Iranian Mathematical Olympiad

Let $B^{\prime }$ and $C^{\prime }$ be the midpoints of $AB$ and $AC$ respectively. $\Gamma$ is known to be the Euler's circle (nine-point circle) of triangle $ABC$, and passes through $H^{\prime }$, the midpoint of $AH$. Thus $\Gamma$ is the circumcircle of triangle $B^{\prime }-H^{\prime }-C^{\prime }$. Now it's easy to see that $\Gamma$ is obtained when $\omega$ is scaled from center $A$ and with scale factor $2$. Because points $B$, $H$ and $C$ are homothetic points of $B^{\prime }$, $H^{\prime }$ and $C^{\prime }$.



Let $Y^{\prime },P^{\prime }$ respectively be the second intersection points of $AX,AP$ with $\omega$, where $P$ lies between $A$ and $P^{\prime }$. Let $X^{\prime }$ be the second intersection point of $XP$ with $\omega$. It suffices to show that $P$ is the midpoint of segment $X{X}^{\prime }$.

Since $\Gamma$ is the homothetic of $\omega$ with scale factor $2$, we get $PY\parallel P^{\prime }{Y}^{\prime }$ (1) and $P^{\prime }{Y}^{\prime }=2PY$ (2).

(1) implies $X{X}^{\prime }{P}^{\prime }{Y}^{\prime }$ is an isosceles trapezoid with $X{X}^{\prime }=P^{\prime }{Y}^{\prime }$ (3); Therefore $$\stackrel{(2),(3)}{⟹}X{X}^{\prime }=2XP,$$ which means $P$ is the midpoint of $X{X}^{\prime }$, hence the claim of the problem. ■