Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $A$ , $B$ , $C$ , and $D$ be points on a circle such that $AB = 11$ and $CD = 19$ . Point $P$ is on segment $AB$ with $AP = 6$ , and $Q$ is on segment $CD$ with $CQ = 7$ . The line through $P$ and $Q$ intersects the circle at $X$ and $Y$ . If $PQ = 27$ , find $XY$ .

Solution

Solution:

Suppose $X$ , $P$ , $Q$ , $Y$ lie in that order. Let $PX = x$ and $QY = y$ . By power of a point from $P$ , $x \cdot (27 + y) = 30$ , and by power of a point from $Q$ , $y \cdot (27 + x) = 84$ . Subtracting the first from the second, $27 \cdot (y - x) = 54$ , so $y = x + 2$ . Now, $x \cdot (29 + x) = 30$ , and we find $x = 1, -30$ . Since $-30$ makes no sense, we take $x = 1$ and obtain $XY = 1 + 27 + 3 = 31$ .

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