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Difference between revisions of "2015 AIME I Problems/Problem 11"

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==Problem==
 
==Problem==
 
Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>.
 
Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>.
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 +
==Solution==
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Let <math>D</math> be the midpoint of <math>\overline{BC}</math>. Then by SAS Congruence, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle ADB = \angle ADC = 90^o</math>.
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Now let <math>BD=x</math>, <math>AB=y</math>, and <math>\angle IBD = \dfrac{\angle ABD}{2} = \theta</math>.
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Then <math>\mathrm{cos}{(\theta)} = \dfrac{x}{8}</math>
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and <math>\mathrm{cos}{(2\theta)} = \dfrac{x}{y} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{x^2-32}{32} \Rightarrow 32x = y(x^2-32)</math>.
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Since <math>x,y>0</math>, <math>x^2-32</math> must be positive, so <math>x > 5.5</math>.
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Additionally, since <math>\triangle IBD</math> is a right triangle with hypotenuse <math>\overline{IB}</math> of length <math>8</math>, <math>BD=x < 8</math>.
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Therefore, given that <math>BC=2x</math> is an integer, the only possible values for <math>x</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>.
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However, only one of these values, <math>x=6</math> yields an integral value for <math>AB=y</math>, so we conclude that <math>x=6</math> and <math>y=\dfrac{32(6)}{(6)^2-32}=48</math>.
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Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2015|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2015|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:12, 20 March 2015

Problem

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$.

Solution

Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.

Now let $BD=x$, $AB=y$, and $\angle IBD = \dfrac{\angle ABD}{2} = \theta$.

Then $\mathrm{cos}{(\theta)} = \dfrac{x}{8}$

and $\mathrm{cos}{(2\theta)} = \dfrac{x}{y} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{x^2-32}{32} \Rightarrow 32x = y(x^2-32)$.

Since $x,y>0$, $x^2-32$ must be positive, so $x > 5.5$.

Additionally, since $\triangle IBD$ is a right triangle with hypotenuse $\overline{IB}$ of length $8$, $BD=x < 8$.

Therefore, given that $BC=2x$ is an integer, the only possible values for $x$ are $6$, $6.5$, $7$, and $7.5$.

However, only one of these values, $x=6$ yields an integral value for $AB=y$, so we conclude that $x=6$ and $y=\dfrac{32(6)}{(6)^2-32}=48$.

Thus the perimeter of $\triangle ABC$ must be $2(x+y) = \boxed{108}$.

See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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