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AMC 12 Mock Exam 5 Problem 12

Problem 12 of 25

In triangle ABCABC, ∠BAC=120∘\angle BAC=120^\circ, AB=3AB=3, and AC=6AC=6. The bisector of ∠BAC\angle BAC meets side BC‾\overline{BC} at DD, as shown (the figure is not necessarily drawn to scale). What is ADAD?

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The bisector splits ∠BAC\angle BAC into two 60∘60^\circ angles, and [ABD]+[ADC]=[ABC][ABD]+[ADC]=[ABC]: 12⋅3⋅ADsin⁡60∘+12⋅6⋅ADsin⁡60∘=12⋅3⋅6sin⁡120∘.\tfrac12\cdot3\cdot AD\sin60^\circ+\tfrac12\cdot6\cdot AD\sin60^\circ=\tfrac12\cdot3\cdot6\sin120^\circ. Since sin⁡60∘=sin⁡120∘\sin60^\circ=\sin120^\circ, this gives 9 AD=189\,AD=18, so AD=2AD=\boxed{2}.

Tagged: angle bisector length via areas