What follows is the supporting mathematical analysis for the GeoGebra construct. This work was an outgrowth of my team's work with the Circle Squaring Problem and is still under review and subject to further analysis.
1. A (0, -1)
2. B (0, 0)
3. Circle c has center A with radius 1. x2 + (y + 1)2 = 1
4. C (-1, -1)
5. Line a is through points C and A and is y = -1.
6. Point D is the intersection of Circle c with Line a. There are two points on Line a intersecting with Circle c: (-1, -1) which is Point C and the other point (1, -1) is Point D.
7. Circle d has center at Point D with a radius of 1. (x – 1)2 + (y + 1)2 = 1
8. Circle e has center at Point C with radius 1. (x + 1)2 + (y + 1)2 = 1
9. Point E (-0.5, (√3)/2 – 1)
10. Point F (0.5, (√3)/2 – 1)
11. Make a square with CD as the base. The corners are C, D, G, and H. Side CD is segment b; DG is segment f; GH is segment g; and HC is segment h.
12. Diagonal DH is segment i. x = -y
13. Diagonal CG is segment j. x = y
14. Circle k is with Center F and passes through Point D. It has a radius 1. (x – 0.5)2 + (y + 1 – (√3)/2)2 = 1
15. Circle p has Center E and passes through Point C with radius 1. (x + 0.5)2 + (y + 1 – (√3)/2)2 = 1
16. Point l is the intersection above X-axis of Circles k and p. Point l is (0, (√3) – 1). The other point of intersection Point J coincides with Point A and is (0, -1)
17. Circle q has center at Point l and passes through Point B. The radius of this circle is the difference between the Y-coordinates of Points l and B and is equal to ((√3) – 1).
x2 + (y - ((√3) – 1))2 = ((√3) – 1)2
18. Point K is the intersection of Segment i and Circle p. Point L is the opposite point of intersection between the diagonal and the circle.
Equation for Circle p is (x + 0.5)2 + (y + 1 – (√3)/2)2 = 1
Equation for Segment i is x = -y
Substituting for x in the equation for Circle p we get:
(-y + 0.5)2 + (y + 1 – (√3)/2)2 = 1
(y2 – y + 0.25) + (y2 + 2 (1 – (√3)/2)y + (1 – (√3)/2)2 ) = 1
y2 – y + 0.25 + y2 + 2 (1 – (√3)/2)y + (1 – (√3)/2)2 = 1
2 y2 + (2 - √3 – 1) y + (1 – (√3)/2)2 +0.25 – 1 = 0
2 y2 + (1 - √3) y + 1 + ¾ - √3 – ¾ = 0
2 y2 + (1 - √3) y + (1 - √3) = 0
2 y2 - ((√3) – 1) y - ((√3) - 1) = 0
The above is a quadratic equation a y2 + b y +c = 0 where
a = 2; b = - ((√3) – 1); and c = - ((√3) - 1)
b2 – 4ac = ((√3) – 1)2 + 8 ((√3) - 1) = 3 + 1 – 2 √3 + 8 √3 – 8 = 6 √3 – 4 = 2((3√3) – 2)
For verification, 2 (3 √3 – 2) = 6.392305
Positive root is (-b + √(b2 – 4ac))/2a = ( 0.732051 + 2.528301)/4 = 0.815088
Hence K(-0.815088, 0.815088)
The negative root is (0.732051 – 2.528301)/4 = -0.449063 giving L(0.449063, -0.449063)
Geogebra gives K(-0.82, 0.82); and L(0.45, -0.45). Verified.
xk = -(((√3) – 1) + √(2((3√3) – 2)))/4
yk = (((√3) – 1) + √(2((3√3) – 2)))/4
Point L is of no interest to us.
19. Point M is the intersection between Circle k and diagonal CG in the 4th quadrant. The point of interest, Point N is the intersection point in the 1st quadrant.
The equation for Circle k is (x – 0.5)2 + (y + 1 – (√3)/2)2 = 1
The equation for diagonal CG is x = y.
Substituting for x in the Circle k equation, we have:
(y – 0.5)2 + (y + 1 – (√3)/2)2 = 1
(y2 – y + 0.25) + (y2 + 2 (1 – (√3)/2) y + (1 – (√3)/2)2) = 1
(y2 – y + 0.25) + (y2 + (2 – √3) y + (1 + ¾ - √3)) = 1
y2 – y + 0.25 + y2 + (2 – √3) y + (1 + ¾ - √3) – 1 = 0
2y2 + (1 – √3) y + (1 - √3) = 0
2y2 – ((√3) – 1) y - ((√3) – 1) = 0.
The above is of the form ay2 + by + c = 0 where
a = 2; b = – ((√3) – 1); and c = - ((√3) – 1).
The solution to the above equation is the same as the one found in step 18.
xN = (((√3) – 1) + √(2((3√3) – 2)))/4
yN = (((√3) – 1) + √(2((3√3) – 2)))/4
Point N is (0.815088, 0.815088)
Geogebra gives M(-0.45, -0.45) and N(0.82, 0.82)
N equations are verified.
20. Line l is the line through Points K and l. Point l is the center of Circle q.
Coordinates of Point K are:
xk = -(((√3) – 1) + √(2((3√3) – 2)))/4
yk = (((√3) – 1) + √(2((3√3) – 2)))/4
Point l is (0, (√3) – 1).
The equation for a line through two points (x1, y1) and (x2, y2) is given by
y = ((y2 – y1) x + (y1 x2 – y2 x1)) / (x2 – x1)
We map Point l as (x1, y1) and Point K as (x2, y2).
The equation for Line l is:
y = (((((√3) – 1) + √(2((3√3) – 2)))/4 – ((√3) – 1)) x + (((√3) – 1) (-(((√3) – 1) + √(2((3√3) – 2)))/4))) / (-(((√3) – 1) + √(2((3√3) – 2)))/4)
For verification we calculate the values of the symbols in the above equation to six digits accuracy.
y = (((((√3) – 1) + √(2((3√3) – 2)))/4 – ((√3) – 1)) x + (((√3) – 1) (-(((√3) – 1) + √(2((3√3) – 2)))/4))) / (-(((√3) – 1) + √(2((3√3) – 2)))/4)
y = (((0.732051 + √(2((3√3) – 2)))/4 – 0.731051) x + (0.732051 (-(0.732051 + √(2((3√3) – 2)))/4))) / (-(0.732051 + √(2((3√3) – 2)))/4)
y = (((0.732051 + √(2(5.196152 – 2)))/4 – 0.731051) x + (0.732051 (-(0.732051 + √(2(5.196152 – 2)))/4))) / (-(0.732051 + √(2(5.196152 – 2)))/4)
y = (((0.732051 + √(2(3.196152)))/4 – 0.731051) x + (0.732051 (-(0.732051 + √(2(3.196152)))/4))) / (-(0.732051 + √(2(3.196152)))/4)
y = (((0.732051 + √(6.392304))/4 – 0.731051) x + (0.732051 (-(0.732051 + √(6.392304))/4))) / (-(0.732051 + √(6.392304))/4)
y = ((3.260352/4 – 0.731051) x + (0.732051 (-3.260352/4))) / (-3.260352/4)
y = ((0.815088 – 0.731051) x + (0.732051 (-0.815088))) / (-0.815088)
y = (0.084037 x + (0.732051 (-0.815088))) / (-0.815088)
y = (0.084037 x + (-0.596686)) / (-0.815088)
y = (0.084037 x - 0.596686) / (-0.815088)
y = (0.596686 - 0.084037 x) / (0.815088)
y = 0.732051 – 0.103102 x
The above six-digit accuracy equation suggests that we could arrive at a simpler equation by expanding the terms.
y = (((((√3) – 1) + √(2((3√3) – 2)))/4 – ((√3) – 1)) x + (((√3) – 1) (-(((√3) – 1) + √(2((3√3) – 2)))/4))) / (-(((√3) – 1) + √(2((3√3) – 2)))/4)
y = (((((√3) – 1) + √(2((3√3) – 2)))/4 – ((√3) – 1))/ (-(((√3) – 1) + √(2((3√3) – 2)))/4)) x + (((√3) – 1) (-(((√3) – 1) + √(2((3√3) – 2)))/4)) / (-(((√3) – 1) + √(2((3√3) – 2)))/4)
y = (((((√3) – 1) + √(2((3√3) – 2)))/4 – ((√3) – 1))/ (-(((√3) – 1) + √(2((3√3) – 2)))/4)) x + ((√3) – 1)
y = (((((√3) – 1) + √(2((3√3) – 2)))/4 – ((√3) – 1))/ (-(((√3) – 1) + √(2((3√3) – 2)))/4)) x + ((√3) – 1)
Let z = (((√3) – 1) + √(2((3√3) – 2)))/4. Then the equation becomes
y = ((z – ((√3) – 1))/ (-z)) x + ((√3) – 1)
y = ((√3) – 1) - ((z – ((√3) – 1))/ (z)) x
y = ((√3) – 1) - ((1 – ((√3) – 1)/z)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/z) x
Substituting for z, we get:
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) x
Rechecking
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2(5.196152 – 2)))/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2(3.196152)))/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(6.392304))/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + 2.528301)/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/(((1.732051 – 1) + 2.528301)/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/(((0.732051) + 2.528301)/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/((0.732051 + 2.528301)/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/((3.260352)/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/(3.260352/4)) x
y = ((√3) – 1) - (1 – ((√3) – 1)/0.815088) x
y = ((√3) – 1) - (1 – (1.732051 – 1)/0.815088) x
y = ((√3) – 1) - (1 – (0.732051)/0.815088) x
y = ((√3) – 1) - (1 – 0.898125) x
y = ((√3) – 1) – 0.101875 x
There is discrepancy in the coefficient for x. We review the earlier computation from the colored equation below.
y = (((0.732051 + √(6.392304))/4 – 0.731051) x + (0.732051 (-(0.732051 + √(6.392304))/4))) / (-(0.732051 + √(6.392304))/4)
y = (((0.732051 + 2.528301)/4 – 0.731051) x + (0.732051 (-(0.732051 + 2.528301)/4))) / (-(0.732051 + 2.528301)/4)
y = (((3.260352)/4 – 0.731051) x + (0.732051 (-(0.732051 + 2.528301)/4))) / (-(0.732051 + 2.528301)/4)
y = ((0.815088 – 0.731051) x + (0.732051 (-(0.732051 + 2.528301)/4))) / (-(0.732051 + 2.528301)/4)
y = (0.084037 x + (0.732051 (-(0.732051 + 2.528301)/4))) / (-(0.732051 + 2.528301)/4)
y = (0.084037 x + (0.732051 (-(3.260352)/4))) / (-(3.260352)/4)
y = (0.084037 x + (0.732051 (-0.815088))) / (-0.815088)
y = (0.084037 x + (-0.596686)) / (-0.815088)
y = (0.084037 x -0.596686) / (-0.815088)
y = (0.596686 - 0.084037 x ) / (0.815088)
y = (0.732051 – 0.103102 x )
The discrepancy persists. Hence we retrace further.
y = (((((√3) – 1) + √(2((3√3) – 2)))/4 – ((√3) – 1)) x + (((√3) – 1) (-(((√3) – 1) + √(2((3√3) – 2)))/4))) / (-(((√3) – 1) + √(2((3√3) – 2)))/4)
y = ((((0.732051) + √(2((3√3) – 2)))/4 – (0.732051)) x + (((√3) – 1) (-((0.732051) + √(2((3√3) – 2)))/4))) / (-((0.732051) + √(2((3√3) – 2)))/4)
Comparing with the reduction performed earlier we find that the second copying of 0.732051 was typed wrongly as 0.731051 and that error was carried through. Hence the correct equation for Line l is:
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) x
21. Point O is a point of intersection between Circle q and Line l. The other point of intersection, Point P is of interest.
The equation for Circle q is:
x2 + (y - ((√3) – 1))2 = ((√3) – 1)2
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) x
Substituting for y in the equation for the circle, we have:
x2 + (((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) x - ((√3) – 1))2 = ((√3) – 1)2
Rearranging the terms shown in color we get:
x2 + (((√3) – 1) - ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) x)2 = ((√3) – 1)2
As the colored terms cancel each other, we get:
x2 + ( - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) x)2 = ((√3) – 1)2
x2 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 x2 = ((√3) – 1)2
(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ) x2 = ((√3) – 1)2
x2 = ((√3) – 1)2/ (1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
x = ± ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
The point of interest P has a positive X-coordinate and hence we take
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
Using the above value for x, the value of y is
y = ((√3) – 1) - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
Interchanging the colored terms we get:
y = ((√3) – 1) - ((√3) – 1) (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
Removing the common factor out
y = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))
Verification:
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2(5.196152 – 2)))/4))2 )
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(6.392304))/4))2 )
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + 2.528301)/4))2 )
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/(((0.732051) + 2.528301)/4))2 )
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((0.732051 + 2.528301)/4))2 )
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((3.260352)/4))2 )
x = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/(0.815088))2 )
x = ((√3) – 1) / √(1 + (1 – (0.732051)/(0.815088))2 )
x = ((√3) – 1) / √(1 + (1 – 0.732051/0.815088)2 )
x = ((√3) – 1) / √(1 + (1 – 0.898125)2 )
x = ((√3) – 1) / √(1 + (0.101875)2 )
x = ((√3) – 1) / √(1 + 0.010379 )
x = ((√3) – 1) / √(1.010379 )
x = ((√3) – 1) / 1.005176
x = (0.732051) / 1.005176
x = 0.728281
y = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))
y = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2(5.196152 – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2(5.196152 – 2)))/4))2 ))
y = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2(3.196152)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2(3.196152)))/4))2 ))
y = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(6.392304))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(6.392304))/4))2 ))
y = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + 2.528301)/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + 2.528301)/4))2 ))
y = ((√3) – 1) (1 - (1 – 0.732051/((0.732051 + 2.528301)/4)) / √(1 + (1 – 0.732051/((0.732051 + 2.528301)/4))2 ))
y = ((√3) – 1) (1 - (1 – 0.732051/(3.260352/4)) / √(1 + (1 – 0.732051/(3.260352/4))2 ))
y = ((√3) – 1) (1 - (1 – 0.732051/0.815088) / √(1 + (1 – 0.732051/0.815088)2 ))
y = ((√3) – 1) (1 - (1 – 0.898125) / √(1 + (1 – 0.898125)2 ))
y = ((√3) – 1) (1 – 0.101875 / √(1 + (0.101875)2 ))
y = ((√3) – 1) (1 – 0.101875 / √(1 + 0.010379))
y = ((√3) – 1) (1 – 0.101875 / √(1.010379))
y = ((√3) – 1) (1 – 0.101875 / 1.005176)
y = ((√3) – 1) (1 – 0.101350)
y = ((√3) – 1) 0.89865
y = (0.732051) 0.89865
y = 0.657858
Point P is (0.728281, 0.657858)
Geogebra gives O(-0.73, 0.81) and P(0.73, 0.66)
The coordinates of Point P are verified. The corresponding equations are:
xP = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
yP = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))
22. Line m is a line through the Points P and F.
Point F (0.5, (√3)/2 – 1)
xF = 0.5
yF = ((√3)/2 – 1)
xP = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
yP = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))
The Point S is the point of intersection between Line m and Line n which in turn is a line joining two points, Points R and N. Since the point of solution S is the intersection of two lines connecting two points each, we prefer to derive the point of intersection of two lines one joining points (x1, y1) and (x2, y2) and the other joining (x3, y3) and (x4, y4) and substitute the coordinates of points F, P, R, and N in the derived equation in step 25 to follow.
23. Point Q is the intersection in 1st quadrant between Circle q and segment g that is side GH of the square. The other point, Point R in the 2nd quadrant is of interest.
The equation for Circle q is x2 + (y - ((√3) – 1))2 = ((√3) – 1)2
The side GH is y = 1. Substituting for y in the equation for circle, we get:
x2 + (1 - ((√3) – 1))2 = ((√3) – 1)2
x2 = ((√3) – 1)2 - (1 - ((√3) – 1))2
x = ±√(((√3) – 1)2 - (1 - ((√3) – 1))2)
Point R is in the 2nd quadrant and thus
xR = - √(((√3) – 1)2 - (1 - ((√3) – 1))2)
yR = 1
For verification we calculate xR.
xR = - √(((√3) – 1)2 - (1 - ((√3) – 1))2)
xR = - √((0.732051)2 - (1 - (0.732051))2)
xR = - √((0.732051)2 - (0.267949)2)
xR = - √(0.535899 – 0.071797)
xR = - √(0.464102)
xR = - 0.68125
Hence the Point R is (-0.68125, 1) and is as given by Geogebra.
Geogebra gives Q(0.68, 1) and R(-0.68, 1)
xR = - √(((√3) – 1)2 - (1 - ((√3) – 1))2)
yR = 1
24. Segment n is the line joining Points R and N.
Point R is (xR, yR)
xR = - √(((√3) – 1)2 - (1 - ((√3) – 1))2)
yR = 1
Point N is (xN, yN)
xN = (((√3) – 1) + √(2((3√3) – 2)))/4
yN = (((√3) – 1) + √(2((3√3) – 2)))/4
As indicated in step 22 we do not derive the equation for Line n now. We derive the generic equation for Point S in step 25 and substitute the coordinate expressions of Points P, F, R, and N to get the equations for Point S.
25. Point S is the point of intersection of Segment n and Line m.
The point of interest Point S is the intersection of two lines each of which are defined by two points that lie on the lines. Let us derive the equation for the coordinates of the point of intersection of line passing through points (x1, y1) and (x2, y2) and another line passing through points (x3, y3) and (x4, y4) and substitute in the resulting equation the coordinate equations of points P, F, R, and N.
The first line is
y = ((y2 – y1) x + (y1 x2 – y2 x1)) / (x2 – x1)
The second line is
y = ((y4 – y3) x + (y3 x4 – y4 x3)) / (x4 – x3)
Substitute for y from the second line in the equation for the first line to get
((y4 – y3) x + (y3 x4 – y4 x3)) / (x4 – x3) = ((y2 – y1) x + (y1 x2 – y2 x1)) / (x2 – x1)
((y4 – y3) x + (y3 x4 – y4 x3)) (x2 – x1) = ((y2 – y1) x + (y1 x2 – y2 x1)) (x4 – x3)
Multiplying we get
(y4 – y3) (x2 – x1) x + (y3 x4 – y4 x3) (x2 – x1) = (y2 – y1) (x4 – x3) x + (y1 x2 – y2 x1) (x4 – x3)
Bringing the x factors on the LHS we get
(y4 – y3) (x2 – x1) x - (y2 – y1) (x4 – x3) x = (y1 x2 – y2 x1) (x4 – x3) - (y3 x4 – y4 x3) (x2 – x1)
((y4 – y3) (x2 – x1) - (y2 – y1) (x4 – x3)) x = (y1 x2 – y2 x1) (x4 – x3) - (y3 x4 – y4 x3) (x2 – x1)
x = ((y1 x2 – y2 x1) (x4 – x3) - (y3 x4 – y4 x3) (x2 – x1)) / ((y4 – y3) (x2 – x1) - (y2 – y1) (x4 – x3))
Substitute the above expression for x in one of the line equations to get the expression for y.
From the equation for the first line,
y = ((y2 – y1) x + (y1 x2 – y2 x1)) / (x2 – x1) becomes
y = ((y2 – y1) (((y1 x2 – y2 x1) (x4 – x3) - (y3 x4 – y4 x3) (x2 – x1)) / ((y4 – y3) (x2 – x1) - (y2 – y1) (x4 – x3))) + (y1 x2 – y2 x1)) / (x2 – x1)
Line m joins points P and F; segment n joins points R and N. The following table gives the different possibilities. Assigning point 1 automatically decides the assignment for point 2; likewise assigning point 3 automatically decides the assignment for point 4.
Sl. No.
Point 1 (Point 2)
Point 3 (Point 4)
1
P (F)
R (N)
2
P (F)
N (R)
3
F (P)
R (N)
4
F (P)
N (R)
5
R (N)
P (F)
6
R (N)
F (P)
7
N (R)
P (F)
8
N (R)
F (P)
Out of the 8 possible assignments we plan to select one by a heuristic and continue.
We select the one that gives the minimum number of characters.
xP has 56 characters; yP has 101 characters; xF has 3 characters; yF has 10 characters; xR has 27 characters; yR has 1 character; xN has 26 characters; and yN also has 26 characters. The expression for x has x1 3 times; y1 2 times; x2 3 times; y2 2 times; x3 3 times; y3 2 times; x4 3 times; and y4 2 times. Likewise the expression for y has x1 5 times; y1 4 times; x2 5 times; y2 4 times; x3 3 times; y3 2 times; x4 3 times; and y4 2 times. Adding the number of occurrences in expression for x and expression for y we get x1 8 times; y1 6 times; x2 8 times; y2 6 times; x3 6 times; y3 4 times; x4 6 times; and y4 4 times. When the total number of characters in the expressions for the x and y coordinates are considered we find that the ith row in the above table gives the same number of characters as the (i+1)th row in the table where I is 1, 3, 5, and 7. Thus the alternatives to be considered are reduced to the following 4.
Sl. No.
Point 1 (Point 2)
Point 3 (Point 4)
1
P (F)
R (N)
2
F (P)
R (N)
3
R (N)
P (F)
4
N (R)
P (F)
A similar reasoning indicates that the resultant number of characters would be the same between rows 1 and 2 of the above table. Likewise the number of characters would be same between rows 3 and 4. Removing rows 2 and 4 of the above table, we get just two possibilities.
Sl. No.
Point 1 (Point 2)
Point 3 (Point 4)
1
P (F)
R (N)
2
R (N)
P (F)
The actual number of characters for the above two possibilities are calculated in the table below.
Sl. No
Characters due to
Total
1
xP
yP (1)
xF
yF (2)
xR
yR (3)
xN
yN (4)
1,564
8 x 56 = 448
6 x 101 = 606
8 x 3 = 24
6 x 10 = 60
6 x 27 = 162
4 x 1 = 4
6 x 26 = 156
4 x 26 = 104
2
xR
yR (1)
xN
yN (2)
xP
yP (3)
xF
yF (4)
1,384
8 x 27 = 216
6 x 1 = 6
8 x 26 = 208
6 x 26 = 156
6 x 56 = 336
4 x 101 = 404
6 x 3 = 18
4 x 10 = 40
There is a difference of 180 characters at the start of the expressions. However we do not know as to the number of characters that would result after simplification if any. We take the second approach of assigning Point R as Point 1, Point N as Point 2, Point P as Point 3 and Point F as Point 4 and derive the equation.
x = ((y1 x2 – y2 x1) (x4 – x3) - (y3 x4 – y4 x3) (x2 – x1)) / ((y4 – y3) (x2 – x1) - (y2 – y1) (x4 – x3))
x1 = xR = (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) Substitute for x1 to get:
x = ((y1 x2 – y2 (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) (x4 – x3) - (y3 x4 – y4 x3) (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – y3) (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) - (y2 – y1) (x4 – x3))
x = ((y1 x2 + y2 (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (x4 – x3) - (y3 x4 – y4 x3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – y3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - (y2 – y1) (x4 – x3))
y1 = yR = 1. Substituting we get:
x = ((x2 + y2 (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (x4 – x3) - (y3 x4 – y4 x3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – y3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - (y2 – 1) (x4 – x3))
x2 = xN = (((√3) – 1) + √(2((3√3) – 2)))/4 = yN = y2. Since x2 = y2, we simplify as below:
x = ((x2 (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) (x4 – x3) - (y3 x4 – y4 x3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – y3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - (y2 – 1) (x4 – x3))
The blue colored parentheses above are redundant and hence are deleted below.
x = (x2 (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (x4 – x3) - (y3 x4 – y4 x3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – y3) (x2 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - (y2 – 1) (x4 – x3))
Substitute for x2 to get
x = (((((√3) – 1) + √(2((3√3) – 2)))/4) (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (x4 – x3) - (y3 x4 – y4 x3) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – y3) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (x4 – x3))
x3 = xP = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )
Replacing x3 gives
x = (((((√3) – 1) + √(2((3√3) – 2)))/4) (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (y3 x4 – y4 (((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – y3) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))
y3 = yP = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))
Replace y3 to get:
x = (((((√3) – 1) + √(2((3√3) – 2)))/4) (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))x4 – y4 (((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))
x4 = xF = 0.5
The equation for x becomes,
x = (((((√3) – 1) + √(2((3√3) – 2)))/4) (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) 0.5 – y4 (((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((y4 – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))
y4 = yF = ((√3)/2 – 1)
Hence the equation for the x-coordinate of Point S, xS is given by
xS = (((((√3) – 1) + √(2((3√3) – 2)))/4) (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) 0.5 – ((√3)/2 – 1) (((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((((√3)/2 – 1) – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))
Let us evaluate the value of xS now. We identify the inner most expression in the expression and replace it with its value and continue reducing the depth eventually arriving at the value desired.
Verification of xS:
With too many parentheses we are liable to get confused. To avoid confusion, we number the level of parentheses. We start with 0, a left parenthesis increases the level by 1, while a right parenthesis decreases the level by 1. The expression should end with 0 level to be free of errors. The parentheses levels are given in blue color to avoid confusion with numbers that are part of the expression.
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 (21+ (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8(9√3)8 – 1)7 + √(82(9(103√3)9 – 2)8)7)6/4)5)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8(9√3)8 – 1)7 + √(82(9(103√3)9 – 2)8)7)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8(9√3)8 – 1)7 + √(82(9(103√3)9 – 2)8)7)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)32 )2)1)0
In evaluating the equation we start with the highest level and reduce the levels one at a time. The highest level is 10. The sub expression (103√3)9 appears thrice in the expression for the x-coordinate of Point S. This is replaced by 5.196152 for a six digit accuracy after the decimal point. We get:
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 (21+ (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8(9√3)8 – 1)7 + √(82(9 5.196152 – 2)8)7)6/4)5)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8(9√3)8 – 1)7 + √(82(9 5.196152 – 2)8)7)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8(9√3)8 – 1)7 + √(82(9 5.196152 – 2)8)7)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8(93√3)8 – 2)7)6)5/4)4)32 )2)1)0
Likewise we reduce in steps. Now the highest level is 9.
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 (21+ (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8 5.196152 – 2)7)6)5/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8 5.196152 – 2)7)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8 1.732051 – 1)7 + √(82 x 3.196152)7)6/4)5)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8 1.732051 – 1)7 + √(82 x 3.196152)7)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8 5.196152 – 2)7)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(7(8 1.732051 – 1)7 + √(82 x 3.196152)7)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(7(8√3)7 – 1)6 + √(72(8 5.196152 – 2)7)6)5/4)4)32 )2)1)0
The highest level is 8. Reducing by one level we get:
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 (21+ (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(71.732051 – 1)6 + √(72 x 3.196152)6)5/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(71.732051 – 1)6 + √(72 x 3.196152)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(70.732051 + √6.392304)6/4)5)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(70.732051 + √6.392304)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(6(71.732051 – 1)6 + √(72 x 3.196152)6)5/4)4)3 / √(41 + (51 – (6(7√3)6 – 1)5/(6(70.732051 + √6.392304)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(6(71.732051 – 1)6 + √(72 x 3.196152)6)5/4)4)32 )2)1)0
The highest level in the above expression is 7.
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 (21+ (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(60.732051 + √6.392304)5/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(60.732051 + √6.392304)5/4)4)3 / √(41 + (51 – (61.732051 – 1)5/(6(70.732051 + √6.392304)6/4)5)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – (61.732051 – 1)5/(6(70.732051 + √6.392304)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(60.732051 + √6.392304)5/4)4)3 / √(41 + (51 – (61.732051 – 1)5/(6(70.732051 + √6.392304)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(60.732051 + √6.392304)5/4)4)32 )2)1)0
The above expression still has level 7 because factors like (70.732051 + √6.392304)6 cannot be reduced until √6.392304 is reduced first. We replace √6.392304 by its 6 digit accurate value of 2.528301 to get the following expression.
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 (21+ (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)3 / √(41 + (51 – (61.732051 – 1)5/(6(70.732051 + 2.528301)6/4)5)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – (61.732051 – 1)5/(6(70.732051 + 2.528301)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)3 / √(41 + (51 – (61.732051 – 1)5/(6(70.732051 + 2.528301)6/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)32 )2)1)0
Reducing the above gives:
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 (21+ (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)3 / √(41 + (51 – (61.732051 – 1)5/(63.260352/4)5)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – (61.732051 – 1)5/(63.260352/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)3 / √(41 + (51 – (61.732051 – 1)5/(63.260352/4)5)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 + (3√(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (5(6√3)5 – 1)4/(5(60.732051 + 2.528301)5/4)4)32 )2)1)0
The above expression has the highest level of 6. We continue the reduction below.
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4)1 (21+ (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (51.732051 – 1)4/(53.260352/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (51.732051 – 1)4/(53.260352/4)4)3 / √(41 + (51 – 0.732051/0.815088)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – 0.732051/0.815088)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 + (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (51.732051 – 1)4/(53.260352/4)4)3 / √(41 + (51 – 0.732051/0.815088)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 + (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (51.732051 – 1)4/(53.260352/4)4)32 )2)1)0
The highest level is reduced to 5 now. Before we remove one more level of parentheses we have to replace 0.732051/0.815088 with 0.898125.
xS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4)1 (21+ (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (51.732051 – 1)4/(53.260352/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (51.732051 – 1)4/(53.260352/4)4)3 / √(41 + (51 – 0.898125)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – 0.898125)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 + (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (51.732051 – 1)4/(53.260352/4)4)3 / √(41 + (51 – 0.898125)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 + (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (51.732051 – 1)4/(53.260352/4)4)32 )2)1)0
Reducing by one more level we get:
xS = 0(1(2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4)1 (21+ (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (51.732051 – 1)4/(53.260352/4)4)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – (51.732051 – 1)4/(53.260352/4)4)3 / √(41 + (51 – 0.898125)42 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(4(5√3)4 – 1)3 / √(41 + (51 – 0.898125)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 + (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – (51.732051 – 1)4/(53.260352/4)4)3 / √(41 + (51 – 0.898125)42 )3)2)1 (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 + (3√(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2)1 - (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – (51.732051 – 1)4/(53.260352/4)4)32 )2)1)0
We have 5 levels now. We continue the reduction.
xS = 0(1(2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4)1 (21+ (3√(40.7320512 – 0.2679492)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – 0.732051/0.815088)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – 0.732051/0.815088)3 / √(41 + 0.1018752 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(41.732051 – 1)3 / √(41 + 0.1018752 )3)2)1 (2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4 + (3√(40.7320512 - 0.2679492)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – 0.732051/0.815088)3 / √(41 + 0.1018752 )3)2)1 (2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4 + (3√(40.7320512 - 0.2679492)3)2)1 - (2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – 0.732051/0.815088)32 )2)1)0
In the 4th level expression above, we need to reduce some factors before the parentheses could be removed.
xS = 0(1(2(3(41.732051 – 1)3 + √(46.392304)3)2/4)1 (21+ (3√(40.535899 – 0.071797)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – 0.898125)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – 0.898125)3 / √(41 + 0.010379 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(41.732051 – 1)3 / √(41 + 0.010379 )3)2)1 (2(3(41.732051 – 1)3 + √(46.392304)3)2/4 + (3√(40.535899 - 0.071797)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – 0.898125)3 / √(41 + 0.010379 )3)2)1 (2(3(41.732051 – 1)3 + √(46.392304)3)2/4 + (3√(40.535899 - 0.071797)3)2)1 - (2(3(41.732051 – 1)3 + √(46.392304)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (41 – 0.898125)32 )2)1)0
We still need to reduce the expressions under square root and square operations to single values.
xS = 0(1(2(3(41.732051 – 1)3 + √(46.392304)3)2/4)1 (21+ (3√(40.464102)3)2)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (40.101875)32 )2)1 - (2(3(4√3)3 – 1)2 (31 - (41 – 0.898125)3 / √(41.010379 )3)2 0.5 – (3(4√3)3/2 – 1)2 (3(41.732051 – 1)3 / √(41.010379 )3)2)1 (2(3(41.732051 – 1)3 + √(46.392304)3)2/4 + (3√(40.464102)3)2)1)0 / (1(2(3(4√3)3/2 – 1)2 – (3(4√3)3 – 1)2 (31 - (41 – 0.898125)3 / √(41.010379 )3)2)1 (2(3(41.732051 – 1)3 + √(46.392304)3)2/4 + (3√(40.464102)3)2)1 - (2(3(41.732051 – 1)3 + √(46.392304)3)2/4 – 1)1 (20.5 – (3(4√3)3 – 1)2 / √(31 + (40.101875)32 )2)1)0
Reducing the parentheses level by 1 we get:
xS = 0(1(2(30.732051 + 2.528301)2/4)1 (21+ (30.681250)2)1 (20.5 – (31.732051 – 1)2 / √(31 + 0.010379 )2)1 - (2(31.732051 – 1)2 (31 – 0.101875 / 1.005176)2 0.5 – (31.732051/2 – 1)2 (30.732051 / 1.005176)2)1 (2(30.732051 + 2.528301)2/4 + (30.681250)2)1)0 / (1(2(31.732051/2 – 1)2 – (31.732051 – 1)2 (31 – 0.101875 / 1.005176)2)1 (2(30.732051 + 2.528301)2/4 + (30.681250)2)1 - (2(30.732051 + 2.528301)2/4 – 1)1 (20.5 – (31.732051 – 1)2 / √(31 + 0.010379 )2)1)0
The parentheses in the above expression are at a maximum level of 3. We reduce some expressions first.
xS = 0(1(2(30.732051 + 2.528301)2/4)1 (21+ (30.681250)2)1 (20.5 – (31.732051 – 1)2 / √1.010379)1 - (2(31.732051 – 1)2 (31 – 0.101350)2 0.5 – (31.732051/2 – 1)2 (30.732051 / 1.005176)2)1 (2(30.732051 + 2.528301)2/4 + 0.681250)1)0 / (1(2(31.732051/2 – 1)2 – (31.732051 – 1)2 (31 – 0.101350)2)1 (2(30.732051 + 2.528301)2/4 + 0.681250)1 - (2(30.732051 + 2.528301)2/4 – 1)1 (20.5 – (31.732051 – 1)2 / √1.010379)1)0
Now we reduce the parentheses level by 1.
xS = 0(1(23.260352/4)1 (21+ 0.681250)1 (20.5 – 0.732051 / √1.010379)1 - (20.732051 x 0.898650 x 0.5 + 0.133974 x 0.728281)1 (23.260352/4 + 0.681250)1)0 / (1(2-0.133974 – 0.732051 x 0.898650)1 (23.260352/4 + 0.681250)1 - (23.260352/4 – 1)1 (20.5 – 0.732051 / √1.010379)1)0
Reducing the expressions leads to
xS = 0(1(23.260352/4)1 (21+ 0.681250)1 (20.5 – 0.728281)1 - (20.328929 + 0.097571)1 (20.815088 + 0.681250)1)0 / (1(2-0.133974 – 0.657858)1 (20.815088 + 0.681250)1 - (20.815088 – 1)1 (20.5 – 0.728281)1)0
The above level 2 expression in parentheses is reduced further.
xS = 0(1 – 0.312829 – 0.426500 x 1.496338)0 / (1– 0.791832 x 1.496338 – 0.184912 x 0.228281)0
Reducing the expressions, we get:
xS = 0(1 – 0.312829 – 0.638188)0 / (1– 1.184848 – 0.042212)0
= 0.951017 / 1.22706
= 0.775037
The X-coordinate of S given by geogebra is 0.78. Thus, the X-coordinate of Point S is verified.
We continue with the derivation for the expression of the Y-coordinate of Point S.
The equation for the Y-coordinate is
yS = ((y2 – y1) (((y1 x2 – y2 x1) (x4 – x3) - (y3 x4 – y4 x3) (x2 – x1)) / ((y4 – y3) (x2 – x1) - (y2 – y1) (x4 – x3))) + (y1 x2 – y2 x1)) / (x2 – x1)
x1 = xR = (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) and thus
yS = ((y2 – y1) (((y1 x2 – y2 (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) (x4 – x3) - (y3 x4 – y4 x3) (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((y4 – y3) (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - (y2 – y1) (x4 – x3))) + (y1 x2 – y2 (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )
y1 = yR = 1. Substituting we get:
yS = ((y2 – 1) (((x2 – y2 (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) (x4 – x3) - (y3 x4 – y4 x3) (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((y4 – y3) (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - (y2 – 1) (x4 – x3))) + (x2 – y2 (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / (x2 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )
x2 = xN = (((√3) – 1) + √(2((3√3) – 2)))/4 = yN = y2.
Substituting for x2 and y2 we get:
yS = (((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) ((((((√3) – 1) + √(2((3√3) – 2)))/4 x (1– (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) ) (x4 – x3) - (y3 x4 – y4 x3) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((y4 – y3) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (x4 – x3))) + ((((√3) – 1) + √(2((3√3) – 2)))/4) x (1 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )
x3 = xP = ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ) Substitute for x3 to get:
yS = (((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) ((((((√3) – 1) + √(2((3√3) – 2)))/4 x (1– (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) ) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (y3 x4 – y4 ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((y4 – y3) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))) + ((((√3) – 1) + √(2((3√3) – 2)))/4) x (1 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )
y3 = yP = ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) Replacing y3 gives:
yS = (((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) ((((((√3) – 1) + √(2((3√3) – 2)))/4 x (1– (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) ) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) x4 – y4 ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((y4 – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (x4 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))) + ((((√3) – 1) + √(2((3√3) – 2)))/4) x (1 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )
x4 = xF = 0.5 Hence,
yS = (((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) ((((((√3) – 1) + √(2((3√3) – 2)))/4 x (1– (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) ) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) x 0.5 – y4 ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((y4 – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))) + ((((√3) – 1) + √(2((3√3) – 2)))/4) x (1 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )
y4 = yF = ((√3)/2 – 1) Substitute for y4 to get the expression
yS = (((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) ((((((√3) – 1) + √(2((3√3) – 2)))/4 x (1– (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) ) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) x 0.5 – ((√3)/2 – 1) ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3)/2 – 1) – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))) + ((((√3) – 1) + √(2((3√3) – 2)))/4) x (1 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )
Indicating the parenthesis levels in red, we have:
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 x (51– (6- √(7(8(9√3)8 – 1)72 - (81 - (9(10√3)9 – 1)8)72)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(9(10(11√3)10 – 1)9 + √(102(11(123√3)11 – 2)10)9)8/4)7)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(9(10(11√3)10 – 1)9 + √(102(11(123√3)11 – 2)10)9)8/4)7)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
Now we are ready to evaluate the expression.
The highest level of parenthesis is 12. We reduce this level now.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 x (51– (6- √(7(8(9√3)8 – 1)72 - (81 - (9(10√3)9 – 1)8)72)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(9(10(11√3)10 – 1)9 + √(102(115.196152 – 2)10)9)8/4)7)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(9(10(11√3)10 – 1)9 + √(102(115.196152 – 2)10)9)8/4)7)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(10(113√3)10 – 2)9)8)7/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
We reduce the next highest level of 11 now.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 x (51– (6- √(7(8(9√3)8 – 1)72 - (81 - (9(10√3)9 – 1)8)72)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(105.196152 – 2)9)8)7/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(105.196152 – 2)9)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(9(101.732051 – 1)9 + √(102 x 3.196152)9)8/4)7)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(105.196152 – 2)9)8)7/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(105.196152 – 2)9)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(9(101.732051 – 1)9 + √(102 x 3.196152)9)8/4)7)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(9(10√3)9 – 1)8 + √(92(105.196152 – 2)9)8)7/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
The above expression has the highest level of 10. We reduce.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 x (51– (6- √(7(8(9√3)8 – 1)72 - (81 - (91.732051 – 1)8)72)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(90.732051 + √6.392304)8/4)7)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(90.732051 + √6.392304)8/4)7)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
The highest level in the expression above is 9. Before reducing one more level of parenthesis, we need to replace some expressions by their values.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 x (51– (6- √(7(8(9√3)8 – 1)72 - (81 - (91.732051 – 1)8)72)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(90.732051 + 2.528301)8/4)7)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)5 / √(61 + (71 – (8(9√3)8 – 1)7/(8(90.732051 + 2.528301)8/4)7)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (8(9√3)8 – 1)7)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(8(91.732051 – 1)8 + √(92 x 3.196152)8)7/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
Reducing by one more level, we get:
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 x (51– (6- √(7(81.732051 – 1)72 - (81 - 0.732051)72)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(80.732051 + √6.392304)7/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(80.732051 + √6.392304)7/4)6)5 / √(61 + (71 – (81.732051 – 1)7/(83.260352/4)7)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(80.732051 + √6.392304)7/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (81.732051 – 1)7)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(7(80.732051 + √6.392304)7/4)6)5 / √(61 + (71 – (81.732051 – 1)7/(83.260352/4)7)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (81.732051 – 1)7)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(7(80.732051 + √6.392304)7/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
We need to reduce some expressions.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 x (51– (6- √(7(80.732051)72 - (80.267949)72)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(73.260352/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(73.260352/4)6)5 / √(61 + (71 – (81.732051 – 1)7/(83.260352/4)7)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(73.260352/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (81.732051 – 1)7)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (7(8√3)7 – 1)6/(73.260352/4)6)5 / √(61 + (71 – (81.732051 – 1)7/(83.260352/4)7)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – (5- √(6(7(8√3)7 – 1)62 - (71 - (81.732051 – 1)7)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(7(83√3)7 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (7(8√3)7 – 1)6/(73.260352/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
We reduce by one more level now.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 x (51– (6- √(70.535899 - 0.071797)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (71.732051 – 1)6/(73.260352/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (71.732051 – 1)6/(73.260352/4)6)5 / √(61 + (71 – 0.732051/0.815088)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (71.732051 – 1)6/(73.260352/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – (5- √(6(71.732051 – 1)62 - (71 - 0.732051)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (71.732051 – 1)6/(73.260352/4)6)5 / √(61 + (71 – 0.732051/ 0.815088)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – (5- √(6(71.732051 – 1)62 - (71 - 0.732051)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (71.732051 – 1)6/(73.260352/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
Reducing some expressions, we get:
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 x (51– (6- √(70.535899 - 0.071797)6)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (71.732051 – 1)6/(73.260352/4)6)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – (71.732051 – 1)6/(73.260352/4)6)5 / √(61 + (70.101875)62 )5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – (71.732051 – 1)6/(73.260352/4)6)52 )4)3 (4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – (5- √(6(70.732051)62 - (70.267949)62)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – (71.732051 – 1)6/(73.260352/4)6)5 / √(61 + (70.101875)62 )5)4)3 (4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – (5- √(6(70.732051)62 - (70.267949)62)5)4 )3 - (4(5(6(7√3)6 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – (71.732051 – 1)6/(73.260352/4)6)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (6(7√3)6 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
The highest level in the above expression is 7. Reduce one level now.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(61.732051 – 1)5 + √(62 x 3.196152)5)4/4 x (51– (6- √0.464102)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – 0.732051/0.815088)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – 0.732051/0.815088)5 / √(61 + 0.010379)5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (61 – 0.732051/0.815088)52 )4)3 (4(5(61.732051 – 1)5 + √(62 x 3.196152)5)4/4 – (5- √(60.535899 - 0.071797)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – 0.732051/0.815088)5 / √(61 + 0.010379)5)4)3 (4(5(61.732051 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – (5- √(60.535899 - 0.071797)5)4 )3 - (4(5(61.732051 – 1)5 + √(62 x 3.196152)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (61 – 0.732051/0.815088)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
The above expression has the depth of 6 in parenthesis. Before we could reduce the parenthesis level, we need to simplify some expressions.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4 – 1)1 (2(3(4(5(61.732051 – 1)5 + √(62 x 3.196152)5)4/4 x (51– (6- √0.464102)5)4 )3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (60.101875)52 )4)3 - (4(5(6√3)5 – 1)4 (51 - (61 – 0.732051/0.815088)5 / √(61 + 0.010379)5)4 x 0.5 – (5(6√3)5/2 – 1)4 (5(6√3)5 – 1)4 / √(51 + (60.101875)52 )4)3 (4(5(61.732051 – 1)5 + √(62 x 3.196152)5)4/4 – (5- √(60.535899 - 0.071797)5)4 )3)2 / (3(4(5(6√3)5/2 – 1)4 – (5(6√3)5 – 1)4 (51 - (61 – 0.732051/0.815088)5 / √(61 + 0.010379)5)4)3 (4(5(61.732051 – 1)5 + √(62(75.196152 – 2)6)5)4/4 – (5- √(60.535899 - 0.071797)5)4 )3 - (4(5(61.732051 – 1)5 + √(62 x 3.196152)5)4/4 – 1)3 (40.5 – (5(6√3)5 – 1)4 / √(51 + (60.101875)52 )4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(5(63√3)5 – 2)4)3)2/4)1 x (21 – (3- √(4(5(6√3)5 – 1)42 - (51 - (61.732051 – 1)5)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (5(6√3)5 – 1)4)32)2)1 )0
Now we reduce the parenthesis level by one.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 – 1)1 (2(3(4(50.732051 + √6.392304)4/4 x (51 + √0.464102)4 )3 (40.5 – (51.732051 – 1)4 / √(51 + 0.010379)4)3 - (4(51.732051 – 1)4 (51 - 0.101875 / √1.010379)4 x 0.5 – (51.732051/2 – 1)4 (51.732051 – 1)4 / √(51 + 0.010379)4)3 (4(50.732051 + √6.392304)4/4 – (5- √0.464102)4 )3)2 / (3(4(51.732051/2 – 1)4 – (51.732051 – 1)4 (51 - 0.101875 / √1.010379)4)3 (4(50.732051 + √6.392304)4/4 – (5- √0.464102)4 )3 - (4(50.732051 + √6.392304)4/4 – 1)3 (40.5 – (51.732051 – 1)4 / √(51 + 0.010379)4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4)1 x (21 – (3- √(4(51.732051 – 1)42 - (51 - 0.732051)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (51.732051 – 1)4)32)2)1 )0
Before we reduce the parenthesis level in the above 5-level equation, we reduce some expressions.
yS = 0(1(2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4 – 1)1 (2(3(4(50.732051 + 2.528301)4/4 x (51 + 0.681250)4 )3 (40.5 – (51.732051 – 1)4 / √(51 + 0.010379)4)3 - (4(51.732051 – 1)4 (51 - 0.101350)4 x 0.5 – (50.866026 – 1)4 (51.732051 – 1)4 / √(51 + 0.010379)4)3 (4(50.732051 + 2.528301)4/4 – (5- 0.681250)4 )3)2 / (3(4(50.866026 – 1)4 – (51.732051 – 1)4 (51 - 0.101350)4)3 (4(50.732051 + 2.528301)4/4 – (5- 0.681250)4 )3 - (4(50.732051 + 2.528301)4/4 – 1)3 (40.5 – (51.732051 – 1)4 / √(51 + 0.010379)4)3)2)1 + (2(3(4(5√3)4 – 1)3 + √(42(55.196152 – 2)4)3)2/4)1 x (21 – (3- √(4(50.732051)42 - (50.267949)42)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(4(53√3)4 – 2)3)2)1/4 – (2- √(3(4(5√3)4 – 1)32 - (41 - (51.732051 – 1)4)32)2)1 )0
Now we reduce the parenthesis level by one.
yS = 0(1(2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4 – 1)1 (2(3(43.260352/4 x 1.681250 )3 (40.5 – 0.732051 / √1.010379)3 - (40.732051 x 0.898650 x 0.5 + 0.133974 x 0.732051 / √1.010379)3 (43.260352/4 + 0.681250)3)2 / (3(4–0.133974 – 0.732051 x 0.898650)3 (43.260352/4 + 0.681250)3 - (43.260352/4 – 1)3 (40.5 – 0.732051 / √1.010379)3)2)1 + (2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4)1 x (21 – (3- √(40.535899 - 0.071797)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(45.196152 – 2)3)2)1/4 – (2- √(3(41.732051 – 1)32 - (41 - 0.732051)32)2)1 )0
The above 4th level expression has some complex factors, which need to be simplified first.
yS = 0(1(2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4 – 1)1 (2(3(43.260352/4 x 1.681250 )3 (40.5 – 0.728281)3 - (40.328929 + 0.097571)3 (40.815088+ 0.681250)3)2 / (3(4–0.133974 – 0.657858)3 (40.815088 + 0.681250)3 - (40.815088 – 1)3 (40.5 – 0.728281)3)2)1 + (2(3(41.732051 – 1)3 + √(42 x 3.196152)3)2/4)1 x (21 – (3- √(40.535899 - 0.071797)3)2 )1)0 / (1(2(3(4√3)3 – 1)2 + √(32(45.196152 – 2)3)2)1/4 – (2- √(3(40.732051)32 - (40.267949)32)2)1 )0
Reducing by one level, we get:
yS = 0(1(2(30.732051 + √6.392304)2/4 – 1)1 (2(31.370367 x (– 0.228281) - 0.426500 x 1.496338)2 / (3(–0.791832) x 1.496338 - (–0.184912) x (– 0.228281))2)1 + (2(30.732051 + √6.392304)2/4)1 x (21 – (3- √0.464102)2 )1)0 / (1(2(31.732051 – 1)2 + √(32 x 3.196152)2)1/4 – (2- √(30.535899 - 0.071797)2)1 )0
We get the following by reducing some complex factors in the above equation with 3 levels of parenthesis.
yS = 0(1(2(30.732051 + 2.528301)2/4 – 1)1 (2(3 –0.312829 –0.638189)2 / (3–1.184848 -0.042212)2)1 + (2(30.732051 + 2.528301)2/4)1 x (21 – (3-0.681250)2 )1)0 / (1(2(31.732051 – 1)2 + √(32 x 3.196152)2)1/4 – (2- √(30.535899 - 0.071797)2)1 )0
Reduce by 1 level to get:
yS = 0(1(23.260352/4 – 1)1 (2–0.951018 / –1.227060)1 + (23.260352/4)1 x (21 + 0.681250 )1)0 / (1(20.732051 + √6.392304)1/4 – (2- √0.464102)1 )0
Simplifying complex factors gives:
yS = 0(1(20.815088 – 1)1 (20.775038)1 + (23.260352/4)1 x (21 + 0.681250 )1)0 / (1(20.732051 + 2.528301)1/4 – (2- 0.681250)1 )0
Reduce by one more level to get:
yS = 0(1-184912 x 0.775038 + 0.815088 x 1.681250)0 / (13.260352/4 + 0.681250)0
yS = 0(1–0.143314 + 1.370367)0 / (10.815088 + 0.681250)0
yS = 1.227053 / 1.496338 = 0.820037
Y-coordinate of Point S verified.
Geogebra gives S(0.78, 0.82)
26. Circle r has center at Point B and passes through Point S. This circle is the result.
27. The area of this circle is calculated to be 4.
28. This area is recorded.
The Point S to 6-digit accuracy is (0.775037, 0.820037)
The square of the radius is 0.7750372 + 0.8200372 = 0.600682 + 0.672460 = 1.273142.
Area of the circle is x 1.273142 = 3.999694
Error = (4 – 3.999694)/4 x 100% = 0.00765%
Accuracy = (100 – 0.00765)% = 99.99235%
Earlier accuracy is 99.8% and by God’s grace the solution by Ed is more accurate.
The value of radius r is of interest to us. It should be equal to 2/(√Π). From this equation we get an algebraic equation for Π.
Π = 4 / r2 = 4 / ((xS – xB)2 + (yS – yB)2)
= 4 / { [(((((√3) – 1) + √(2((3√3) – 2)))/4) (1+ (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) 0.5 – ((√3)/2 – 1) (((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2)))) / ((((√3)/2 – 1) – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 + (√(((√3) – 1)2 - (1 - ((√3) – 1))2))) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))]2 + [(((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) ((((((√3) – 1) + √(2((3√3) – 2)))/4 x (1– (- √(((√3) – 1)2 - (1 - ((√3) – 1))2))) ) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) - (((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) x 0.5 – ((√3)/2 – 1) ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3)/2 – 1) – ((√3) – 1) (1 - (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4)) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 ))) ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) ) - ((((√3) – 1) + √(2((3√3) – 2)))/4 – 1) (0.5 – ((√3) – 1) / √(1 + (1 – ((√3) – 1)/((((√3) – 1) + √(2((3√3) – 2)))/4))2 )))) + ((((√3) – 1) + √(2((3√3) – 2)))/4) x (1 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )) / ((((√3) – 1) + √(2((3√3) – 2)))/4 – (- √(((√3) – 1)2 - (1 - ((√3) – 1))2)) )]2}
The above expression for is algebraic involving one irrational number √3 only.
A philosophical query:
We shifted the origin of the figure identified by Ed in the hope that such a shift might produce a simpler expression. In one approach to proving the correctness of programs with loops, we identified ‘loop invariants’ which helped us to eliminate the need to identify the number of times a loop would be executed. Are we suggesting that “a geometric construction for an algebraic equation is ‘expression invariant’ of the relationship between the RHS and LHS of the algebraic equation?”
