AMC 12 Mock Exam 3 Problem 15

Problem 15 of 25

The incircle of $\triangle ABC$ is tangent to side $\overline{BC}$ at $D$ , where $BD=3$ and $DC=5$ , as shown (the figure is not necessarily drawn to scale). If $\angle BAC=60^\circ$ , what is the area of $\triangle ABC$ ?

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Solution

Equal tangent lengths: if the incircle touches $\overline{AB}$ and $\overline{AC}$ at points at distance $x$ from $A$ , then $AB=x+3$ , $AC=x+5$ , and $BC=8$ . The law of cosines with $\angle A=60^\circ$ gives $64=(x+3)^2+(x+5)^2-(x+3)(x+5)=x^2+8x+19,$ so $x^2+8x=45$ . The area is $\tfrac12\,AB\cdot AC\sin60^\circ=\tfrac{\sqrt3}{4}(x+3)(x+5)=\tfrac{\sqrt3}{4}\left(x^2+8x+15\right)=\tfrac{\sqrt3}{4}\cdot60=\boxed{15\sqrt3}.$ (Here $x=\sqrt{61}-4>0$ , so the triangle exists; there is no need to compute $x$ .)

Tagged: incircle tangent lengths with a 60-degree angle

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