Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 6.6 National olympiad Prove it Austria

Let $ABC$ be a triangle. Its incircle meets the sides $BC$ , $CA$ and $AB$ in the points $D$ , $E$ and $F$ , respectively. Let $P$ denote the intersection point of $ED$ and the line perpendicular to $EF$ and passing through $F$ , and similarly let $Q$ denote the intersection point of $EF$ and the line perpendicular to $ED$ and passing through $D$ .

Prove that $B$ is the mid-point of the segment $PQ$ .

Solution

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

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