2018 AIME I Problems/Problem 13
Problem
Let 
have side lengths 
, 
, and 
. Point 
lies in the interior of 
, and points 
and 
are the incenters of 
and 
, respectively. Find the minimum possible area of 
as varies along .
Solution
First note that ![\[\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2\]](http://latex.artofproblemsolving.com/d/8/9/d895bacce5e0371314efdba3ffe2ade30afd6bd6.png)
is a constant not depending on , so by ![$[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$](http://latex.artofproblemsolving.com/a/9/9/a994396308758994303ccaf33f2ce05d1dbe7be5.png)
it suffices to minimize 
. Let 
, 
, 
, and 
. Remark that ![\[\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.\]](http://latex.artofproblemsolving.com/3/f/3/3f3355faf6a6684f8fbdf32c3bc4246e61265a64.png)
Applying the Law of Sines to 
gives ![\[\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.\]](http://latex.artofproblemsolving.com/8/5/3/8531fde275164f46427ecf2df964a35015a8c5d9.png)
Analogously one can derive 
, and so ![\[[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,\]](http://latex.artofproblemsolving.com/1/2/1/1210e62bc0a338f092d2a8728344b66d198952dc.png)
with equality when 
, that is, when is the foot of the perpendicular from 
to . In this case the desired area is 
. To make this feasible to compute, note that ![\[\sin\frac A2=\sqrt{\frac{1+\cos A}2}=\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.\]](http://latex.artofproblemsolving.com/4/a/c/4ac7fb1dfae71b47e79d5d48311c1b0572fc4d7a.png)
Applying similar logic to 
and 
and simplifying yields a final answer of 
