Let $ABC$ be a triangle with $\angle ACB=90$ and $AC>BC.$ Let $\Omega$ be the circumcircle of $ABC.$ Point $ D $ is the midpoint of small arc $AC$ of $\Omega.$ Point $ M $ is symmetric with $ A$ with respect to $D.$ Point $ N$ is the midpoint of $MC.$ Line $AN$ intersects $\Omega$ in point $ P $ and line $BP$ intersects line $DN$ in point $Q.$ Prove that line $QM$ passes through the midpoint of $AC.$
Problem
Source: Moldova TST 2023
Tags: geometry
TheMathBob
09.04.2023 16:06
$1.$ See that $DN \parallel AC$, so if $QM$ passes through the midpoint of $AC$, then $Q$ is the midpoint $DN$. $2.$ $M,C,B$ are collinear, because $AD=DC=DM$, so $\angle ACM = 90^\circ = \angle ACB$. $3.$ $DN$ is tangent to $\Omega$, because if $O$ is the center of $\Omega$, then $DN \parallel AC \perp DO$, because $D$ is the midpoint of $AC$. $4.$ $(BPN)$ is tangent to $DN$ at $N$, because $\angle NPB = 90^\circ$, so the center of $(BPN)$ is on $BN$ which is perpendicular to $DN$. $5.$ So the radical axis of $\Omega$ and $(BPN)$ is $BP$ and their common tangent is $DN$, so $DQ^2 = NQ^2$, hence $Q$ is the midpoint of $DN$, which from $1.$ gives us what we wanted.
Orestis_Lignos
11.04.2023 10:30
It suffices to prove that $Q$ is the midpoint of $ND$. Let $K$ be the midpoint of $BC$. By the Ratio Lemma, this reduces to $\dfrac{\sin \angle NBQ}{\sin \angle QBD}=\dfrac{BD}{BN}$. Note that $\dfrac{\sin \angle NBQ}{\sin \angle QBD}=\dfrac{\sin \angle CBN}{\sin \angle NAD}=\dfrac{AD}{ND}=\dfrac{2AD}{AC}=\dfrac{2\sin \dfrac{\angle B}{2}}{\sin \angle B}=\dfrac{1}{\cos \dfrac{\angle B}{2}},$ and $\dfrac{BD}{BN}=\dfrac{BD}{BC+KD}=\dfrac{1}{\dfrac{\cos \angle B}{\cos \dfrac{\angle B}{2}}+\dfrac{\sin \dfrac{\angle B}{2}}{\cot \dfrac{\angle B}{2}}}=\dfrac{\cos \dfrac{\angle B}{2}}{\cos \angle B+\sin^2 \dfrac{\angle B}{2}}=\dfrac{\cos \dfrac{\angle B}{2}}{\cos^2 \dfrac{\angle B}{2}}=\dfrac{1}{\cos \dfrac{\angle B}{2}},$ and so the two fractions are equal, as desired.