Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Circles and triangles.
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 \(a+\frac{b}{c}d^\frac{1}{2}\) where a,b,c,d are positive integers b and c are relative prime and d is not divisible by the square of any prime, find a+b+c+d.
Angles
Trigonometry
Triangles
Answer: is 41.
AIME I, 2012, Question 13
Geometry Revisited by Coxeter
In triangle ABC AO=3, BO=4, CO=5 let AB-BC=CA=s [ABC]=\(\frac{s^{2}3^\frac{1}{2}}{4}\)
\(s^{2}=3^{2}+4^{2}-2(3)(4)cosAOB\)=25-24cosAOB then [ABC]=\(\frac{25(3)^\frac{1}{2}}{4}-6(3)^\frac{1}{2}cosAOB\)
of the required form for angle AOB=150 (in degrees) then [ABC]=\(\frac{25(3)^\frac{1}{2}}{4}+9\) then a+b+c+d=25+3+4+9=41.
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