4 people solved it during the test.

Official solution: Let $\triangle PQR$ be the medial triangle of $\triangle ABC$ and $\triangle XYZ$ be the excentral triangle of $\triangle ABC$. Let $I$ be the incenter of $\triangle PQR$ and $I'$ be the incenter of $\triangle ABC$. We claim that $I$ is the radical center of the three excircles of $\triangle ABC$. By homothety, $AI' \parallel PI$. But, as $AI' \perp YZ$, so $PI \perp YZ$. Thus it suffices to show that $P$ lies on the radical axis of $B$ and $C$-excircle. But this is obvious as the length of the tangent from $B$ to the $C$-excircle is equal to the semi perimeter of $\triangle ABC$, which in turn is equal to the length of the tangent from $C$ to the $B$-excircle. Subtracting $BP=CP$ from both sides of the above equality, we get the desired result. Return to the problem at hand. By the above claim, $I$ is the incenter of $\triangle ABC$. Now, WLOG assume that $D$ lies on $\overline{CP}$. We claim that $E$ lies on $\overline{AQ}$ and $F$ lies on $\overline{BR}$. Assume to the contrary that $E$ lies on $\overline{CQ}$. Then the inradius of $\triangle PQR$ is smaller than that of $\triangle DEF$. Now, WLOG assume that $F$ lies on $\overline{AR}$. Then the inradius of $\triangle PQR$ is larger than that of $\triangle DEF$. Thus we arrive at a contradiction. Hence, $D$ lies on $\overline{CP}$, $E$ lies on $\overline{AQ}$ and $F$ lies on $\overline{BR}$. Now, As $S_{PQR} = \frac{1}{4}S_{ABC}$, it suffices to show that $S_{PQR} \leq S_{DEF}$. Now fix $E$ and $F$, and vary $D$ over $\overline{CP}$. Then for all such positions of $D$, we have that $d(P,EF) \leq d(D,EF)$ $\Rightarrow$ $S_{PEF} \leq S_{DEF}$. By a similar argument we can show that $S_{PQR} \leq S_{PQF} \leq S_{PEF}$. Thus we get the desired result. REMARK: $I$ is known as the Speiker Center of $\triangle ABC$.

WizardMath wrote: The vertices of a triangle $DEF$ lie on different sides of a triangle $ABC$. The lengths of the tangents from the incenter of $DEF$ to the excircles of $ABC$ are equal. Prove that $4S_{DEF} \ge S_{ABC}$. Note: By $S_{XYZ}$ we denote the area of triangle $XYZ$. Let $D'E'F'$ be the medial triangle of $\triangle ABC$. Let $J$ be the incenter of $\triangle D'E'F'$. Let $I_aI_bI_c$ be the excentral triangle of $\triangle ABC$. Then $I_bI_c \perp D'J$ and $D'$ bisects a common tangent between excircles $\odot(I_b), \odot(I_c)$ hence $D'J$ is their radical axis. Consequently, $J$ is the incenter of $\triangle DEF$. Note that upto cyclic cases, we have only two distinct possibilities: Case 1. Suppose $E$ and $F$ lie on segments $AE'$ and $AF'$ respectively, while $D$ lies on segment $BD'$. Note that $d(J, E'F') \le d(J, EF)$ while $d(J, DE) \ge d(J, D'E')$ contradicting that $J$ is the incenter of $\triangle D'E'F'$. Case 2. Suppose $D, E, F$ lie on segments $BD', CE', AF'$ respectively. Note that $FE$ meets ray $BC$ extended beyond $C$; so $d(D', FE) \le d(D, FE)$. We obtain the chain $$[D'E'F'] \ge [DE'F'] \ge [DEF']=[DEF]$$proving the claim.

anantmudgal09 wrote: Case 2. Suppose $D, E, F$ lie on segments $BD', CE', AF'$ respectively. Note that $F'E'$ meets ray $BC$ extended beyond $C$; so $d(M, F'E') \le d(D', F'E')$. We obtain the chain $$[D'E'F'] \ge [DE'F'] \ge [DEF']=[DEF]$$proving the claim. First of all, what's $M$? Also I think you meant that $FE$ meets ray $BC$ beyond $C$, because $F'E' \parallel BC$.