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

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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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==Solution 1 (No Trig)==
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Let <math>AB=x</math> and the foot of the altitude from <math>A</math> to <math>BC</math> be point <math>E</math> and <math>BE=y</math>. Since ABC is isosceles, <math>I</math> is on <math>AE</math>. By pythagorean theorem, <math>AE=\sqrt{x^2+y^2}</math>. Let <math>IE=a</math> and <math>IA=b</math>. By angle bisector theorem, <math>\frac{y}{a}=\frac{x}{b}</math>. Also, <math>a+b=\sqrt{x^2+y^2}</math>. Solving for <math>a</math>, we get <math>a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}</math>. Then, using pythagorean on <math>BEI</math> we have <math>y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64</math>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64</math>. Adding 1 to the fraction and simplifying, we have <math>\frac{y^2x(x+y)}{(x+y)^2}=32</math>. Crossing out the <math>x+y</math> and multiplying the denominator yields <math>32y = x(y^2-32)</math>. Then, we continue as solution 2.
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Note: In solution 2, the variables for <math>x</math> and <math>y</math> are switched.
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==Solution 2 (Trig)==
  
 
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>.
 
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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Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.
 
Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.
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-john garett
 
  
 
==See Also==
 
==See Also==

Revision as of 00:31, 12 September 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 1 (No Trig)

Let $AB=x$ and the foot of the altitude from $A$ to $BC$ be point $E$ and $BE=y$. Since ABC is isosceles, $I$ is on $AE$. By pythagorean theorem, $AE=\sqrt{x^2+y^2}$. Let $IE=a$ and $IA=b$. By angle bisector theorem, $\frac{y}{a}=\frac{x}{b}$. Also, $a+b=\sqrt{x^2+y^2}$. Solving for $a$, we get $a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}$. Then, using pythagorean on $BEI$ we have $y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64$. Simplifying, we have $y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64$. Factoring out the $y^2$, we have $y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64$. Adding 1 to the fraction and simplifying, we have $\frac{y^2x(x+y)}{(x+y)^2}=32$. Crossing out the $x+y$ and multiplying the denominator yields $32y = x(y^2-32)$. Then, we continue as solution 2.

Note: In solution 2, the variables for $x$ and $y$ are switched.

Solution 2 (Trig)

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

Cross-multiplying yields $32x = y(x^2-32)$.

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

Additionally, since $\triangle IBD$ has 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
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