Track / Stage 4 / 241 of 340 #501 of 1964

Problem 501

AMC 12 late, AIME early

Geometry Difficulty 4.8 Find the answer

If $S_{\triangle A B C}=\frac{1}{4} b^{2} \tan B$ , where $b$ is the side opposite to $\angle B$ , then $\frac{\cot B}{\cot A+\cot C}=$ $\qquad$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

$\begin{array}{l}\text { II. 1. } \frac{1}{2} . \\ S_{\triangle A B C}=\frac{1}{4} b^{2} \tan B \Leftrightarrow \frac{1}{2} a c \sin B=\frac{1}{4} b^{2} \tan B \\ \Leftrightarrow 2 \cot B=\frac{\sin (A+C)}{\sin A \cdot \sin C}=\cot A+\cot C \\ \Leftrightarrow \frac{\cot B}{\cot A+\cot C}=\frac{1}{2} .\end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.