Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

$ABCD$ is a convex quadrilateral. $P$ , $Q$ are points on the sides $AD$ , $BC$ respectively such that $AP / PD = BQ / QC = AB / CD$ . Show that the angle between the lines $PQ$ and $AB$ equals the angle between the lines $PQ$ and $CD$ .

Solution

Solution:

If $AB$ is parallel to $CD$ , then it is obvious that $PQ$ is parallel to both. So assume $AB$ and $CD$ meet at $O$ . Take $O$ as the origin for vectors. Let $\mathbf{e}$ be a unit vector in the direction $OA$ and $\mathbf{f}$ a unit vector in the direction $OC$ . Take the vector $OA$ to be $a\mathbf{e}$ , $OB$ to be $b\mathbf{e}$ , $OC$ to be $c\mathbf{f}$ , and $OD$ to be $d\mathbf{f}$ . Then $OP$ is $((d-c)a\mathbf{e} + (a-b)d\mathbf{f})/(d-c+a-b)$ and $OQ$ is $((d-c)b\mathbf{e} + (a-b)c\mathbf{f})/(d-c+a-b)$ . Hence $PQ$ is $(c-d)(a-b)(\mathbf{e}+\mathbf{f})/(d-c+a-b)$ . But $\mathbf{e}$ and $\mathbf{f}$ are unit vectors, so $\mathbf{e}+\mathbf{f}$ makes the same angle with each of them and hence $PQ$ makes the same angle with $AB$ and $CD$ .

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