Harder Temple Geometry Problems
171
But ΔOKF and ΔO′K′F are congruent, and so their contributions cancel out. With the side of the equilateral triangle = 2a, the above expression amounts to 1 1 a 3a = r 3r + r (2a − 3r ) + ra. 2 2
Simplifying gives 3r 2 − 6ar + 3a 2 = 0, or (r − 3a)2 = 2a 2 , which immediately yields r = ( 3 − 2 )a .
Notice that the method of solution is essentially the same as in problem 9.
Problem 11 The answer is b=
a + 6 ac + c . 4
To solve the problem we realize that the diagram is symmetrical around the centerline and so we only need to concentrate on one side, as in figure 5.33. From the drawing, convince yourself that v + w = x + y.
c w y b
v x
a
Figure 5.33. The auxiliary lines help find b in terms of a and c.