Question

Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt b,$ where $a$ and $b$ are integers. Find $a+b.$

Answer

Answer: 616

Concepts:trigonometric identity isosceles triangle trigonometry

Difficulty rating: 2170

Small Hint:

Let $BM=x$ and let half the apex angle be $\alpha$

Big Hint:

Use $\tan\alpha=\frac{x}{11},$ $DM=1,$ and the triple-angle formula for tangent

Solution:

Let $BM=x$ and $\alpha=\angle BAM,$ so $\tan\alpha=\frac{x}{11}.$ Because $DM=AM-AD=1,$ symmetry gives $\angle BDC=2\arctan x,$ while $\angle BAC=2\alpha.$ Hence $\arctan x=3\alpha.$ Put $t=\tan\alpha=\frac{x}{11}.$ Then $\frac{3t-t^3}{1-3t^2}=11t,$ so $t^2=\frac{1}{4}$ and $x=\frac{11}{2}.$ Thus $BC=11$ and $AB=\frac{11\sqrt5}{2},$ making the perimeter $11+11\sqrt5=11+\sqrt{605}.$ Therefore $a+b=11+605=616.$

1995 AIME Problem 9

Answer: 616

Difficulty rating: 2170

Small Hint:

Big Hint:

Solution:

Problem 9 in Other Years