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2018/19 AHSMC Part 1 Problem 15

Problem 15 of 16

In triangle ABCABC, DD is a point on BCBC such that ∠BAD=30∘\angle BAD = 30^\circ and ∠DAC=15∘\angle DAC = 15^\circ. If AB=32AB = 3\sqrt{2} and AC=6AC = 6, the length of ADAD is

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Solution

Since Area[ABD]+Area[ADC]=Area[ABC] \mathrm{Area}[ABD] + \mathrm{Area}[ADC] = \mathrm{Area}[ABC] then 32⋅AD⋅sin⁡30∘+AD⋅6⋅sin⁡15∘=182sin⁡45∘. 3\sqrt{2} \cdot AD \cdot \sin 30^\circ + AD \cdot 6 \cdot \sin 15^\circ = 18\sqrt{2} \sin 45^\circ. Notice that sin⁡15∘=sin⁡(45∘−30∘)=sin⁡45∘cos⁡30∘−sin⁡30∘cos⁡45∘=6−24. \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}. Hence, AD=182⋅2232⋅12+6⋅6−24=26. AD = \frac{18\sqrt{2} \cdot \frac{\sqrt{2}}{2}}{3\sqrt{2} \cdot \frac{1}{2} + 6 \cdot \frac{\sqrt{6}-\sqrt{2}}{4}} = 2\sqrt{6}. The answer is (A).

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