Let $AB$ be a chord on a circle $O$, $M$ be the midpoint of the smaller arc $AB$. From a point $C$ outside the circle $O$ draws two tangents to the circle $O$ at the points $S$ and $T$. Suppose $MS$ intersects with $AB$ at the point $E$, $MT$ intersects with $AB$ at the point $F$. From $E,F$ draw a line perpendicular to $AB$ that intersects with $OS,OT$ at the points $X,Y$, respectively. Draw another line from $C$ which intersects with the circle $O$ at the points $P$ and $Q$. Let $R$ be the intersection point of $MP$ and $AB$. Finally, let $Z$ be the circumcenter of triangle $PQR$. Prove that $X$,$Y$ and $Z$ are collinear.
Problem
Source: 2016 Taiwan TST 1st IMO Mock P1
Tags: geometry
ThE-dArK-lOrD
08.07.2016 19:00
Note that for all point $D$ on major arc $\overarc{BC}$ , if $MD\cap AB=D'$, then $MD\cdot MD'=MA^2$ (This can be proved by angle-chasing) Then since $OM\perp AB$ and $EX\perp AB$, so $EX\parallel OM$, this mean $\triangle{SXE}\sim \triangle{SOM}$, so $XS=XE$ Similarly $YF=YT$ Let $\omega_1,\omega_2$ are circle center at $X,Y$ with radius $XE,YF$ respectively, we get that $\omega_1,\omega_2$ are tangent to $(O)$ Since $CS^2=CT^2$ and $ME\cdot MS=MT\cdot MF$ so $C,M$ lie on radical axis of $\omega_1,\omega_2$, so $XY \perp CM$ Since $CP\cdot CQ=CT^2$ and $MR\cdot MQ=MF\cdot MT$ so $C,M$ lie on radical axis of $\omega_2,(PQR)$, so $YZ \perp CM$ So $X,Y,Z$ collinear as we want $\centerline{Best regards,}$ $\centerline{ThE-dArK-lOrD}$ P.S.Can you post all of Taiwan TST $?$
va2010
08.07.2016 20:12
This is China TST 2007 Problem 1
Willie
09.07.2016 06:33
va2010 wrote: This is China TST 2007 Problem 1 Haha, Taiwan pirate often the problems of china, for example: http://artofproblemsolving.com/community/c6h1113618p5087311 ,and the resource of this problem is a magazine.
K6160
22.09.2016 09:35
Notice that $MR\cdot MP = MA^2$ since $\angle RAM = \angle RPA \implies MA$ is tangent to $\odot (AMP)$. Also $CQ\cdot CP = CS^2$. Therefore, $MZ^2-CZ^2$ is fixed which implies that $Z$ lies on a line. Taking the degenerate cases when $P=Q=S$ and $P=Q=T$ show that this line passes through $X$ and $Y$. $\square$