Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 4.3 AIME Prove it Turkey

Let $O$ be the circumcenter of an acute triangle $ABC$ . A line perpendicular to $AO$ intersects the line segments $[AC]$ and $[AB]$ at $D$ and $E$ , respectively. Let $K$ be a point on $[BC]$ which is not on the line $AO$ . The line $AK$ intersects the circumcircle of triangle $ADE$ again at $L$ . Let $M$ be the point symmetric to $A$ with respect to the line $DE$ . Show that the points $K$ , $L$ , $M$ , $O$ are concyclic.

Solution

Let $AM$ and $DE$ intersect at $N$ , and $F$ be the foot of the perpendicular line from $O$ to $AC$ . Since $AO \perp DE$ and $\angle BAO = 90^\circ - \angle ACB$ we get $\angle ALD = \angle AED = \angle ACB$ and hence the points $K$ , $C$ , $D$ , $L$ are concyclic. Using power equations we obtain
$AL \cdot AK = AD \cdot AC. \qquad (1)$

Since $F$ is the midpoint of $[AC]$ and $N$ is the midpoint of $[AM]$ , we get
$AC = 2AF \quad \text{and} \quad AM = 2AN. \qquad (2)$

On the other hand, $\angle DNO = \angle OFD = 90^\circ$ implies that the points $D$ , $N$ , $O$ , $F$ are also concyclic. Using power equations again, we get
$AN \cdot AO = AD \cdot AF. \qquad (3)$
Using (1), (2) and (3), we conclude that
$AL \cdot AK = AD \cdot AC = 2AF \cdot AD = 2AN \cdot AO = AM \cdot AO$
which shows that the points $K$ , $L$ , $M$ , $O$ are concyclic.

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