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

(→‎Solution 1 (No Trig))
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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>.
 
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>.
+
Now let <math>BD=y</math>, <math>AB=x</math>, and <math>\angle IBD = \dfrac{\angle ABD}{2} = \theta</math>.
  
−
Then <math>\mathrm{cos}{(\theta)} = \dfrac{x}{8}</math>
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Then <math>\mathrm{cos}{(\theta)} = \dfrac{y}{8}</math>
  
−
and <math>\mathrm{cos}{(2\theta)} = \dfrac{x}{y} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{x^2-32}{32}</math>.
+
and <math>\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}</math>.
  
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Cross-multiplying yields <math>32x = y(x^2-32)</math>.
+
Cross-multiplying yields <math>32y = x(y^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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Since <math>x,y>0</math>, <math>y^2-32</math> must be positive, so <math>y > 5.5</math>.
  
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Additionally, since <math>\triangle IBD</math> has hypotenuse <math>\overline{IB}</math> of length <math>8</math>, <math>BD=x < 8</math>.
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Additionally, since <math>\triangle IBD</math> has hypotenuse <math>\overline{IB}</math> of length <math>8</math>, <math>BD=y < 8</math>.
  
−
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>.
+
Therefore, given that <math>BC=2y</math> is an integer, the only possible values for <math>y</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>.
  
−
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>.
+
However, only one of these values, <math>y=6</math>, yields an integral value for <math>AB=x</math>, so we conclude that <math>y=6</math> and <math>x=\dfrac{32(6)}{(6)^2-32}=48</math>.
  
 
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>.

Revision as of 12:41, 8 June 2017

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 solving for $x$ yields $32y = x(y^2-32)$. Then, we continue as solution 2 does.

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

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

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

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

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

Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y < 8$.

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

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

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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