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

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(→‎Solution 2 (Trig))
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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>.
 +
 +
 +
==Solution 3 (More Trig)== -ccx09 (Someone please help clean this up. Thanks)
 +
 +
Let x be the measure of <ABI. Thus, <ABI=<CBI=<ACI=<BCI=x, and x<45 degrees. Note that the angle bisector of <BAC will pass through I and since AB=AC, will also be the perpendicular bisector of BC. Call the point of intersection of segments AI and BC point M.
 +
Let BM=CM=a, and AB=AC=b. Thus, we want to minimize p=2(a+b), or minimize a+b.
 +
 +
We now consider triangle BAM. We note that cos(x)=BM/BI=a/8, and cos(2x)=a/b.
 +
We can now simplify as follows. cos(2x)=2(cos x)(cos x)-1, and since cos(x)=a/8, we have
 +
a/b = 2(a/8)^2-1 = a^2/32-1 = (a^2-32)/32. Therefore b=32a/(a^2-32).
 +
And thus a + b = a + 32a / (a^2-32). Substituting 8cos(x) for a, we obtain
 +
a + b = 8cos(x) + 32(8cos x) / ((8cos x)^2)-32).
 +
 +
Since a and b are integers, we must have cos(x) = 0, 1/8, 2/8, 3/8, 4/8, 5/8. 6/8, 7/8, or 1.
 +
However, since x<45 degrees, we must have cos(x)>=6/8. To minimize a + b, we thus let cos(x)=6/8. Plugging cos(x)=6/8 into the previous expression of a + b yields a + b = 54.
 +
This is the minimum possible value of a + b.
 +
 +
Thus, the minimum perimeter of triangle ABC is 54*2 = 108.
  
 
==See Also==
 
==See Also==

Revision as of 20:29, 18 August 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 Theorem on $\triangle BEI$ 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.


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 $2(x+y) = \boxed{108}$.


==Solution 3 (More Trig)== -ccx09 (Someone please help clean this up. Thanks)

Let x be the measure of <ABI. Thus, <ABI=<CBI=<ACI=<BCI=x, and x<45 degrees. Note that the angle bisector of <BAC will pass through I and since AB=AC, will also be the perpendicular bisector of BC. Call the point of intersection of segments AI and BC point M. Let BM=CM=a, and AB=AC=b. Thus, we want to minimize p=2(a+b), or minimize a+b.

We now consider triangle BAM. We note that cos(x)=BM/BI=a/8, and cos(2x)=a/b. We can now simplify as follows. cos(2x)=2(cos x)(cos x)-1, and since cos(x)=a/8, we have a/b = 2(a/8)^2-1 = a^2/32-1 = (a^2-32)/32. Therefore b=32a/(a^2-32). And thus a + b = a + 32a / (a^2-32). Substituting 8cos(x) for a, we obtain a + b = 8cos(x) + 32(8cos x) / ((8cos x)^2)-32).

Since a and b are integers, we must have cos(x) = 0, 1/8, 2/8, 3/8, 4/8, 5/8. 6/8, 7/8, or 1. However, since x<45 degrees, we must have cos(x)>=6/8. To minimize a + b, we thus let cos(x)=6/8. Plugging cos(x)=6/8 into the previous expression of a + b yields a + b = 54. This is the minimum possible value of a + b.

Thus, the minimum perimeter of triangle ABC is 54*2 = 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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