place by three points and three circles: i.e. AG and CD are 2 ¼” circles that intersect at the point of intersection with the visible inner circle which is also 2 ¼”. All of this means that two vertices of the tetrahedron also touch the outer sphere in this hemisphere, since they share two vertices on this side of the cube. The four cube vertices that touch the outer sphere in the opposite hemisphere also include the other two vertices of the tetrahedron of course.

Cubing the Tetrahedron: The tetrahedron has four triangular, equal, planes; (three sides plus a base). Thus it also has six linear edges and, in this case, each one has a length of 2 ¼”. The cube has six planal ,equal, square sides that are 1 19/32” squared; and each of the six squares has a diagonal of 2 ¼”. Thus we have the cube unfolded with all of its side squares and diagonals; which also denote the placement of the six linear sides of the tetrahedron inside the cube when it is folded back up.

The math has already been done. When the cube is folded back together, the four vertices of the tetrahedron, (A, B, C, and D,) are perfectly aligned at the proper angles as the intersection at vertex C shows in Fig. R above.

We may now take any size sphere and very quickly find its corresponding inner cube and tetrahedron. Likewise; if we have the mere measurement of one single line, of a simple Equilateral Triangle; we may find its tetrahedron, cube, and-or sphere. Truly the Lord did stretch a line upon the Earth!                   Let’s say now that we have a 4 1/8” sphere:

The hypotenuse is a given at 4 1/8” or 4.125” Side b is one side of the equilateral tetrahedron and also a diagonal of the cube. Side a is one vertical, linear, side measurement of the cube. Sine 55° = 0.819152 Sine 35° = 0.5735764 Since sine 55° x hypotenuse = opposite side (b), then: b = 0.819152 x 4.125 = 3.379002 3.379 x 16 = 54.064 b = 54/16ths = 3 3/8” Since sine 35° x hypotenuse = opposite side (a), then: a = 0.5735764 x 4.125 = 2.366002 2.366 x 16 = 37.856 a = 37.8/16ths = 2 11/32”

As for the Foundations of the Earth: I see four equal, triangular planes. And as for the Pillars of the Earth: I count six great columns all equal in stature. And as for the stretching of a line? I too stretched a line in every diagram on this page. And when it comes to the Four Corners of the Earth: There are absolutely four corners, and the Chief Corner-Stone is Christ Yeshua. "Then YHWH answered Job out of the whirlwind, and said, Who is this that darkened counsel by words without knowledge? Gird up now thy loins like a man; for I will demand of thee, and answer thou me. Where wast thou when I laid the foundations of the earth? declare, if thou hast understanding. Who hath laid the measures thereof, if thou knowest? or who hath stretched the line upon it? Whereupon are the foundations thereof fastened? or who laid the corner stone thereof; When the morning stars sang together, and all the sons of 'Elohiym shouted for joy?" What more is light than the stretching of a line? What more is the beam of light than a measuring rod in the hand of the Almighty? Oh YHWH, holy and true are thy ways! Just and upright are all thy thoughts! I want to spend the remainder of days with you, and when all of my days are gone, I want to spend eternity with you. You alone are Holy... There is none other, No not one... "The LORD maketh poor, and maketh rich: he bringeth low, and lifteth up. He raiseth up the poor out of the dust, and lifteth up the beggar from the dunghill, to set them among princes, and to make them inherit the throne of glory: for the pillars of the earth are the LORD's, and he hath set the world upon them." "And after these things I saw four angels standing on the four corners of the earth, holding the four winds of the earth, that the wind should not blow on the earth, nor on the sea, nor on any tree." And this leads to another topic: The attributes and dimensions of the Holy City New Jerusalem given in Revelation Chapter 21 ~ "On the east three gates; on the north three gates; on the south three gates; and on the west three gates. And the wall of the city had twelve foundations, and in them the names of the twelve apostles of the Lamb. And he that talked with me had a golden reed to measure the city, and the gates thereof, and the wall thereof."                 And the city lieth tetragonos: FOUR CORNERED ...

PT II 17-153-969 Temple of Yeshua PT III The Great Pyramid Internal Tetrahedron Capstone PT IV Ezekiel Temple: Temple of a Man

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