Track / Stage 8 / 23 of 180 #1723 of 1964

Problem 1723

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.1 Prove it Balkan Mathematical Olympiad Shortlist · Balkan Mathematical Olympiad

Let $ABCD$ be a trapezium inscribed in a circle $k$ with diameter $AB$ . A circle with center $B$ and radius $BE$ , where $E$ is the intersection point of the diagonals $AC$ and $BD$ meets $k$ at points $K$ and $L$ . If the line, perpendicular to $BD$ at $E$ , intersects $CD$ at $M$ , prove that $KM \perp DL$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since $AB \parallel CD$ , we have that $ABCD$ is isosceles trapezium. Let $O$ be the center of $k$ and $EM$ meets $AB$ at point $Q$ . Then, from the right angled triangle $BEQ$ , we have $BE^2 = BO \cdot BQ$ . Since $BE = BK$ , we get $BK^2 = BO \cdot BQ$ (1). Suppose that $KL$ meets $AB$ at $P$ . Then, from the right angled triangle $BAK$ , we have $BK^2 = BP \cdot BA$ (2).

From (1) and (2) we get $\dfrac{BP}{BQ} = \dfrac{BO}{BA} = \dfrac{1}{2}$ , and therefore $P$ is the midpoint of $BQ$ (3). However, $DM \parallel AQ$ and $MQ \parallel AD$ (both are perpendicular to $DC$ ). Hence, $AQMD$ is parallelogram and thus $MQ = AD = BC$ . We conclude that $QBCM$ is isosceles trapezium. It follows from (3) that $KL$ is the perpendicular bisector of $BQ$ and $CM$ , that is, $M$ is symmetric to $C$ with respect to $KL$ . Finally, we get that $M$ is the orthocenter of the triangle $DLK$ by using the well-known result that the reflection of the orthocenter of a triangle to every side belongs to the circumcircle of the triangle and vice versa.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.