[Geometry]
Problem: In triangle $A B C, A B=B C$. A trisector of $\angle B$ intersects $\overline{A C}$ at $D$. If $A B, A C$, and $B D$ are integers and $A B-B D=7$, compute $A C$.

Solution: Let $E$ be the point where the other trisector of $\angle B$ intersects side $\overline{A C}$. Let $A B=B C=a$, and let $B D=B E=d$. Draw $X$ on $\overline{B C}$ so that $B X=d$. Then $C X=7$.

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The placement of point $X$ guarantees that $\triangle B E X \cong \triangle B D E$ by Side-Angle-Side. Therefore $\angle B X E \cong \angle B E X \cong \angle B D E$, and so $\angle C X E \cong \angle A D B \cong \angle C E B$. By Angle-Angle, $\triangle C E X \sim \triangle C B E$. Let $E X=c$ and $E C=x$. Then comparing ratios of corresponding sides yields

$$
\frac{c}{d}=\frac{7}{x}=\frac{x}{d+7}
$$

Using the right proportion, $x^{2}=7(d+7)$. Because $d$ is an integer, $x^{2}$ is an integer, so either $x$ is an integer or irrational. The following argument shows that $x$ cannot be irrational. Applying the Angle Bisector Theorem to $\triangle B C D$ yields $D E=c=\frac{d}{d+7} \cdot x$. Then $A C=2 x+c=$ $x\left(2+\frac{d}{d+7}\right)$. Because the expression $\left(2+\frac{d}{d+7}\right)$ is rational, $A C$ will not be an integer if $x$ is irrational.

Hence $x$ is an integer, and because $x^{2}$ is divisible by $7, x$ must also be divisible by 7 . Let $x=7 k$ so that $d=c k$. Rewrite the original proportion using $7 k$ for $x$ and $c k$ for $d$ :

$$
\begin{aligned}
\frac{c}{d} & =\frac{x}{d+7} \\
\frac{c}{c k} & =\frac{7 k}{c k+7} \\
7 k^{2} & =c k+7 \\
7 k & =c+\frac{7}{k} .
\end{aligned}
$$



Because the left side of this last equation represents an integer, $7 / k$ must be an integer, so either $k=1$ or $k=7$. The value $k=1$ gives the extraneous solution $c=0$. So $k=7$, from which $c=48$. Then $d=336$ and $A C=2 x+c=2 \cdot 49+48=\mathbf{1 4 6}$.

Final answer: 146