Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Philippines

Problem:

In square $A B C D$ , $P$ and $Q$ are points on sides $C D$ and $B C$ , respectively, such that $\angle A P Q = 90^{\circ}$ . If $A P = 4$ and $P Q = 3$ , find the area of $A B C D$ .

Solution

Solution:

Note that triangles $A D P$ and $P C Q$ are similar, so $A D / P C = A P / P Q = 4 / 3$ . Let $A D = 4x$ and $P C = 3x$ . Since $A B C D$ is a square, $P D = x$ . Applying Pythagorean theorem on triangle $A D P$ , we have $x^{2} + (4x)^{2} = 16$ , so that $x^{2} = 16 / 17$ . Hence, the area of square $A B C D$ is $16 x^{2} = 256 / 17$ .

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