2025 Caucasus MO Seniors P7

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Source: Caucasus MO

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#1 h Mar 26, 2025, 12:50 AM • 1 Y

Y by sami1618

From a point $O$ lying outside the circle $\omega$ , two tangents are drawn touching $\omega$ at points $M$ and $N$ . A point $K$ is chosen on the segment $MN$ . Let points $P$ and $Q$ be the midpoints of segments $KM$ and $OM$ respectively. The circumcircle of triangle $MPQ$ intersects $\omega$ again at point $L$ ( $L \neq M$ ). Prove that the line $LN$ passes through the centroid of triangle $KMO$ .

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#2 h Mar 30, 2025, 6:24 PM • 2 Y

Y by Maksat_B, sami1618

Let $OP\cap NL=G$ , it's enough to prove $\frac{OG}{GP} = 2$ . Let $\angle LNM= \angle LMQ= a$ and $\angle LNO= \angle LMN= b$ . Ratio lemma at $\triangle ONP$ yields $\frac{OG}{GP} = \frac{ON}{NP} \cdot \frac{\sin \angle b}{\sin \angle a}$ . We have $\frac{\sin \angle b}{\sin \angle a} = \frac{LP}{LQ}$ because $MPLQ$ cyclic. We have $ON=OM=2QM$ and by angle chasing $\triangle LQM \sim \triangle LPN$ gives us $\frac{QM}{QL} = \frac{PN}{PL}$ . And finally pu these two ratios to the ratio lemma equality, we get that $\frac{OG}{GP} = 2$ . The result follows. :cool:

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This post has been edited 3 times. Last edited by X.Luser, Mar 30, 2025, 6:40 PM

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#3 h Apr 1, 2025, 10:33 PM • 1 Y

Y by X.Luser

Let $S$ be the reflection of $N$ about $P$ and let $T$ be the midpoint of $OS$ . Since segments $MK$ and $SN$ share a midpoint the centroids of triangles $KMO$ and $NSO$ coincide. Thus it suffices to show that $LN$ passes through the centroid of $NSO$ , or that $LN$ passes through $T$ .

We first show that $T$ lies on $w$ which follows from $\angle TQM=\angle QMP=\angle ONS=\angle TPM$ . Then to show that $L$ , $N$ , and $T$ are collinear, notice $\angle PTL=\angle NML=\angle LNO=\angle PTN$ .

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