National Competition

Let $H$ be the common point of $PF$ and $QD$, as can be seen in Figure 3. Since $\angle EDH$ and $\angle HFE$ are both right angles, $HE$ is a diameter of the incircle of $ABC$. Now let $X$ denote the common point of $EH$ and $PQ$. We see that $H$ is the orthocenter of the triangle $EPQ$, and $X$, $D$ and $F$ are the feet of the altitudes in this triangle. The incenter $I$ of $ABC$ is also the mid-point of an altitude segment. It follows that points $I$, $F$, $X$ and $D$ all lie on the nine-point circle of $EPQ$.

Because of the right angles in $F$ and $D$, we know that $I$, $F$, $D$ and $B$ lie on a common circle. This circle is the nine-point circle of $EPQ$. For the same reason, $B$ is the diametrically opposed point to $I$ on the nine-point circle of $EPQ$.

It is well known that the mid-point of each altitude segment lies diametrically opposed to the mid-point of the corresponding side of the triangle. (Note the right angle in $X$.) We therefore see that $B$ must be the mid-point of $PQ$, as we had set out to show.

(Sara Kropf) ☐

Figure 3: Problem 2