Let $\triangle ABC$ be an isosceles triangle such that $AB=AC$. Let $\omega$ be a circle centered at $A$ with a radius strictly less than $AB$. Draw a tangent from $B$ to $\omega$ at $P$, and draw a tangent from $C$ to $\omega$ at $Q$. Suppose that the line $PQ$ intersects the line $BC$ at point $M$. Prove that $M$ is the midpoint of $BC$.
Problem
Source: 2021 Thailand MO P1
Tags: geometry
jasperE3
29.11.2021 06:06
Some TMO problems were posted already.
puntre
02.12.2021 05:10
Let $M'$ be the midpoint $\overline{BC}$. One might see that $A,Q,M,C$ and $A,P,B,M$ are concyclic. Then we wish to show that $\angle BMP=\angle BMQ$. Note that $\triangle APB\cong\triangle AQC$, So\[\angle BM'P=\angle BAP=\angle CAQ=\angle BM'Q.\]Thus $M'=M$ and We are done.
guptaamitu1
19.06.2022 06:49
Let $BP_0,CQ_0$ be the other tangents from $B,C$ to $\omega$, and $\ell$ be the perpendicular bisector of segment $BC$. Let $\mathcal R$ denote reflection in $\ell$. Note $\mathcal R$ swaps, $$ A \to A ~,~ B \to C ~,~ P \to Q_0 ~,~ Q \to P_0 $$Let $M' = PQ \cap P_0Q_0$. Note $\mathcal R$ fixes $M'$, so $M' \in \ell$. It suffices to show $M' \in BC$. This is true since points (we are looking at circle $\omega$) $$ B = PP \cap P_0P_0 ~,~ C = QQ \cap Q_0Q_0 ~,~ M' = PQ \cap P_0Q_0 $$are collinear (they all lie on polar of $PP_0 \cap QQ_0$ due to Brokard's Theorem). $\blacksquare$