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

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$\textrm{(A)} \ 9 \qquad \textrm{(B)} \ 10 \qquad \textrm{(C)} \ 11 \qquad \textrm{(D)} \ 12 \qquad \textrm{(E)} \ 13$

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 $\frac {3 + 19}{2} = 11 \Longrightarrow \mathrm{(C)}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1999 AHSME Problems/Problem 19

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