Difference between revisions of "1995 AIME Problems/Problem 9"

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== Problem ==
 
== Problem ==
Triangle <math>\displaystyle ABC</math> is isosceles, with <math>\displaystyle AB=AC</math> and altitude <math>\displaystyle AM=11.</math>  Suppose that there is a point <math>\displaystyle D</math> on <math>\displaystyle \overline{AM}</math> with <math>\displaystyle AD=10</math> and <math>\displaystyle \angle BDC=3\angle BAC.</math>  Then the perimeter of <math>\displaystyle \triangle ABC</math> may be written in the form <math>\displaystyle a+\sqrt{b},</math> where <math>\displaystyle a</math> and <math>\displaystyle b</math> are integers.  Find <math>\displaystyle a+b.</math>
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Triangle <math>ABC</math> is [[isosceles triangle|isosceles]], with <math>AB=AC</math> and [[altitude]] <math>AM=11.</math>  Suppose that there is a point <math>D</math> on <math>\overline{AM}</math> with <math>AD=10</math> and <math>\angle BDC=3\angle BAC.</math>  Then the perimeter of <math>\triangle ABC</math> may be written in the form <math>a+\sqrt{b},</math> where <math>a</math> and <math>b</math> are integers.  Find <math>a+b.</math>
  
[[Image:AIME_1995_Problem_9.png]]
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<asy>
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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;
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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);
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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);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy>
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<!-- [[Image:AIME_1995_Problem_9.png|center]] -->
  
== Solution ==
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== Solution 1 ==
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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>
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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>\angle CAM=x</math>, <math>\angle CDM=3x</math>. Then, <math>(tan 3x)/(tan x)=(CM/1)/(CM/11)=11</math>. Expanding tan 3x using the angle sum formula gives <math>tan 3x=(3tan x-tan^3x)/(1-3tan^2x)</math>. Thus, <math>(3-tan^2x)/(1-3tan^2x)=11</math>. Solving, we get <math>tan x=1/2</math>. Hence, <math>CM=11/2</math> and AC=<math>11\sqrt{5}/2</math> by Pythag. The total perimeter is double the sum of these, which is <math>\sqrt{605}+11</math>. The answer is then <math>616</math>.
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== Solution 2 ==
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In a similar fashion, we encode the angles as complex numbers, so if <math>BM=x</math>, then <math>\angle BAD=\text{Arg}(11+xi)</math> and <math>\angle BDM=\text{Arg}(1+xi)</math>. So we need only find <math>x</math> such that <math>\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)</math>. This will happen when <math>\frac{363x-x^3}{1331-33x^2}=x</math>,  which simplifies to <math>121x-4x^3=0</math>. Therefore, <math>x=\frac{11}{2}</math>. By the Pythagorean Theorem, <math>AB=\frac{11\sqrt{5}}{2}</math>, so the perimeter is <math>11+11\sqrt{5}=11+\sqrt{605}</math>, giving us our answer, <math>\boxed{616}</math>.
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== Solution 3 ==
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Let <math>\angle BAD=\alpha</math>, so <math>\angle BDM=3\alpha</math>, <math>\angle BDA=180-3\alpha</math>, and thus <math>\angle ABD=2\alpha.</math> We can then draw the angle bisector of <math>\angle ABD</math>, and let it intersect <math>\overline{AM}</math> at <math>E.</math> Since <math>\angle BAE=\angle ABE</math>, <math>AE=BE.</math> Let <math>AE=x</math>. Then we see by the Pythagorean Theorem, <math>BM=\sqrt{BM^2-ME^2}=\sqrt{x^2-(11-x)^2}=\sqrt{22x-121}</math>, <math>BD=\sqrt{BM^2+1}=\sqrt{22x-120}</math>, <math>BA=\sqrt{BM^2+121}=\sqrt{22x}</math>, and <math>DE=10-x.</math> By the angle bisector theorem, <math>BA/BD=EA/ED.</math> Substituting in what we know for the lengths of those segments, we see that <cmath>\frac{\sqrt{22x}}{\sqrt{22x-120}}=\frac{x}{10-x}.</cmath> multiplying by both denominators and squaring both sides yields <cmath>22x(10-x)^2=x^2(22x-120)</cmath> which simplifies to <math>x=\frac{55}{8}.</math> Substituting this in for x in the equations for <math>BA</math> and <math>BM</math> yields <math>BA=\frac{\sqrt{605}}{2}</math> and <math>BM=\frac{11}{2}.</math> Thus the perimeter is <math>11+\sqrt{605}</math>, and the answer is <math>\boxed{616}</math>.
  
 
== See also ==
 
== See also ==
* [[1995_AIME_Problems/Problem_8|Previous Problem]]
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{{AIME box|year=1995|num-b=8|num-a=10}}
* [[1995_AIME_Problems/Problem_10|Next Problem]]
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* [[1995 AIME Problems]]
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Trigonometry Problems]]
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{{MAA Notice}}

Revision as of 00:51, 28 November 2015

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 $\boxed{616}$.

See also

1995 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AIME Problems and Solutions

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