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

Revision as of 10:58, 11 July 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 $\mathrm{cos}{(2\theta)} = \dfrac{x}{y} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{x^2-32}{32}$ .

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

-john garett

@@ Line 23: / Line 23: @@
 Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.
+-john garett
 ==See Also==

2015 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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Difference between revisions of "2015 AIME I Problems/Problem 11"

Revision as of 10:58, 11 July 2015

Problem

Solution

See Also