Solution to Problem #24

Answer: r = (sqrt[7]-sqrt[3])a/8 = .114212563 a
Solution:
Define the length of the inner equilateral triangle, A'B'C', as d.
Then:
0) The height of A'B'C' is 3r.
1)              d = 2r*sqrt(3)

Consider triangle ABA'.  Define AB = a, BA' = b, AA' = c, A to the
point of tangency with the inscibed circle = x, B to the circle = y,
A' to the circle = z.  Then:
2)              a = x+y
3)              b = y+z
4)              c = x+z
5)              d = c-b = x-y

6) (1->5)       x = y + 2r*sqrt(3)

There are two small right triangles with base z and height r, splitting
the angle BA'A.  Angle BA'A is 120 degrees (it's next to the 60 degree
angle B'A'C') so:
7)              z/r = tan(30)  -->  z = r/sqrt(3)

The area of ABA' can be found by the formula A=sqrt[s(s-a)(s-b)(s-c)], where
s=(a+b+c)/2.  It can also be found by adding up the six small right triangles,
each with height r:
8)              s = (a+b+c)/2 = x+y+z
8a)             A = sqrt[s(s-a)(s-b)(s-c) = sqrt[xyz(x+y+z)]
9)              A = r(x+y+z)
10) (8a=9)      xyz = r^2(x+y+z)
11) (6,7->10)   (y+2r*sqrt(3))(y)(r/sqrt(3)) = r^2(y+2r*sqrt(3)+y+r/sqrt(3))

Simplifying:
11a)            y^2/sqrt(3) + 2ry = 2ry + r^2(2sqrt(3) + 1/sqrt(3))
11b)            y^2 = sqrt(3)*r^2*(2sqrt(3) + 1/sqrt(3))
11c)            y^2 = 7r^2
11d)            y = r*sqrt(7)
12) (11d->6)    x = r(sqrt(7) + 2sqrt(3))

Total area of ABC:
13)             A = a*(a*sqrt3)/2)*(1/2) = a^2*sqrt(3)/4
14)             A = 3*[ABA' Area] + d^2*sqrt(3)/4
14a)              = 3*[r(x+y+z)] + 3r^2*sqrt(3)
14b)              = 3*[r[r(sqrt(7) + 2sqrt(3)) + r*sqrt(7) + r/sqrt(3)]]
			+ 3r^2*sqrt(3)
14c)              = 3r^2*[2*sqrt(7) + 3*sqrt(3) + 1/sqrt(3)]
14d)              = r^2*[6sqrt(7) + 10sqrt(3)]

15) (13=14)     a^2*sqrt(3)/4 = r^2*[6sqrt(7) + 10sqrt(3)]
15a)            a^2 = r^2*[24sqrt(7/3) + 40]

Solve for r in terms of a:
16)             r = a/sqrt[24sqrt(7/3) + 40]
		  = ((sqrt[7]-sqrt[3])/2)*a
Numerically,    r = a/8.7556043 
		  = .114212563 a