AoPS - Problem

Source: Mexico National Olympiad 2021 Problem 2

Tags: geometry

jampm

11.11.2021 01:34

Let $ABC$ be a triangle with $\angle ACB > 90^{\circ}$, and let $D$ be a point on $BC$ such that $AD$ is perpendicular to $BC$. Consider the circumference $\Gamma$ with with diameter $BC$. A line $\ell$ passes through $D$ and is tangent to $\Gamma$ at $P$, cuts $AC$ at $M$ (such that $M$ is in between $A$ and $C$), and cuts the side $AB$ at $N$. Prove that $M$ is the midpoint of $DP$ if and only if $N$ is the midpoint of $AB$.

Let $AM$ intersect $\Gamma$ at $E\neq C$ and let $AD$ intersect $\Omega$ at $F\neq D$, where $\Omega$ is the circumcircle of triangle $DCE$. Let $X$ be the intersection of $\Gamma$ and segment $BC$. Note that $\angle BEC = 90^\circ$ and $\angle CEF= \angle ADC=90^\circ$ so $B, E,$ and $F$ are collinear. Then since $BC\perp AF$ and $AE\perp BF$, it follows that $C$ is the orthocentre of $\triangle ABF$, implying that $B, C,$ and $X$ are collinear. Now as $\angle ADB=90^\circ$, and $BXDF$ is cyclic, \begin{align*} NA=NB&\iff ND=NB \\ &\iff \angle NDB=\angle XBC \\ &\iff \angle NDB = \angle XFD \end{align*}which is equivalent to $MD$ being tangent to $\Omega$, or $MD = \sqrt{MC\cdot ME} = MP$ as needed.