October Mensa Math Puzzle

Can't figure out how to post an image just yet.
Call angles of triangle including the circle A,B,C (CCW from lower left)
A=45 (degrees)
B=atan (2) .... B is also angle at midpoint of top for right triangle with horizontal side (1/2) and vertical right side (1)
C=180-A-B=135-B

Then
sin(A) = 1/sqrt(2)
sin(B) = 2/sqrt(5).....cos(B) = 1/sqrt(5)
sin(C) = sin(135-B)=sin(135)cos(B)-cos(135)sin(B) = 3/sqrt(10)

Call sides opposite A,B,C = a,b,c

Law of sines
a/sin(A) = b/sin(B) = c/sin(C)

c=1. .... bottom of square
b=sin(B)/sin(C) = 2sqrt(2)/3
a=sin(A)/sin(C) = sqrt(5)/3

Break sides a,b,c into two parts at the tangent point....with

x+y = c
y+z = a
x+z = b

NOTE: the two tangent segments from the triangle vertex to circle are the same by symmetry

Solve
b+c-a = (x+z)+(x+y)-(y+z)=2x

x=(b+c-a)/2 = sqrt(2)/3+1/2-sqrt(5)/6

Then bisect angle A to the center of circle and draw radius to tangent on the bottom

tan(22.5) = R/x

R = x tan(22.5)

Area = pi*R^2
=pi* ( (sqrt(2)/3+1/2-sqrt(5)/6)*tan(22.5) )^2
=0.19322152214633

Using vector math, I find a radius of 2 / [3 + 2sqrt(2) + sqrt(5)].


The unit vector pointing from (0,0) to (1,0) is i. And the unit vector pointing from (0,0) to (1,1) is (i + j) / sqrt(2).

Any vector in the direction from (0,0) to the center of the circle would be proportional to the sum the above two unit vectors. One such vector is M = [1+sqrt(2)]i + j.

Likewise, any vector in the direction from (1,0) to the center of the circle would be proportional to the sum of these two unit vectors: -i and (-i + 2j) / sqrt(5). One such vector would be N = [1+sqrt(5)]i - 2j. (actually this vector points in the exactly opposite direction, but it doesn't matter for the calculation.)

The position vector of the center of the circle would be the product of a constant (call it 'a') and M which is also equal to i + bN. That is: aM = i + bN.

This vector equation is equivalent to these two equations:
a[1+sqrt(2)] = 1 + b[1+sqrt(5)]
a = -2b

Substituting b = -a/2 into the first equation yields:
a[1+sqrt(2)] = 1 - a[1+sqrt(5)]/2

Dividing by a yields:
1+sqrt(2) = 1/a - [1+sqrt(5)]/2

Therefore:
1/a = 1+sqrt(2) + 1/2 + sqrt(5)/2

Or:
a = 2 / [3 + 2sqrt(2) + sqrt(5)]

And remember that the position vector of the circle's center (x,y) is aM, so y is simply a. The radius of the circle is equal to y, so r = 2 / [3 + 2sqrt(2) + sqrt(5)].

For some reason I thought it asked for area....not too Mensa-like to read the question wrong.

From my previous answer (ignoring the area calculation) I had

R=x tan (22.5)
=(sqrt(2)/3+1/2-sqrt(5)/6)*tan(22.5)

If use half angle for tan(22.5) = (1-cos(45))/sin(45) = sqrt(2) - 1

So without trig function in answer I get

R = (sqrt(2)/3+1/2-sqrt(5)/6) * (sqrt(2) - 1)
= (1+sqrt(2)+sqrt(5)-sqrt(10))/6
=0.248000646617418

Which matches Chesterdog, via calculation. Haven't figured out how my method matches it yet though.

I am getting the same answer everyone else seems to be getting.

Let ABCD be the square such that A is the upper left corner and B the upper right
Let M be the midpoint of AB, X the point where BD intersects CM, and O the center of the circle
Define 2t = the measure of BDC = 45 degrees, and 2p = the measure of MCD = arctan 2
Let E be on side CD such that OE is perpendicular to CD
tan 2t = 1, so tan t = sqrt(2) - 1
tan 2p = 2, so tan p = (sqrt(5) - 1) / 2
r = x tan t = (1 - x) tan p
x (sqrt(2) - 1) = (1 - x) (sqrt(5) - 1) / 2
x = (sqrt(5) - 1) / (sqrt(5) + 2 sqrt(2) - 3)
r = x tan t = (sqrt(5) - 1) / (sqrt(5) + 2 sqrt(2) - 3) * (sqrt(2) - 1)
= (sqrt(10) - sqrt(5) - sqrt(2) + 1) / (sqrt(5) + 2 sqrt(2) - 3)
= about 0.24800065

That result can be reduced:
(sqrt(10) - sqrt(5) - sqrt(2) + 1) / (sqrt(5) + 2 sqrt(2) - 3)
= (sqrt(5) - 1) (sqrt(2) - 1) / (sqrt(5) + 2 sqrt(2) - 3)
= (sqrt(5) - 1) (sqrt(2) - 1) (sqrt(5) - (2 sqrt(2) - 3)) / ((sqrt(5) + (2 sqrt(2) - 3)) (sqrt(5) - (2 sqrt(2) - 3)))
= (sqrt(5) - 1) (sqrt(2) - 1) (sqrt(5) - 2 sqrt(2) + 3)) / (5 - (2 sqrt(2) - 3)^2)
= (sqrt(5) - 1) (sqrt(2) - 1) (sqrt(5) - 2 sqrt(2) + 3)) / (5 - 8 + 12 sqrt(2) - 9)
= (sqrt(5) - 1) (sqrt(2) - 1) (sqrt(5) - 2 sqrt(2) + 3)) / (12 (sqrt(2) - 1))
= (sqrt(5) - 1) (sqrt(2) - 1) (sqrt(2) + 1) (sqrt(5) - 2 sqrt(2) + 3)) / (12 (sqrt(2) - 1) (sqrt(2) + 1))
= (sqrt(5) - 1) (sqrt(5) - 2 sqrt(2) + 3)) / 12
= (5 - 2 sqrt(10) + 3 sqrt(5) - sqrt(5) + 2 sqrt(2) - 3) / 12
= (1 - sqrt(10) + sqrt(5) + sqrt(2)) / 6

The top corner of the triangle is clearly (2/3,2/3) by solving the system of equations y=x and y=-2(x-1). So the triangle has base 1 and height 2/3 and thus area 1/3.

But we can also use the area formula K = rs, where r is the inradius and s is the semiperimeter. Since we have the coordinates of all three vertices, s = 1 + sqrt(5)/3 + sqrt(8)/3, and r = K/s = 2/(3+sqrt(5)+sqrt(8)).

r = 2/(3+sqrt(5)+sqrt(8)) * [3+sqrt(5)-sqrt(8)]/[3+sqrt(5)-sqrt(8)]
r = (6+4rt(2)-2rt(5)) / (12+12rt(2)) * (rt(2)-1)/(rt(2)-1)
r = (1+rt(2)+rt(5)-rt(10)) / 6

(1+sqrt(2)+sqrt(5)+sqrt(50))/12