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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

$O K R A$ is a trapezoid with $O K$ parallel to $R A$ . If $O K=12$ and $R A$ is a positive integer, how many integer values can be taken on by the length of the segment in the trapezoid, parallel to $O K$ , through the intersection of the diagonals?

Solution

Solution:

Let $R A = x$ . If the diagonals intersect at $X$ , and the segment is $P Q$ with $P$ on $K R$ , then $\triangle P K X \sim \triangle R K A$ and $\triangle O K X \sim \triangle R A X$ (by equal angles), giving $R A / P X = A K / X K = 1 + A X / X K = 1 + A R / O K = (x+12)/12$ , so $P X = 12x/(12+x)$ . Similarly $X Q = 12x/(12+x)$ also, so $P Q = 24x/(12+x) = 24 - \frac{288}{12+x}$ . This has to be an integer. $288 = 2^{5} 3^{2}$ , so it has $(5+1)(3+1) = 18$ divisors. $12 + x$ must be one of these. We also exclude the 8 divisors that don't exceed 12, so our final answer is 10.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.