Mensa Angle Geometry Problem Solution

 

Redrawn with highlighted circles, angles

Solution:

Reconstruct the geometry as shown in the diagram above, which includes the more detailed information given below. Then draw a line from the center of each circle to the nearest  45 ^o  vertex AND to either point of tangency with the triangle (A/B, a/b).   These lines and the sides of the triangle make right triangles with angles: 22.5 ^o , 67.5 ^o, 90 ^o  (22.5 ^o  is the 45 ^o  vertex  bisected.)

Therefore, the intercepted minor arcs between the points where the circle touches the triangle (at AB  and ab) measure 135 ^o (67.5 ^o + 67.5 ^o).  

Now, draw a diameter through the center of each circle to the point of tangency with the circle's hypotenuse.  Let these diameters be AC and ac. Since they are perpendicular to the same line they are perpendicular to each other.  The measure of arcs AC and ac are both  180 ^o  so: BC = bc = AC - AB = ac - ab =

180 ^o -   135 ^o  =  45 ^o 

Now, draw a line between the centers of both circles. It passes through the point of tangency between both circles (X), which is also the vertex of <x. Note the line joining the centers of both circles intersects parallel lines AC and ac. By the alternate angle theorem, the pairs of alternate interior angles (u and v) are congruent. Hence, u + v = 180 ^o .

Note BOX and boX are both isosceles triangles. Let the interior angles of BOX be: t,  v + 45 ^o , and t. Let the interior angles of triangle boX be: s,  u + 45 ^o , and s.

We have:

 t+ v + 45 ^o  + t = 180 ^o ® 2t  + v  = 135 ^o 

s+ u + 45 ^o  + s = 180 ^o ® 2s  + u  = 135 ^o 

Adding both equations together yields:

2t + 2s + v + u =   270 ^o 

Substitute  180 ^o for v + u:   2t + 2s + 180 ^o  =   270 ^o 

Then:  2t + 2s =  90 ^o  ® t  + s  = 45 ^o 

Since the line connecting the centers of the circles (O and o) is straight:

t + x + s =  180 ^o

Substituting 45 ^o for t + s:    45 ^o + x =  180 ^o  

Then:  x =  180 ^o  - 45 ^o  = 135 ^o