Let △ABC have side lengths AB=13, BC=14, and CA=15. Triangle △A′B′C′ is obtained by rotating △ABC about its circumcenter so that A′C′ is perpendicular to BC, with A′ and B not on the same side of line B′C′. Find the integer closest to the area of hexagon AA′CC′BB′.
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The idea is that we will sum the areas of the six central triangles AOA′, A′OC,COC′,C′OB,BOB′, and B′OA. Since AC and BC form an angle of C, and A′C′ is perpendicular to the side, the rotation must be 90∘−C, and so:
∠AOA′=∠COC′=∠BOB′=90∘−C
It is also known that ∠AOC=2B and ∠COB=2A and ∠BOA=2C, so:
∠A′OC=2B+C−90∘∠C′OB=2A+C−90∘∠B′OA=3C−90∘
Hence, the area of hexagon is:
[AA′CC′BB′]=21R2(3⋅sin(90∘−C)+sin(2B+C−90∘)+sin(2A+C−90∘)+sin(3C−90∘))=21R2(3cosC−cos(2B+C)−cos(2A+C)−cos3C)
Note that:
cos(2A+C)+cos(2B+C)=2cos(A+B+C)cos(A−B)=−2cos(A−B)
and so:
[AA′CC′BB′]=21R2(3cosC−cos3C+2cos(A−B))
Now, note that:
sinA=6556sinB=1312sinC=54
cosA=6533cosB=135cosC=53
Hence:
cos(A−B)=cosAcosB+sinAsinB=6533⋅135+6556⋅1312=845837
and:
cos3C=4cos3C−3cosC=4(53)3−3(53)=−125117
Now, note that by Heron's formula, the area of △ABC is 84, so:
R=4⋅Areaabc=4⋅8413⋅14⋅15=865
Therefore:
[AA′CC′BB′]=21(865)2(3⋅53+125117+2⋅845837)=1281⋅652⋅(125342+8451674)=640132⋅342+52⋅1674=320169⋅171+25⋅837=3201702−1+20925=32028899+20925=32049824=101557
The integer closest to 155.7 is 156.
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