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1995 AIME Problems/Problem 9: Difference between revisions

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


== See also ==
== See also ==

Revision as of 14:16, 8 September 2012

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

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 \[\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^2x}{1-3\tan^2x}.\] 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 $a+b=\boxed{616}$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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