Library / Japan competition problems /4 of 38

Japan competition problems, 2007

Geometry Difficulty 4.7 AIME Prove it Japan

A quadrilateral $ABCD$ that satisfies $AB = 5$ , $BC = 7$ , $CD = 6$ is given. And $AC$ and $BD$ are perpendicular to each other. Find the length of $DA$ .

Solution

Define $P$ as the intersection point of $AC$ and $BD$ . Then from the Pythagorean theorem, $AB^2 = AP^2 + BP^2$ , $BC^2 = BP^2 + CP^2$ , $CD^2 = CP^2 + DP^2$ , $DA^2 = DP^2 + AP^2$ . Then $DA^2 = AB^2 + CD^2 - BC^2 = 5^2 + 6^2 - 7^2 = 12$ . So $DA = 2\sqrt{3}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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