SOLITU Part H:  The Hiram Key or Self-Reciprocal Construction


Copyright 2015, John Manimas Medeiros


I am not now and never have been a member of the Free and Accepted Masons.  This is my work and my work alone.  I have never received any geometric knowledge or secret knowledge from any Freemason or Rosicrucian or anyone else.  Anyone may prefer to call this construction the "Self-Reciprocal Construction" rather than the "Hiram Key."  It may be the Hiram Key because it is the proper way to begin any natural geometric construction, and it does enable construction of a straight line equal to pi exactly in proportion to the natural 1.


The Hiram Key Drawings (Constructions) and Brief Explanation


The conditions of the construction - in detail for clarity:

1)  Construct a circle within a tangent outer square, then vertical and horizontal diameters dividing the square into four equal quadrants.




2)  Line X is a diagonal diameter extending from an upper-left (EU) end point downward to the right to a lower-right (ER) end point.  Note that the construction (or drawing) is "right-handed" in its horizontal orientation.  If inverted horizontally, it would be "left-handed" with the upper end point EU to the right and the lower end point ER to the left.


3)  The upper end point EU is at the intersection of the slanted diameter of the circle with the upper left circumference of the circle, and the lower end point ER is on the extension of the slanted diameter of the circle outside of the circumference and is at the intersection of the extended slanted diameter with the lower horizontal side of the square in the lower right quadrant.  This is the line length of X (from EU to ER).  When we perform this construction, we do not know the measured length of X or any of the lines in the construction.  We do not and will not measure any lines.  We will not know the value of X except in special cases.  We will discuss three special cases, 45 degrees, 60 degrees and the construction of the pentagon.  The Hiram Key is in the angle, not the lines.


Seeing the proportional 1 and the reciprocal of X:

A)  I am going to show that there is something more in this construction than what we usually see.  What we usually see is a right triangle and in this case the right triangle is of course constructed in the lower right quadrant of the square.  In our first drawing we have only three element labels:  EU for the upper end point, ER for the lower end point, and X for the straight line constructed from EU to ER. 





B)  In our second drawing which is the same construction, I have added labels for the vertical Altitude A, horizontal Base B, and Hypotenuse H of the right triangle.  We do not know the measured lengths of X or H or A or B.  The Altitude A is equal to the radius of the circle and one-half of the side of the square.  The Base B is the length that extends from the intersection of the vertical diameter with the lower side of the square to the lower end point of line X, ER.  Obviously line H, the Hypotenuse, lies on line X, and line X has a length equal to 2 times the radius of the circle plus the section of line X that extends outside of the circumference of the circle to the lower end point ER.


C)  Next I will change the labels for the lines to show how in fact the line we have previously named B for Base of the right triangle is one (1) in proportion to line X and all other lines in the construction, AND IS 1 in proportion to X and can have no other proportional value.  This establishes the crucial new understanding that the line length B is 1 BY NATURE and not by arbitrary human designation.  This change in the labels for the lines is based upon using the quadratic equation (ax^2 + bx + c = 0) as the quadratic construction.  The two solutions to the quadratic construction are then both valid: 

A) (larger):  x = - b -  sqrt[b^2 - 4ac]/ 2a, which is a line length (sign does not matter)

B) (smaller):  x = - b + sqrt[b^2 - 4ac]/ 2a, which is a line length (sign does not matter)

If you are not following this right now, let it go.  It is explained next.




We change the line labels as follows: 

            1) Altitude A is changed to B;

            2)  The Base B is changed to S, which is comprised of CL and CS;

            3)  The section of X that extends outside of the circle to ER is now RP, and the RP label stands for reciprocal because RP = 1/X.

            4)  We can see that H = B + RP and X = (2*B) + RP


But what is most important is that the next condition D of the construction is fixed and applies for all constructions that meet the conditions we have listed thus far:


D)        B =      X  -  .   1  .  which is the radius and altitude of the right triangle

                        2       2*X


            H =      X  +  .   1  .  which is the hypotenuse of the right triangle

                        2        2*X


            S = 1 and can only be 1 in proportion to B and X

            S = CL + CS, which required values can be easily computed because the small lower right triangle is similar to the larger right triangle.


The absolutely crucial jump in mental observation that is required here is to focus on the fact that we do not know the length of X and the proportions of the lines and all of the inherent characteristics of this construction are ACTUAL and not representational, and all of the inherent characteristics of this construction are determined by THE ANGLE and will NOT be observed clearly by assigning a numerical value to X.  HOWEVER, we can confirm and verify the proportional value of S being 1 BY NATURE by examining a set of constructions where we assign a numerical value to B, the radius and altitude of the right triangle.  We will then see that the fixed, natural  proportional relationships apply to B, H, S, X and RP.  This is the KEY that makes the construction "natural" and provides us with a line length of 1 by Nature and not by human designation.  (Note:  the drawings may become distorted when printed on paper.)


Examination of four angles through the eyes of the Hiram Key:


45 degrees:      We know that the Altitude A = 1 and the Base B = 1 and the Hypotenuse H = sqrt(2) or 1.414213562.  So we note that the tangent = 1, the same value of A of course.  So let us look at our values for X, 2X, (1/2X), and (X/2). 

X = H + A or sqrt(2) plus 1 = 2.414213562

2*X = 4.828427125,   and (1/2X) = 0.207106781,  and (X/2) = 1.207106781


Therefore, our algorithm does apply:

A =      X  -  .   1  .  which is the radius and altitude of the right triangle

            2       2*X


H =      X  +  .   1  .  which is the hypotenuse of the right triangle

            2        2*X


A =  1.207106781  - 0.207106781 = 1

H =  1.207106781 + 0.207106781 = 1.414213562


The algorithm expressions can also be written as:

X^2 -1 = A      and X^2 + 1 = H         which one sees checks out the same.

  2X                             2X


60 degrees:  our values are:

X = H + A or sqrt(3) + 2 = 3.732050808

2*X = 7.464101615, and (1/2X) = 0.133974596, and (X/2) = 1.866025404

and the values we get for A and H are:

1.866025404 - 0.133974596 = 1.732050808  [sqrt(3)]  =  A

1.866025404 + 0.133974596 = 2 =  H


63.43494882…degrees, the tangent=2 right triangle (the pentagon construction):

X = H + A or sqrt(5) + 2 = 4.236067977

2*X = 8.472135955, and (1/2X) = 0.118033988, and (X/2) = 2.118033988


and the values we get for A and H are:

2.118033988 - 0.118033988 = 2 =  A

2.188033988 + 0.118033988 =  2.236067977 = sqrt(5) =  H


81 degrees (or any angle from 0< to <90 ):

X = H + A  as follows:  A = tangent G or tangent 81

tangent 81 = 6.3137515146750430989794642447682,  or 6.313751514 = A

H = sqrt( [tangent 81]^2 + 1) =  6.392453221 = H

same as:  6.3924532214996615470422157340739

same as sqrt(40.863458189061400972849371646108), therefore:

X = H + A =  6.392453221 + 6.313751514 = 12.706204736, and then

2*X =  25.412409472, and (1/2X) =  0.039350853, and X/2) = 6.353102368

and the values we get for A and H are:

6.353102368 - 0.039350853 =  6.313751514  = A

6.353102368 + 0.039350853 =  6.392453221 =  H


THEREFORE:  we see that the algorithm is valid and works always so long as the conditions for the construction are met.  IT IS ALSO IMPORTANT TO REMEMBER AT ALL TIMES THAT WE CANNOT CONSTRUCT THE 81 DEGREE RIGHT TRIANGLE BY CONSTRUCTING LINE B FIRST AND ARBITRARILY DESGINATING THAT LINE B AS HAVING A LENGTH OF 1.  THE LINE LENGTHS FOR B ARE 1 IN OUR EXAMPLES ONLY IN ORDER TO SHOW HOW THE ALGORITHM WORKS, BUT THE CONSTRUCTION IS NEVER PERFORMED BY MEASUREMENT OF ANY LINES.  WHEN THE CONDITIONS OF THE CONSTRUCTION ARE MET, THE PROPORTIONAL RELATIONSHIPS APPLY WHETHER WE KNOW THE LINE LENGTHS OR NOT.  AND, THE PRIMARY PURPOSE OF THIS EXERCISE IS TO SHOW THAT THE TANGENT = 2 RIGHT TRIANGLE, WHICH WE USE TO BEGIN CONSTRUCTION OF A PENTAGON, IS A SPECIAL CASE OF THE GENERIC HIRAM KEY OR SELF-RECIPROCAL CONSTRUCTION.  ALSO, if we made any of the right triangles larger but similar, the proportional value of the Base B would still be 1 in proportion to the other lines.  The reason for the proposed change in naming the lines B, H, CS, etc. in place of A, H, B is for the discovery of other interesting relationships that can be seen and explored in a longer version of this material.  Also, to be courteous to the Masons, who support the story of Hiram Abiff and the history, or legend, of the lost Hiram Key, anyone can choose to call this the "Self-Reciprocal Construction" rather than the Hiram Key.  It may be that this construction alone is not the Hiram Key, but that this construction plus the Unification Construction is the Hiram Key.  Or, maybe not.


A longer explanation of the Hiram Key or Self-Reciprocal Construction would involve a detailed explanation of the many geometric facts that are revealed when we treat the quadratic equation as the quadratic construction:


The Quadratic Equation Viewed as the Quadratic Construction:

ax^2 + bx + c = 0 … or … ax^2 + bx – c = 0 … or … ax^2 + bx = c

The formula for finding two solutions to this equation is as follows:

1) (two minus signs) negative solution (NS larger):  - b – sqrt[ b^2 – 4ac ] / 2a

2) (one minus sign) positive solution (PS smaller):  - b + sqrt[ b^2 – 4ac ] / 2a

I concluded, based upon what I feel are rather obvious historical facts, that to the ancient Pythagoreans, this was not an algebraic formula devised to solve “quadratic” problems, but was in fact originally the Quadratic Construction.  The terms of the equation represent line values and line lengths.  It is most important to notice that when we consider this piece of ancient wisdom to be the description of a construction, rather than strictly algebra, we then see that the minus signs and plus signs do not mean – as taken in algebra – plus and minus values, but rather each value is the length of a line that can be added or subtracted using the compass and straightedge.  Look at what I call the “determinant” enclosed in the box-like brackets.  The square root of [ b^2 – 4ac ] is similar to the square root of [ b^2 + a^2] if we make c into minus c ( - c) and then make 4ac the substitute for a^2.  This is a simple replacement formula if we make a = 1 and c = -1.  Then the divisor is simply 2, and we can divide any line in half with the compass and straightedge.  Notice also that if a = (1/2) and c = - (1/2) then the value of (- 4ac) is 1, and the value of 2*a also equals 1. 


If our quadratic construction is:  (1/2)X^2 + bX + (- 1/2 )


Then  - b +/- sqrt[ b^2 – (4*(1/2) * (- 1/2 ) ]  = - b +/- sqrt[b^2 + 1]


And, if b is a line length, and 1 is a line length, each the perpendicular sides of a right triangle, then the determinant is H with line length = sqrt[b^2 + 1] and the two solutions are line lengths:


#1 Solution:  - b - sqrt[b^2 + 1]  ……….  and #2 Solution: - b + sqrt[b^2 + 1]


Where the negative Solution #1 (NS) turns out to be the longer line, actually a sum of two lines, and the positive Solution #2 (PS) turns out to be the shorter line, essentially subtracting the shorter line from the longer line (H – B), and, key to the path toward the Self-Reciprocal Construction is the fact that these two solutions are reciprocals.  Try it.  Assign any value to b.


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