2018 AIME I Problems/Problem 13

Problem

Let $\triangle ABC$ have side lengths $AB=30$ , $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ .

Solution

First note that $\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2$ is a constant not depending on $X$ , so by $[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$ . Let $a = BC$ , $b = AC$ , $c = AB$ , and $\alpha = \angle AXB$ . Remark that $\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.$ Applying the Law of Sines to $\triangle ABI_1$ gives $\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.$ Analogously one can derive $AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}$ , and so $[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,$ with equality when $\alpha = 90^\circ$ , that is, when $X$ is the foot of the perpendicular from $A$ to $\overline{BC}$ . In this case the desired area is $bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2$ . To make this feasible to compute, note that $\sin\frac A2=\sqrt{\frac{1+\cos A}2}=\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.$ Applying similar logic to $\sin \tfrac B2$ and $\sin\tfrac C2$ and simplifying yields a final answer of $\begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}.\end{align*}$

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2018 AIME I Problems/Problem 13

Problem

Solution