Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it United States

Problem:
Let ABCABC be a triangle with ∠A=120∘\angle A = 120^\circ. The bisector of ∠A\angle A meets side BCBC at DD. Prove that
1AD=1AB+1AC \frac{1}{AD} = \frac{1}{AB} + \frac{1}{AC}

Solution

Solution:
The area of △ABC\triangle ABC is the sum of the areas of triangles ABDABD and ADCADC, so
12AB⋅AC⋅sin⁡120∘=12AB⋅AD⋅sin⁡60∘+12AD⋅AC⋅sin⁡60∘12AB⋅AC⋅32=12AB⋅AD⋅32+12AD⋅AC⋅32AB⋅AC=AB⋅AD+AD⋅AC. \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⋅AC⋅ADAB \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.