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2023/24 AHSMC Part 1 Problem 16

Problem 16 of 16

In △ABC\triangle ABC with ∠A=60∘\angle A = 60^\circ, DD is a point on the segment BCBC such that ∠BAD=15∘\angle BAD = 15^\circ. If AB=4AB = 4 and AD=6AD = \sqrt{6} then ACAC is equal to

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Solution

Since Area(△ABD)+Area(△ADC)=Area(△ABC)\mathrm{Area}(\triangle ABD) + \mathrm{Area}(\triangle ADC) = \mathrm{Area}(\triangle ABC), 46 sin⁡15∘+AC⋅6 sin⁡45∘=4 AC sin⁡60∘, 4\sqrt{6}\,\sin 15^\circ + AC\cdot\sqrt{6}\,\sin 45^\circ = 4\,AC\,\sin 60^\circ, so AC=46 sin⁡15∘4sin⁡60∘−6 sin⁡45∘=46(32⋅22−12⋅22)4⋅32−6⋅22=23(3−1)3=23−2. AC = \frac{4\sqrt{6}\,\sin 15^\circ}{4\sin 60^\circ - \sqrt{6}\,\sin 45^\circ} = \frac{4\sqrt{6}\bigl(\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} - \frac{1}{2}\cdot\frac{\sqrt{2}}{2}\bigr)}{4\cdot\frac{\sqrt{3}}{2} - \sqrt{6}\cdot\frac{\sqrt{2}}{2}} = \frac{2\sqrt{3}(\sqrt{3}-1)}{\sqrt{3}} = 2\sqrt{3}-2. The answer is (C).

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