# 2012 AMC 12A Problems/Problem 18

## Problem

Triangle $ABC$ has $AB=27$, $AC=26$, and $BC=25$. Let $I$ be the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$? $\textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3}$

## Solution 1

Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$, $BC$ at $R$, and $AC$ at $S$. Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$(also $I$) of circle $C$. Let $x=QB$, $y=RC$ and $z=AS$. Note that $BR=x$, $SC=y$ and $AQ=z$. Hence $x+z=27$, $x+y=25$, and $z+y=26$. Subtracting the last 2 equations we have $x-z=-1$ and adding this to the first equation we have $x=13$.

By Heron's formula for the area of a triangle we have that the area of triangle $ABC$ is $\sqrt{39(14)(13)(12)}$. On the other hand the area is given by $(1/2)25r+(1/2)26r+(1/2)27r$. Then $39r=\sqrt{39(14)(13)(12)}$ so that $r^2=56$.

Since the radius of circle $O$ is perpendicular to $BC$ at $R$, we have by the pythagorean theorem $BO^2=BI^2=r^2+x^2=56+169=225$ so that $BI=\boxed{\textbf{(A) } 15}$.

## Solution 2

We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label $A$ with a mass of $25$, $B$ with $26$, and $C$ with $27$. We also label where the angle bisectors intersect the opposite side $A'$, $B'$, and $C'$ correspondingly. It follows then that point $B'$ has mass $52$. Which means that $\overline{BB'}$ is split into a $2:1$ ratio. We can then use Stewart's to find $\overline{BB'}$. So we have $25^2\frac{27}{2} + 27^2\frac{25}{2} = \frac{25 \cdot 26 \cdot 27}{4} + 26\overline{BB'}^2$. Solving we get $\overline{BB'} = \frac{45}{2}$. Plugging it in we get $\overline{BI} = 15$. Therefore the answer is $\boxed{\textbf{(A) } 15}$

-Solution by arowaaron

## Solution 3

We can use POP(Power of a point) to solve this problem. First, notice that the area of $\triangle ABC$ is $\sqrt{39(39 - 27)(39 - 26)(39 - 25)} = 78\sqrt{14}$. Therefore, using the formula that $sr = A$, where $s$ is the semi-perimeter and $r$ is the length of the inradius, we find that $r = 2\sqrt{14}$.

Draw radii to the three tangents, and let the tangent hitting $BC$ be $T_1$, the tangent hitting $AB$ be $T_2$, and the tangent hitting $AC$ be $T_3$. Let $BI = x$. By the pythagorean theorem, we know that $BT_1 = \sqrt{x^2 - 56}$. By POP, we also know that $BT_2$ is also $\sqrt{x^2 - 56}$. Because we know that $BC = 25$, we find that $CT_1 = 25 - \sqrt{x^2 - 56}$. We can rinse and repeat and find that $AT_2 = 26 - (25 - \sqrt{x^2 - 56}) = 1 + \sqrt{x^2 - 56}$. We can find $AT_2$ by essentially coming in from the other way. Since $AB = 27$, we also know that $AT_3 = 27 - \sqrt{x^2 - 56}$. By POP, we know that $AT_2 = AT_3$, so $1 + \sqrt{x^2 - 56} = 27 - \sqrt{x^2 - 56}$.

Let $\sqrt{x^2 - 56} = A$, for simplicity. We can change the equation into $1 + A = 27 - A$, which we find $A$ to be $13$. Therefore, $\sqrt{x^2 - 56} = 13$, which further implies that $x^2 - 56 = 169$. After simplifying, we find $x^2 = 225$, so $x = \boxed{\textbf{(A) } 15}$

~EricShi1685

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