Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Circles and triangles.

## Circles and triangles – AIME I, 2012

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.

- is 107
- is 41
- is 840
- cannot be determined from the given information

**Key Concepts**

Angles

Trigonometry

Triangles

## Check the Answer

Answer: is 41.

AIME I, 2012, Question 13

Geometry Revisited by Coxeter

## Try with Hints

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.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s