Inequality in circle configuration

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Inequality in circle configuration

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Source: Mexico National Olympiad 2018 Problem 1

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322 posts

#1 h Nov 5, 2018, 9:26 PM • 2 Y

Y by Adventure10, Mango247

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 $MX > XY$

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#3 h Nov 6, 2018, 10:10 AM • 2 Y

Y by XbenX, Adventure10

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.

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#4 h Nov 8, 2018, 2:45 AM • 3 Y

Y by Justpassingby, Adventure10, Mango247

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.
Official Solution

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Feb 4, 2022, 3:49 PM

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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.

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#6 h Aug 10, 2022, 8:19 PM

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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 $MX>XY$ . $\square$

This post has been edited 2 times. Last edited by enrique17, Jan 9, 2023, 1:16 AM

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