This is a nice problem, compliments to your friend! First, we prove that the circumferences with centres X,Y and radii XS and YT touch the line AB and the main circumference. This picture explains why, whatch it. Invalid URL Now, here is a simplified picture to solve the problem: Invalid URL First, we prove that the quadrilaterals $TPRF,SPRE$ are cyclic. To prove this, consider the homotety with centre T bringing F to M, it brings AB to the tangent t at M, and using directed angles it immidiately follows that $\angle RFT = \angle (AB,MT) = \angle (t,MT) = \angle MPT = \angle RPT$, the same for the other circles. Now, let's observe that $CT^{2}= CP \cdot CQ = CS^{2}$ because of power of points, and also $MF \cdot MT = MR \cdot MP = ME \cdot MS$ because of power of points and the ciclicy proven above. But then both C,M lie on the radical axis of all the three red circles taken in pairs, and then the three circles are coaxial, and their centres (X,Y,Z) are coollinear!

It is problem 1 of China TST 2007. It is proposed by XiongBin.

Attachments:
icmo2007s.pdf (477kb)

edriv wrote: This is a nice problem, compliments to your friend! First, we prove that the circumferences with centres X,Y and radii XS and YT touch the line AB and the main circumference. This picture explains why, whatch it. Don't get it... how can the picture explain?

Ok, I'll explain why, then. I forgot to name the points: let M be the midpoint of the arc determined by r, $C = MP \cap r$, D the intersection of OP with the perpendicular to r through C. Note that OM is perpendicular to r, and also DC is. Therefore, OM and DC are parallel. Then, triangles $PDC$ and $POM$ are similar. But $POM$ is isosceles! Then $PDC$ is isosceles, too: $PD = DC$. Therefore there is a circle centered at D that passes through P and C. But since P,D,O are on a line, this circle is tangent to the big circle. And DC is perpendicular to r: our circle touches also r. Is it clear?

April wrote: Points $A$ and $B$ lie on the circle with center $O.$ Let point $C$ lies outside the circle; let $CS$ and $CT$ be tangents to the circle. $M$ be the midpoint of minor arc $AB$ of $(O).$ $MS,\,MT$ intersect $AB$ at points $E,\,F$ respectively. The lines passing through $E,\,F$ perpendicular to $AB$ cut $OS,\,OT$ at $X$ and $Y$ respectively. A line passed through $C$ intersect the circle $(O)$ at $P,\,Q$ ($P$ lies on segment $CQ$). Let $R$ be the intersection of $MP$ and $AB,$ and let $Z$ be the circumcentre of triangle $PQR.$ Prove that: $X,\,Y,\,Z$ are collinear. I don't know Chinese but I think the solution's to seem like: Proof. We have: $\blacktriangleright\quad \begin{cases}XE\parallel OM\\ \triangle OMS\;\text{is isosceles}\\ \overline{OXS}\perp CS\end{cases}\Longrightarrow\begin{cases}E,S\subset (X,\,XS) \\ (X)\;\text{is tangent to}\; (O)\end{cases}$ $\blacktriangleright \quad \angle MAB=\angle MPA=\angle MSA\Longrightarrow\begin{cases}\triangle MAR\sim\triangle MPA\\ \triangle MAE\sim \triangle MSA\end{cases}\Longrightarrow \begin{cases}MA^{2}=MP\cdot MR\\ MA^{2}=MS\cdot ME\end{cases}$ $\Longrightarrow MP\cdot MR=MS\cdot ME\Longrightarrow M\;\text{lies on the radical axis of two circles}\; (PQR)\;\text{and}\; (X)\quad (1)$ $\blacktriangleright\quad CP\cdot CQ=CS^{2}\Longrightarrow C\;\text{lies on the radical axis of two circles}\; (PQR)\;\text{and}\; (X)\quad (2)$ From $(1)$ and $(2)$ we have: $MC\perp XZ$ Similar $MC\perp YZ$ And hence $X,\,Y,\,Z$ are collinear

Let $\mathcal{P}(X, \omega)$ denote the power of a point $X$ w.r.t. a circle $\omega.$ Animate the point $P$ on $(O)$ and let $\Gamma\left(O_1\right)$ be the circumcircle of $\triangle PQR.$ We claim that as $P$ varies, $O_1$ varies along a fixed line. To see this, note that the inversion with center $M$ and radius $MA = MB$ swaps $AB$ and $(O).$ Thus $P, R$ are swapped under this inversion, implying that $\mathcal{P}(M, \Gamma) = MP \cdot MR = MA^2$ is fixed. Meanwhile, note that $\mathcal{P}(C, \Gamma) = CP \cdot CQ = \mathcal{P}(C, (O))$ is fixed as well. Hence, it follows from the Power of a Point Formula that $\mathcal{P}(M, \Gamma) - \mathcal{P}(C, \Gamma) = O_1M^2 - O_1C^2$ is fixed. Since $M$ and $C$ are fixed, it is then well-known the locus of $O_1$ is a line. To determine this line, we will consider two special cases. First, note that $OM \perp AB$ because $M$ is the midpoint of arc $\widehat{AB}.$ Therefore, $OM \parallel XE$, implying that $\triangle SOM \sim \triangle SXE \implies XS = XE.$ Then when $P \equiv S$, we have $Q \equiv S$ as well, implying that $\Gamma$ and $(O)$ are tangent at $S.$ Hence $O_1 \in OS$, but we also have $O_1S = O_1E$, whence we deduce that $O_1 \equiv X.$ Similarly, when $P \equiv T$ we find that $O_1 \equiv Y.$ Therefore, $XY$ is the fixed line upon which $O_1$ varies. $\square$

We can consider that $M,A,B,S$ and the circle are fixed and $P$ varies along the circle. It's clearly shown that $X=Z$ and $Y=Z$ when $P=S$ and $P=T$ respectively. It suffice to show that locus of $Z$ is a line as $P$ varies along the circle. Let $N=MQ \cap AB$. $MQ \cdot MN=MP \cdot MR$. Hence, $P,Q,R,N$ are concyclic. Let $P_1,P_2$ are points lie on the circle. We define $Q_1,Q_2,R_1,R_2,N_1,N_2,Z_1,Z_2$ similarly. Since, $CQ_1 \cdot CP_1=CQ_2 \cdot CP_2$ and $MQ_1 \cdot MN_1=MQ_2 \cdot MN_2$. $MC$ is radical axis of $\odot P_1Q_1R_1$ and $\odot P_2Q_2R_2$. Hence, $Z_1Z_2\bot MC$. Locus of $Z$ is a line perpendicular to $MC$.

Solved with VJ27 Let $\omega$ be the circle centered at $X$ with radius $XE$ ; $\gamma$ be the circle centered at $Y$ with radius $YF$ and $MQ$ intersect $AB$ at $D$. $S$ and $T$ are the touchpoints of $(ABPQ)$ & $\omega$ ; $(ABPQ)$ & $\gamma$ respectively coz of homothety. $DRPQ$ is cyclic coz $MD*MQ=MB^2=MR*MP$. $M$ lies has equal power w.r.t. $\omega$, $DRPQ$ & $\gamma$ since $MT*MF$=$MB^2$=$ME*MS$=$MD*MQ$ .Also $C$ has equal power with respect to the 3 circles coz of radax theorem on $ABPQ$ , $DRPQ$ & $\omega$ and $CS=CT$. So the three circles are coaxial having radax $CM$.Hence $X$ , $Y$ and $Z$ are collinear.

Nice problem! Let $\overline{AB}$ and $\overline{MQ}$ intersect at $R'$. By the shooting lemma, $PQR'R$ is cyclic. Since $\triangle SOM \sim \triangle SXE$ and $\triangle TOM \sim \triangle TYR$, we have $XE=XS$ and $YR=YT$. Let $\omega$ be the circumcircle of $PQR'R$, let $\omega_1$ be the circle centered at $X$ passing through $E$, and let $\omega_2$ be the circle centered at $Y$ passing through $F$. Notice that $CS^2=CT^2=CP \cdot CQ$, $C$ has equal power with respect to $\omega$, $\omega_1$, and $\omega_2$. By the shooting lemma, $M$ has equal power with respect to $\omega$, $\omega_1$, and $\omega_2$. Thus, $\omega$, $\omega_1$, and $\omega_2$ are coaxial, so $X$, $Y$, and $Z$ are collinear.