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: It's clear that $\angle I_{1}AI_{2}=\frac{1}{2}\angle BAX+\frac{1}{2}\angle CAX=\frac{1}{2}\angle A$ . Thus \begin{align*} AI_{1}I_{2}&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\angle I_{1}AI_{2} \\ &=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right).\end{align*} By the Law of Sines on $\triangle AI_{1}B$ \[\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}.\] Similarly, \[\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}.\] It is well known that \[\angle AI_{1}B=90+\frac{1}{2}\angle AXB~~~\text{and}~~~\angle AI_{2}C=90+\frac{1}{2}\angle AXC.\] Denote $\alpha=\frac{1}{2}\angle AXB$ and $\theta=\frac{1}{2}\angle AXC$ , with $\alpha+\theta=90^{\circ}$ . Thus $\sin\alpha=\cos\theta$ and \begin{align*}\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}\Longrightarrow\frac{AB}{\sin\left(90^{\circ}+\alpha\right)}\Longrightarrow\frac{AB}{\cos\alpha} \\\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}\Longrightarrow\frac{AC}{\sin\left(90^{\circ}+\theta\right)}\Longrightarrow\frac{AC}{\cos\theta}\Longrightarrow\frac{AC}{\sin\alpha}.\end{align*} Thus \[AI_{1}=\frac{AB\sin\left(\frac{1}{2}\angle B\right)}{\cos\alpha}~~~\text{and}~~~AI_{2}=\frac{AC\sin\left(\frac{1}{2}\angle C\right)}{\sin\alpha}\] so \begin{align*}[AI_{1}I_{2}]&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right) \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{2\sin\alpha\cos\alpha} \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{\sin(2\alpha)}.\end{align*} We intend to minimize this expression, which is equivalent to maximizing $\sin(2\alpha)$ , and that occurs when $\alpha=45^{\circ}$ , or $\angle AXB=90^{\circ}$ . Ergo, $X$ is the foot of the altitude from $A$ to $\overline{BC}$ . In that case, we intend to compute \[AB\cdot AC\cdot\sin\left(\frac{1}{2}\angle B\right)\cdot\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right).\] Recall that \[\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\cos\angle B}{2}}\] and similarly for angles $C$ and $A$ . Applying the Law of Cosines to each angle of $\triangle ABC$ gives \begin{align*}\angle B&:\cos\angle B=\frac{30^{2}+32^{2}-34^{2}}{2\cdot 30\cdot 32}=\frac{2}{5} \\ \angle C&:\cos\angle C=\frac{32^{2}+34^{2}-30^{2}}{2\cdot 32\cdot 34}=\frac{10}{17} \\ \angle A&:\cos\angle A=\frac{30^{2}+34^{2}-32^{2}}{2\cdot 30\cdot 34}=\frac{43}{85}.\end{align*} Thus \begin{align*}\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\tfrac{2}{5}}{2}}=\sqrt{\frac{3}{10}} \\ \sin\left(\frac{1}{2}\angle C\right)=\sqrt{\frac{1-\tfrac{10}{17}}{2}}=\sqrt{\frac{7}{34}} \\ \sin\left(\frac{1}{2}\angle A\right)=\sqrt{\frac{1-\tfrac{43}{85}}{2}}=\sqrt{\frac{21}{85}}.\end{align*} Thus the answer is \begin{align*} & 30\cdot 34\cdot\sqrt{\frac{3}{10}\cdot\frac{7}{34}\cdot\frac{21}{85}} \\ =&~30\cdot 34\cdot\sqrt{\frac{3}{2\cdot 5}\cdot\frac{7}{2\cdot 17}\cdot\frac{3\cdot 7}{5\cdot 17}} \\ =&~30\cdot 34\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~(2\cdot 3\cdot 5)\cdot(2\cdot 17)\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~2\cdot 3^{2}\cdot 7 \\ =&~\boxed{126}

Answer:  126