s54 - TAV is j/2 (note that triangles SBU and SBR are...

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Solution to Problem 54 Congratulations to this week’s winner Ray Kremer Correct solutions were also received from Burkart Venzke, William Webb, Philippe Fondanaiche. Consider the following diagram. [IMAGE] 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 q and angle CAB by j so that angle SBU is q/2 and
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Unformatted text preview: 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 2 r + r ( c + a )/ b + r ( c + b )/ a = c . A little algebra (using the fact that a 2 + b 2 = c 2 ) gives that r = a b c ( a + b )( a + b + c )...
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s54 - TAV is j/2 (note that triangles SBU and SBR are...

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