2015 AIME I Problems/Problem 11


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

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 2 (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 1 does.

Solution 3

Let $AB=x$, call the midpoint of $BC$ point $E$, call the point where the incircle meets $AB$ point $D$,

and let $BE=y$. We are looking for the minimum value of $2(x+y)$. $AE$ is an altitude because the triangle

is isosceles. By Pythagoras on $BEI$, the inradius is $\sqrt{64-y^2}$ and by Pythagoras on $ABE$, $AE$ is

$\sqrt{x^2-y^2}$. By equal tangents, $BE=BD=y$, so $AD=x-y$. Since $ID$ is an inradius, $ID=IE$ and using pythagoras on $ADI$ yields $AI=$$\sqrt{x^2-2xy+64}$. $ADI$ is similar to $AEB$ by $AA$, so we

can write $\frac{x-y}{\sqrt{x^2-2xy+64}}=\frac{\sqrt{x^2-y^2}}{x}$. Simplifying, $\frac{x}{\sqrt{x^2-2xy+64}}=\sqrt{\frac{x+y}{x-y}}$.

Squaring, subtracting 1 from both sides, and multiplying everything out, we get $yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2$, which turns into $32y=x(y^2-32)$. Finish as in Solution 1.

Solution 4

Angle bisectors motivate trig bash. Define angle $IBC = x$. Foot of perpendicular from $I$ to $BC$ is point $P$. $\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N$, where $N$ is an integer. Thus, $\cos(x) = \frac{N}{16}$. Via double angle, we calculate $\overline{AB}$ to be $\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}$. This is to be an integer. We can bound $N$ now, as $N > 11$ to avoid negative values and $N < 16$ due to triangle inequality. Testing, $N = 12$ works, giving $\overline{AB} = 48, \overline{BC} = 12$. Our answer is $2 * 48 + 12 = \boxed{108}$. - whatRthose

Solution 5

2015 AIME I 11.png

Let $M$ be midpoint $BC, BM = x, AB = y, \angle IBM = \alpha.$

$BI$ is the bisector of $\angle ABM$ in $\triangle ABM.$ $BI = \frac {2 xy \cos \alpha}{x+y} = 8, \cos \alpha = \frac {x}{8} \implies \frac {x^2 y}{x+y} = 32.$ \[y = \frac {32 x} {x^2 - 32}.\] $BC = 2x$ is integer, $5.5^2 < 32 \implies x \ge 6.$ $BM < BI \implies x =\{ 6, 6.5, 7, 7.5 \}.$

If $x > 6$ then $y$ is not integer. \[x = 6 \implies y = 48 \implies 2(x+y) =  \boxed{\textbf{108}}.\] vladimir.shelomovskii@gmail.com, vvsss

Video Solution



See Also

2015 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png