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Berkeley Math Circle Monthly Contest 7

Geometry Difficulty 4.4 AIME Prove it United States

Problem:
Let $ABC$ be a triangle with $\angle A = 120^\circ$ . The bisector of $\angle A$ meets side $BC$ at $D$ . Prove that
$\frac{1}{AD} = \frac{1}{AB} + \frac{1}{AC}$

Solution

Solution:
The area of $\triangle ABC$ is the sum of the areas of triangles $ABD$ and $ADC$ , so
$\begin{aligned} \frac{1}{2} AB \cdot AC \cdot \sin 120^\circ & = \frac{1}{2} AB \cdot AD \cdot \sin 60^\circ + \frac{1}{2} AD \cdot AC \cdot \sin 60^\circ \\ \frac{1}{2} AB \cdot AC \cdot \frac{\sqrt{3}}{2} & = \frac{1}{2} AB \cdot AD \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} AD \cdot AC \cdot \frac{\sqrt{3}}{2} \\ AB \cdot AC & = AB \cdot AD + AD \cdot AC. \end{aligned}$
Dividing through by $AB \cdot AC \cdot AD$ gives the desired result.

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