A Solution to BMO1 2025, Problem 4

BMO1 2025, Problem 4

Proposed by Gerry Leversha

In an acute triangle $ABC$ with $AB>AC$ , let $M$ be the midpoint of $BC$ . A circle through $M$ is tangent to $AB$ at $B$ , and another circle through $M$ is tangent to $AC$ at $C$ . The circles meet again at $D$ . Prove that

MA\cdot MD=MB\cdot MC.

Acute triangle ABC with M the midpoint of BC. Two circles through M are tangent to AB at B and AC at C and meet again at D. The circumcircle of ABCD meets line DM again at H, the reflection of A across the perpendicular bisector of BC.

Solution

Proof

Working with directed angles modulo $180^\circ$ , the tangent–chord theorem on the two given circles, together with the collinearity of $B,M,C$ , gives

\begin{aligned} \measuredangle BDC &=\measuredangle BDM+\measuredangle MDC\\ &=\measuredangle ABM+\measuredangle MCA\\ &=\measuredangle BAC. \end{aligned}

Hence $A,B,C,D$ lie on one circle $\Omega$ .

Extend $DM$ through $M$ to meet $\Omega$ again at $H$ . Since $D,M,H$ and $B,M,C$ are collinear, the tangent–chord theorem on the circle through $B,M,D$ gives

\measuredangle BDH=\measuredangle BDM=\measuredangle ABM=\measuredangle ABC.

The first and last angles intercept the corresponding arcs $BH$ and $AC$ on the major arc $BC$ , so

BH=AC.

Since $M$ is the midpoint of $BC$ , reflection in the perpendicular bisector of $BC$ fixes $M$ and interchanges these corresponding equal chords. It therefore sends $H$ to $A$ , giving

MH=MA.

Finally, the intersecting-chords theorem at $M$ gives

MB\cdot MC=MD\cdot MH=MD\cdot MA.

This proves the required identity.

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Solution

Diagrams and complete source