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


But try the problem first…

Answer: is 41.

Source
Suggested Reading

AIME I, 2012, Question 13

Geometry Revisited by Coxeter

Try with Hints


First hint

In triangle ABC AO=3, BO=4, CO=5 let AB-BC=CA=s [ABC]=\(\frac{s^{2}3^\frac{1}{2}}{4}\)

Second Hint

\(s^{2}=3^{2}+4^{2}-2(3)(4)cosAOB\)=25-24cosAOB then [ABC]=\(\frac{25(3)^\frac{1}{2}}{4}-6(3)^\frac{1}{2}cosAOB\)

Final Step

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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