Library / Berkeley Math Circle Monthly Contest 5 /6 of 15

Berkeley Math Circle Monthly Contest 5

Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $ABCD$ be a quadrilateral whose diagonals are perpendicular and intersect at $P$ . Let $h_{1}, h_{2}, h_{3}, h_{4}$ be the lengths of the altitudes from $P$ to $AB, BC, CD, DA$ . Show that
$\frac{1}{h_{1}^{2}}+\frac{1}{h_{3}^{2}}=\frac{1}{h_{2}^{2}}+\frac{1}{h_{4}^{2}}$

Solution

Solution:

The area of triangle $ABP$ is equal to $h_{1} \cdot AB / 2$ and also to $AP \cdot BP / 2$ . Hence
$\frac{1}{h_{1}^{2}}=\frac{AB^{2}}{AP^{2} \cdot BP^{2}}=\frac{AP^{2}+BP^{2}}{AP^{2} \cdot BP^{2}}=\frac{1}{BP^{2}}+\frac{1}{AP^{2}}$
Applying the same transformation to all the terms of (1) yields
$\left(\frac{1}{AP^{2}}+\frac{1}{BP^{2}}\right)+\left(\frac{1}{CP^{2}}+\frac{1}{DP^{2}}\right)=\left(\frac{1}{BP^{2}}+\frac{1}{CP^{2}}\right)+\left(\frac{1}{DP^{2}}+\frac{1}{AP^{2}}\right)$
a triviality.

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.