HMMT November, 2013

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Let $ABC$ be a triangle and $D$ a point on $BC$ such that $AB = \sqrt{2}$ , $AC = \sqrt{3}$ , $\angle BAD = 30^{\circ}$ , and $\angle CAD = 45^{\circ}$ . Find $AD$ .

Solution

Solution:

$\boxed{\dfrac{\sqrt{6}}{2}}$ OR $\dfrac{\sqrt{3}}{\sqrt{2}}$

Note that $[BAD] + [CAD] = [ABC]$ . If $\alpha_1 = \angle BAD$ , $\alpha_2 = \angle CAD$ , then we deduce
$\frac{\sin(\alpha_1 + \alpha_2)}{AD} = \frac{\sin \alpha_1}{AC} + \frac{\sin \alpha_2}{AB}$
upon division by $AB \cdot AC \cdot AD$ . Now
$AD = \frac{\sin(30^{\circ} + 45^{\circ})}{\frac{\sin 30^{\circ}}{\sqrt{3}} + \frac{\sin 45^{\circ}}{\sqrt{2}}}$
But $\sin(30^{\circ} + 45^{\circ}) = \sin 30^{\circ} \cos 45^{\circ} + \sin 45^{\circ} \cos 30^{\circ} = \sin 30^{\circ} \frac{1}{\sqrt{2}} + \sin 45^{\circ} \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}\left(\frac{\sin 30^{\circ}}{\sqrt{3}} + \frac{\sin 45^{\circ}}{\sqrt{2}}\right)$ , so our answer is $\boxed{\frac{\sqrt{6}}{2}}$ .

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