# 1999 AHSME Problems/Problem 19

## Problem

Consider all triangles $ABC$ satisfying in the following conditions: $AB = AC$, $D$ is a point on $\overline{AC}$ for which $\overline{BD} \perp \overline{AC}$, $AC$ and $CD$ are integers, and $BD^{2} = 57$. Among all such triangles, the smallest possible value of $AC$ is $[asy] pair A,B,C,D; A=(5,12); B=origin; C=(10,0); D=(8.52071005917,3.55029585799); draw(A--B--C--cycle); draw(B--D); label("A",A,N); label("B",B,SW); label("C",C,SE); label("D",D,NE); [/asy]$ $\textrm{(A)} \ 9 \qquad \textrm{(B)} \ 10 \qquad \textrm{(C)} \ 11 \qquad \textrm{(D)} \ 12 \qquad \textrm{(E)} \ 13$

## Solution

Thus $AD = AC - CD$ and $AB = AC$ are integers. By the Pythagorean Theorem, $$AD^2 + 57 = AB^2 \Longrightarrow 1 \cdot 57 = 3 \cdot 19 = (AB - AD)(AB + AD).$$

Thus $AC = AB = \frac {1 + 57}{2} = 29$ or $\frac {3 + 19}{2} = 11 \Longrightarrow \mathrm{(C)}$.

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