Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

$O K R A$ 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?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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 $P X=12 x /(12+x)$ . Similarly $X Q=12 x /(12+x)$ also, so $P Q=24 x /(12+x)=24-\frac{288}{12+x}$ . 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: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.