Problem:
Three concentric circles have radii 3,4, and 5. An equilateral triangle with one vertex on each circle has side length s. The largest possible area of the triangle can be written as a+cbd, where a,b,c, and d are positive integers, b and c are relatively prime, and d is not divisible by the square of any prime. Find a+b+c+d.
Solution:
Let O be the common center of the circles and let the triangle be ABC, where OA=3,OB=4, and OC=5. If O lies outside △ABC, then the triangle is contained in a semicircle of radius 5. Because ∠ACB=60∘ has its vertex C on the arc of the semicircle, the altitude of the equilateral triangle is limited by 5, so the side of the triangle is less than or equal to 310. If O lies inside △ABC, consider the rotation R of 60∘ centered at A that sends B to C. Let R(O)=P. Then △AOP is equilateral, R(△ABO)=△ACP, and these two triangles are congruent. In △OPC, OP=3,PC=4 and CO=5, implying that ∠OPC=90∘ and ∠APC=∠APO+∠OPC=150∘. Applying the Law of Cosines to △APC gives s2=AC2=AP2+PC2−2AC⋅PCcos∠APC=32+42+3⋅43=25+123. The area is (25+123)⋅43=9+4253, and the requested sum is 9+25+4+3=41.
OR
Place △ABC in a coordinate plane with A=(−2s,0),B=(2s,0), and C=(0,23s). The point O=(x,y) satisfies (x+2s)2+y2=9,(x−2s)2+y2=16, and x2+(y−23s)2=25. Subtracting the second equation from the first gives x=−2s7, and subtracting the third from the first gives sx+3sy=−16+2s2, from which y=23ss2−25. Then
9=OA2=(−2s7+2s)2+(23ss2−25)2
from which s4−50s2+193=0, so s2=25+123, and the result follows.