Problem
GEO-B3-M05-P016 Bound for the Brocard Angle
#16
★★★☆☆ Level 3 of 5
Let \(\varphi\) be the Brocard angle of triangle \(ABC\). Using \(\operatorname{ctg}\varphi=\operatorname{ctg}A+\operatorname{ctg}B+\operatorname{ctg}C\), prove that \(\varphi\le 30^\circ\).
Inspired by Prasolov special points geometry method
C. Hint 1. Express the sum of cotangents through sides and area.
D. Hint 2. Use \(a^2+b^2+c^2\ge 4\sqrt3 S\).
We have \(\operatorname{ctg}A=\frac{b^2+c^2-a^2}{4S}\), and cyclically. Adding gives \(\operatorname{ctg}A+\operatorname{ctg}B+\operatorname{ctg}C=\frac{a^2+b^2+c^2}{4S}\).
By Weitzenbock's inequality, \(a^2+b^2+c^2\ge 4\sqrt3 S\). Hence \(\operatorname{ctg}\varphi\ge \sqrt3\). Since \(\varphi\) is acute, \(\varphi\le 30^\circ\).
The Brocard angle formula can be proved separately by areas; here it is used as a tool.