Let $P$ be a point in the plane and let $\gamma$ be a circle which does not contain $P$. Two distinct variable lines $\ell$ and $\ell'$ through $P$ meet the circle $\gamma$ at points $X$ and $Y$, and $X'$ and $Y'$, respectively. Let $M$ and $N$ be the antipodes of $P$ in the circles $PXX'$ and $PYY'$, respectively. Prove that the line $MN$ passes through a fixed point. Mihai Chis
Problem
Source: Romania TST 4 (All Geometry) 2010, Problem 1
Tags: geometry, parallelogram, geometric transformation, homothety, rectangle, geometry proposed
Luis González
25.08.2012 20:20
Let $O,U,V$ be the centers of $\gamma,$ $\odot(PXX')$ and $\odot(PYY'),$ respectively. $OU \perp XX'$ and $OV \perp YY'$ are the perpendicular bisectors of $\overline{XX'}$ and $\overline{YY'}.$ Since $XX'$ and $YY'$ are antiparallel WRT $PX,PX',$ it follows that $PU \perp YY'.$ Similarly, $PV \perp XX'.$ Hence $PU \parallel OV$ and $PV \parallel OU$ $\Longrightarrow$ $PUOV$ is a parallelogram $\Longrightarrow$ line $UV$ passes through the midpoint of $\overline{PO}.$ Then, the image $MN$ of $UV$ under the homothety $(P,2)$ goes through $O.$
ACCCGS8
26.08.2012 10:18
Assume without loss of generality that $P, X, Y$ and $P, X', Y'$ lie in that order. Let $W$ be the midpoint of $XY$, $U$ be the foot of the perpendicular from $M$ to $YN$, $T$ be the point on $MN$ such that $TW \perp XY$ and $S$ be the intersection of $MU$ and $WT$. Note that $XM, YN \perp XY$. $XW=YW \implies MS=US \implies MT=NT$ as $XMUY$ is a rectangle and Midpoint Theorem. Similarly, if $Z$ is the midpoint of $X'Y'$ and $V$ is the point on $MN$ such that $ZV \perp X'Y'$, $MV=NV$. Thus $T=V$ so the perpendicular bisectors of $XY$ and $X'Y'$ concur on $MN$ so $MN$ passes through $O$, the circumcentre of $\gamma$. Done.
Jul
08.11.2014 18:10
My solution : Let $\left \{ P,T \right \}=(PXX')\cap \left ( PYY' \right )$. Let $J=XX'\cap YY'$. Easily, we get $P,T,J$ are conlinear. From the problem 5, IMO 1985 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366594&sid=3824115ccf232b168958a0149e7b5313#p366594 We have $OT\perp PJ$. But $MT\perp PJ,NT\perp PJ$. Hence, $M,N,O$ are conlinear.
TelvCohl
08.11.2014 18:50
My solution: Let $ L $ be the polar of $ P $ WRT $ \gamma $ . Let $ Z $ be the projection of $ P $ on $ L $ and $ O $ be the center of $ \gamma $. Let $ P_x, P_y $ be the projection of $ P $ on $ XX', YY' $, respectively . From $ XX' \cap YY' \in L $ we get $ (PP_xP_y) $ pass through $ Z $ . Invert with center $ P $ and factor $ PX \cdot PY=PX' \cdot PY' $ we get $ MN $ passes through $ O $ . Q.E.D
P1kachu
10.09.2020 13:10
Let lines $XM$ and $YN$ intersect the $\gamma$ again at points $A$ and $B$ respectively. Since $\angle PXM$ and $\angle PYN$ are right angles, $XABY$ must be a rectangle. Similarly, by intersecting lines $X'M$ and $Y'N$ with $\gamma$ at points $A'$ and $B'$ reveals another rectangle $X'A'B'Y'$.
Since $\gamma\cup XABY\cup X'A'B'Y'$ is symmetric with respect to $O$, this means $M$ and $N$ must also be symmetric with respect to $O$, so $MN$ must pass through $O$.