Congratulations to this week's winner
Ray Kremer
Correct solutions were also received from Burkart Venzke, William Webb,
Philippe Fondanaiche.
Consider the following diagram.
The triangle has vertices at A, B, C,
with sides of length a, b, c. The centers of
the circles are at U, V, with points of tangency at S,
T, respectively. Label angle ABC by
qand angle CAB by
j so that angle SBU is
q/2 and TAV is
j/2 (note that triangles SBU and
SBR are congruent). Using a half angle formula and the labels
on the graph this gives
r/x = tan(q/2)
= sin q/(1 + cos q),
and x = r (c + a) /
b
r/y = tan(j/2)
= sin j/(1 + cos j),
and y = r (c + b) / a.
Therefore
2r + r(c + a)/b
+ r(c + b)/a = c.
A little algebra (using the fact that a2
+ b2 = c2) gives that
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