Let $A$ and $B$ be two points on a line $\ell$, $M$ the midpoint of $AB$, and $X$ a point on segment $AB$ other than $M$. Let $\Omega$ be a semicircle with diameter $AB$. Consider a point $P$ on $\Omega$ and let $\Gamma$ be the circle through $P$ and $X$ that is tangent to $AB$. Let $Q$ be the second intersection point of $\Omega$ and $\Gamma$. The internal angle bisector of $\angle PXQ$ intersects $\Gamma$ at a point $R$. Let $Y$ be a point on $\ell$ such that $RY$ is perpendicular to $\ell$. Show that $MX > XY$
Problem
Source: Mexico National Olympiad 2018 Problem 1
Tags: inequalities, geometry, angle bisector
Tsukuyomi
06.11.2018 13:10
Let $N$ be the centre of $\Gamma$. Since $PQ$ is the radical axis of $\Omega$ and $\Gamma$, the points $M,N,R$ are collinear. Observe that $$\angle{MRX}=\angle{NXR}=\angle{XRY},$$where the last equality holds as $NX\parallel RY$- they are both perpendicular to $\ell$. Thus $RX$ is the angle bisector of $\angle{MRY}$, and therefore by the angle bisector theorem we obtain $$\dfrac{MX}{XY}=\dfrac{RM}{RY}>1,$$as $RM$ is the hypotenuse of $\triangle{RMY}$, thereby giving us our desired result.
plagueis
08.11.2018 05:45
Tsukuyomi wrote: Let $N$ be the centre of $\Gamma$. Since $PQ$ is the radical axis of $\Omega$ and $\Gamma$, the points $M,N,R$ are collinear. Observe that $$\angle{MRX}=\angle{NXR}=\angle{XRY},$$where the last equality holds as $NX\parallel RY$- they are both perpendicular to $\ell$. Thus $RX$ is the angle bisector of $\angle{MRY}$, and therefore by the angle bisector theorem we obtain $$\dfrac{MX}{XY}=\dfrac{RM}{RY}>1,$$as $RM$ is the hypotenuse of $\triangle{RMY}$, thereby giving us our desired result. Nice, I didn't notice the angle bisector when I proposed this.
Note that $R$ is the midpoint of the arc $PQ$ of $\Gamma$ which does not contain $X$.Let $O$ be the center of $\Gamma$. As $P$ and $Q$ are the intersections of the circumference with diameter $AB$, which clearly has $M$ as center, with $\Gamma$. Then $R, O, M$ are collinear since they all lie on the perpendicular bisector of $PQ$. Therefore, as $RY\parallel OX\perp AB$, we must have $XY/MX=OR/OM<1$, since $OR$ is equal to the radius of $\Gamma$ and $OM$ is greater than that because $M$ is outside $\Gamma$. Thus $MX>XY$, as desired.
Mahdi_Mashayekhi
04.02.2022 18:49
Let S be center of $\Gamma$. R lies on midpoint of arc PQ and PQ is radical axis so M,S,R are collinear. MX/XY = MS/SR = MS/SX which MS > SX because ∠SXM = 90. we're Done.
enrique17
10.08.2022 23:19
As $M$ is the center of $\Omega$ we have $MP=MQ$, also as R is midpoint of arc $PQ$, then $RP=RQ$. Then by $LLL$ we have $\triangle PRM \sim \triangle QRM$. In particular $RM$ is angle bisector of $\angle PRQ$ and since $PR=RQ$, then $\angle PDM = 90°$. With $D= PQ \cap RM$ Let $E=RM \cap PX$ and $ F=RY \cap QX$. $\angle QPX = \angle QXB$ due to the tangency. So $\angle REX = \angle QPX + 90° = \angle QXB + 90° = \angle RFX$. So by $AA$, $\triangle REX$ and $\triangle RFX$ are similar, which implies $RX$ is angle bisector of $\angle MRY$ Finally, by angle bisector theorem: $MX/XY = MR/RY > 1 $ last inequality holds since $\angle RYM=90°$ It then follows that $MX>XY$. $\square$