# 1995 AIME Problems/Problem 9

## Problem

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 $a+b.$ $[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("11",(1,1.63),W); dot((1,3),ds); label("A",(1,3),N); dot((0,0),ds); label("B",(0,0),SW); dot((2,0),ds); label("C",(2,0),SE); dot((1,0),ds); label("M",(1,0),S); dot((1,0.7),ds); label("D",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

## Solution 1

Let $x=\angle CAM$, so $3x=\angle CDM$. Then, $\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11$. Expanding $\tan 3x$ using the angle sum identity gives $$\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.$$ Thus, $\frac{3-\tan^2x}{1-3\tan^2x}=11$. Solving, we get $\tan x= \frac 12$. Hence, $CM=\frac{11}2$ and $AC= \frac{11\sqrt{5}}2$ by the Pythagorean Theorem. The total perimeter is $2(AC + CM) = \sqrt{605}+11$. The answer is thus $a+b=\boxed{616}$.

## Solution 2

In a similar fashion, we encode the angles as complex numbers, so if $BM=x$, then $\angle BAD=\text{Arg}(11+xi)$ and $\angle BDM=\text{Arg}(1+xi)$. So we need only find $x$ such that $\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)$. This will happen when $\frac{363x-x^3}{1331-33x^2}=x$, which simplifies to $121x-4x^3=0$. Therefore, $x=\frac{11}{2}$. By the Pythagorean Theorem, $AB=\frac{11\sqrt{5}}{2}$, so the perimeter is $11+11\sqrt{5}=11+\sqrt{605}$, giving us our answer, $\boxed{616}$.

## Solution 3

Let $\angle BAD=\alpha$, so $\angle BDM=3\alpha$, $\angle BDA=180-3\alpha$, and thus $\angle ABD=2\alpha.$ We can then draw the angle bisector of $\angle ABD$, and let it intersect $\overline{AM}$ at $E.$ Since $\angle BAE=\angle ABE$, $AE=BE.$ Let $AE=x$. Then we see by the Pythagorean Theorem, $BM=\sqrt{BM^2-ME^2}=\sqrt{x^2-(11-x)^2}=\sqrt{22x-121}$, $BD=\sqrt{BM^2+1}=\sqrt{22x-120}$, $BA=\sqrt{BM^2+121}=\sqrt{22x}$, and $DE=10-x.$ By the angle bisector theorem, $BA/BD=EA/ED.$ Substituting in what we know for the lengths of those segments, we see that $$\frac{\sqrt{22x}}{\sqrt{22x-120}}=\frac{x}{10-x}.$$ multiplying by both denominators and squaring both sides yields $$22x(10-x)^2=x^2(22x-120)$$ which simplifies to $x=\frac{55}{8}.$ Substituting this in for x in the equations for $BA$ and $BM$ yields $BA=\frac{\sqrt{605}}{2}$ and $BM=\frac{11}{2}.$ Thus the perimeter is $11+\sqrt{605}$, and the answer is $\boxed{616}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 