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Japan Mathematical Olympiad

Geometry Difficulty 5.8 AIME, harder Prove it Japan

A square $ABCD$ is given. Points $P$ , $Q$ , $R$ , $S$ lie on the sides $AB$ , $BC$ , $CD$ , $DA$ , of this square, respectively, in such a way that the lines $PR$ and $BC$ are parallel and so are the lines $SQ$ and $AB$ . Let $Z$ be the point of intersection of the lines $PR$ and $SQ$ . If $BP = 7$ , $BQ = 6$ , $DZ = 5$ , determine the length of a side of the square $ABCD$ . Here we are denoting the length of a line segment $XY$ also by $XY$ .

Solution

Let $x$ be the length of a side of the square $ABCD$ . Then, we have $ZS = x - 7$ and $DS = x - 6$ . Applying the Pythagorean theorem to the right triangle $ZSD$ , we get $(x - 7)^2 + (x - 6)^2 = 5^2$ , from which we get $x = 3, 10$ . But from $x > BP$ we conclude that $x = 10$ , and it is easy to check that $x = 10$ does satisfy the requirement of the problem.

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