, 2018

Solution:

Let $AB = z$, $BC = x$, $CA = y$. Let $X$, $Y$, $Z$, $O$, $N$ be the circumcenter of $AEF$, $BFD$, $CDE$, $ABC$, $DEF$ respectively. Note that $N$ is the nine-point center of $ABC$, and $X$, $Y$, $Z$ are the midpoints of $OA$, $OB$, $OC$ respectively, and thus $XYZ$ is the image of homothety of $ABC$ with center $O$ and ratio $\frac{1}{2}$, so this triangle has side lengths $\frac{x}{2}$, $\frac{y}{2}$, $\frac{z}{2}$. Since $NX$ perpendicularly bisects $EF$, which is parallel to $BC$ and thus $YZ$, we see that $N$ is the orthocenter of $XYZ$. Moreover, $O_{1}$ lies on $YN$ and $O_{1}X$ is perpendicular to $XY$.

To compute the area of $O_{1}O_{2}O_{3}$, it suffices to compute $\left[N O_{1} O_{2}\right] + \left[N O_{2} O_{3}\right] + \left[N O_{3} O_{1}\right]$. Note that $O_{1}X$ is parallel to $NO_{2}$, and $O_{2}Y$ is parallel to $XN$, so $\left[N O_{1} O_{2}\right] = \left[N X O_{2}\right] = [NXY]$. Similarly the other two triangles have equal area as $[NYZ]$ and $[NZX]$ respectively, so the desired area is simply the area of $[XYZ]$, which is
$$\frac{1}{4} \frac{\sqrt{(x+y+z)(x+y-z)(x-y+z)(-x+y+z)}}{4} = \frac{\sqrt{21 \cdot 9 \cdot 5 \cdot 7}}{16} = \frac{21 \sqrt{15}}{16}.$$