Problem 1210

Let us first assume that triangle $ABC$ is not a right triangle – then the circumcenter $O$ of the triangle $ABC$ does not coincide with $D$, $E$, $F$ (see fig. 1). As the circumcenter is in the point of intersection of perpendicular bisectors of the sides,

$\angle AEO = 90^\circ = \angle AFO$, due to which $A, E, F, O$ are concyclic, so $O$ is located on the circumcircle of $AEF$. Analogously $O$ is also located on the circumcircles of $BFD$ and $CDE$. Therefore $O$ is the point we are looking for.

In the end let us also look at the case where $ABC$ is a right triangle – without loss of generality let $\angle ACB = 90^\circ$ (see fig. 2). The circumcircles of triangles $AEF$ and $BFD$ obviously pass through $F$. As $DF \parallel AC$ and $EF \parallel BC$ by midline property, we have $DF \perp BC$ and $EF \perp AC$. Therefore also $\angle EFD = 90^\circ$. Since $\angle DCE = 90^\circ$, the line segment $DE$ is the diameter of the circumcircle of $CDE$, due to which it also passes through $F$. Therefore $F$ is the point we are looking for.

Figure 1

Figure 2

Since $DE$, $EF$ and $FD$ are the midsegments of triangle $ABC$, triangles $AEF$, $FDB$ and $ECD$ are congruent. Therefore their circumcircles also have radii of equal length. Let that length be $r$.

Let the circumcenters of $AEF$, $BFD$ and $CDE$ be $G$, $H$ and $I$, respectively. The circumcenter of a triangle is located in the point of intersection of perpendicular bisectors of the sides, therefore $G$ is located on the perpendicular bisector of $AF$ and $H$ on the perpendicular bisector of $FB$. As triangles $AEF$ and $FDB$ are congruent, points $G$ and $H$ are also located at equal distance from $AB$, due to which the distance between $G$ and $H$ is equal to the distance between the perpendicular bisectors of $AF$ and $FB$. In conclusion

$$|GH| = \frac{1}{2}|AF| + \frac{1}{2}|FB| = \frac{1}{2}(|AF| + |FB|) = \frac{1}{2}|AB| = |AF| = |FB| = |ED|.$$

Analogously $|HI| = |BD| = |DC| = |FE|$ and $|IG| = |CE| = |EA| = |DF|$. Hence the triangle $GIH$ is congruent to triangles $AEF$, $FDB$ and $ECD$ and the radius of the circumcircle of $GIH$ is $r$. The circumcenter $X$ of triangle $GHI$ therefore satisfies $|XG| = |XH| = |XI| = r$, so $X$ is located on the circumcircles of $AEF$, $BFD$ and $CDE$.

Figure 1

Figure 2

A homothetic transformation with $A$ being the homothetic center and with scaling factor $\frac{1}{2}$ takes point $B$ to $F$ and $C$ to $E$, therefore the circumcircle of the triangle $ABC$ goes to the circumcircle of triangle $AFE$. Due to factor $\frac{1}{2}$ the circumcircle of $AFE$ passes through the circumcenter $O$ of triangle $ABC$. Analogously the circumcircles of $BFD$ and $CDE$ also pass through $O$.