HMMT February

Solution:

Let $I$ be the incenter of $ABC$. We claim that $I$ is the circumcenter of $DEF$.

To prove this, let the incircle touch $AB$, $BC$, and $AC$ at $X$, $Y$, and $Z$, respectively. Noting that $XB = BY = 2$, $YC = CZ = 4$, and $ZA = AX = 3$, we see that $XD = YE = ZF = 9$. Thus, since $IX = IY = IZ = r$ (where $r$ is the inradius) and $\angle IXD = \angle IYE = \angle IZF = 90^{\circ}$, we have three congruent right triangles, and so $ID = IE = IF$, as desired.

Let $s = \frac{5 + 6 + 7}{2} = 9$ be the semiperimeter. By Heron's formula, $[ABC] = \sqrt{9(9-5)(9-6)(9-7)} = 6\sqrt{6}$, so $r = \frac{[ABC]}{s} = \frac{2\sqrt{6}}{3}$. Then the area of the circumcircle of $DEF$ is

$$ID^2 \pi = (IX^2 + XD^2) \pi = (r^2 + s^2) \pi = \frac{251}{3} \pi$$

Solution:

Let $D'$ be a point on ray $CB$ beyond $B$ such that $BD' = 7$, and similarly define $E'$, $F'$. Noting that $DA = E'A$ and $AF = AF'$, we see that $DE'F'F$ is cyclic by power of a point. Similarly, $EF'D'D$ and $FD'E'E$ are cyclic. Now, note that the radical axes for the three circles circumscribing these quadrilaterals are the sides of $ABC$, which are not concurrent. Therefore, $DD'FF'EE'$ is cyclic. We can deduce that the circumcenter of this circle is $I$ in two ways: either by calculating that the midpoint of $D'E$ coincides with the foot from $I$ to $BC$, or by noticing that the perpendicular bisector of $FF'$ is $AI$. The area can then be calculated the same way as the previous solution.

$$ID^2 \pi = (IX^2 + XD^2) \pi = (r^2 + s^2) \pi = \frac{251}{3} \pi$$

Remark. The circumcircle of $DEF$ is the Conway circle of $ABC$.