1999 AMC 12 Problem 19

Below is the professionally curated solution for Problem 19 of the 1999 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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Concepts:isosceles triangle Pythagorean Theorem Diophantine Equation

Difficulty rating: 1810

19.

Consider all triangles $ABC$ satisfying the following conditions: $AB = AC,$ $D$ is a point on $\overline{AC}$ for which $\overline{BD} \perp \overline{AC},$ $AD$ and $CD$ are integers, and $BD^2 = 57.$ Among all such triangles, the smallest possible value of $AC$ is

$9$

$10$

$11$

$12$

$13$

Solution:

Let $AD = n$ and $CD = m.$ Since $\triangle ADB$ is right-angled at $D,$ $AB^2 = n^2 + 57.$ Also $AB = AC = m + n,$ so $(m + n)^2 = n^2 + 57,$ which simplifies to $m(m + 2n) = 57.$

The positive integer solutions are $m = 1, n = 28$ (giving $AC = 29$ ) and $m = 3, n = 8$ (giving $AC = 11$ ). The smallest possible value of $AC$ is $11.$

Thus, the correct answer is C.

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