Something very similar to this is going on here, in the sense that $BPXQ$ is a parallelogram, but I'll have to give it some more thought.

This is related to Artzt parabolas. When X moves on AC, the line PQ envelopes the parabola touching BA at A and BC at C; so, the circle BPQ goes through the focus of the parabola (=B-vertex of the second Brocard triangle = projection of O upon the B-symedian) Of course, there is probanly a direct and easier proof. Kind regards. Jean-Pierre

Does anybody know how many contestants solved this problem on the contest? I, personally, needed 5 hours for it, and if I hadn't seen Grobber's observation about the parallelogram BPXQ, it would probably take me 3 more hours. So here is my solution. The first straightforward step is to remove some spam from the problem: The points M and N are not necessary; all we need to know about them is that the line MN is the common chord of our circle which passes through X and touches the side AC (let's call this circle k) with the circumcircle of triangle ABC, i. e. the radical axis of these two circles. So the problem can be stated more nicely in the following way: Given a triangle ABC. Let X be a point on its side AC. Some circle k passes through the point X and touches the side AC at this point X. We assume that the radical axis of the circle k and the circumcircle of triangle ABC bisects the segment BX. Let this radical axis intersect the lines AB and BC at the points P and Q. Prove that the circumcircle of triangle PBQ passes through a fixed point different from B. The next thing is to localize this fixed point. In fact, this can be done by checking the limiting cases X = A and X = C. What you obtain is: Construct the circle passing through the vertices A and B of triangle ABC and touching the side BC at the point B, and construct the circle passing through the vertices B and C and touching the side AB at the point B. These two circles intersect at B; let their second point of intersection be called F. Then, this F is the required fixed point. So what remains to prove is that the circumcircle of triangle PBQ always passes through the point F. [By the way, this point F is, of course, the B-vertex of the 2nd Brocard triangle of triangle ABC. This is just the way the 2nd Brocard triangle is defined.] In the following, we will use directed angles modulo 180° and directed segments. The point F has some trivial but nice properties which will be of use. According to its definition, the point F lies on the circle through the vertices A and B of triangle ABC and touching the side BC at the point B; thus, the line BC is the tangent to this circle at the point B. Hence, by the tangent-chordal angle theorem, < (FB; BC) = < FAB. In other words, < FBC = < FAB, what rewrites as < FAB = < FBC. Similarly, < FBA = < FCB. Hence, the triangles FAB and FBC are directly similar. Now, assume for a moment that we have shown that $\frac{AP}{PB}=\frac{BQ}{QC}$. Then, it follows that the points P and Q are corresponding points in these directly similar triangles FAB and FBC (since they lie on the corresponding sides AB and BC and divide these sides in the same ratio). Since corresponding points in directly similar triangles make equal angles, it thus follows that < APF = < BQF. In other words, < BPF = < BQF. Thus, the circumcircle of triangle PBQ passes through the point F, and the problem is solved. So, in order to solve the problem, it remains to show that $\frac{AP}{PB}=\frac{BQ}{QC}$. Now, I think, the most nontrivial idea of the solution comes: We will prove that PX || BC and QX || AB (in other words, we will prove that the quadrilateral BPXQ is a parallelogram). Once this is shown, we can conclude from Thales that $\frac{AP}{PB}=\frac{AX}{XC}$ and $\frac{BQ}{QC}=\frac{AX}{XC}$, and thus it follows that $\frac{AP}{PB}=\frac{BQ}{QC}$, so the problem is solved. Thus, it remains to prove that PX || BC and QX || AB. We will only establish PX || BC, since the proof of QX || AB is analogous. We will show that PX || BC by an indirect argument: Let the parallel to the line BC through the point X meet the line AB at a point $P_1$; then, we will try to show that the point $P_1$ has equal powers with respect to the circle k and the circumcircle of triangle ABC. Once this will be shown, it will follow that the point $P_1$ lies on the radical axis of the circle k and the circumcircle of triangle ABC, so it is the point of intersection of this radical axis with the line AB; but we know that the point of intersection of this radical axis with the line AB is P, and thus it will follow that $P_1=P$, so that $P_1X\parallel BC$ will become PX || BC, and the problem will be solved. So we have to show that the point $P_1$ has equal powers with respect to the circle k and the circumcircle of triangle ABC. In order to show this, let's compute these powers. Obviously, the power of the point $P_1$ with respect to the circumcircle of triangle ABC is $P_1A\cdot P_1B$. In order to obtain the power of $P_1$ with respect to the circle k, we denote by R the point of intersection of the line $XP_1$ with the circle k (apart from X); then, the power of the point $P_1$ with respect to the circle k is $P_1X\cdot P_1R$. So, in order to show that the point $P_1$ has equal powers with respect to the two circles, we have to show that $P_1X\cdot P_1R=P_1A\cdot P_1B$. By the converse of the intersecting chords theorem, this is equivalent to the assertion that the points A, B, X and R lie on one circle. So it remains to prove that the points A, B, X and R lie on one circle. This is, of course, equivalent to < ABR = < AXR. Since $XP_1\parallel BC$, we have $\measuredangle\left(AC;\;XP_1\right)=\measuredangle\left(AC;\;BC\right)$, or, equivalently, < AXR = < ACB, and thus, proving < ABR = < AXR comes down to proving < ABR = < ACB. But still, we are not directly able to do this since we don't know how to identify the angle < ABR. We try to identify this angle by the a trick we already used above: Find some similar triangles and apply the fact that corresponding points in similar triangles make equal (or oppositely equal if the triangles are inversely similar) angles. Yet some work has to be done in order to find such similar triangles. First, it's time to involve the assumption that the radical axis of the circle k and the circumcircle of triangle ABC bisects the segment BX. In other words, if M is the midpoint of the segment BX, then this point M has equal powers with respect to the circle k and to the circumcircle of triangle ABC. Again, let's compute these powers: If the line BX meets the circle k at a point U (apart from X) and the circumcircle of triangle ABC at a point T (apart from B), then the power of the point M with respect to the circle k is $MX\cdot MU$, and the power of the point M with respect to the circumcircle of triangle ABC is $MB\cdot MT$. Since the two powers of M are equal, we thus have $MX\cdot MU=MB\cdot MT$. In other words, $MX\cdot MU=BM\cdot TM$. But since M is the midpoint of the segment BX, we have MX = BM, and thus MU = TM. Hence, BU = BM + MU = MX + TM = TX, so that $\frac{XB}{BU}=\frac{XB}{TX}=\frac{BX}{XT}$. [By the way, this yields a nice construction of the circle k from the point X.] This ratio has a good chance of turning out useful. In fact, we can easily chase the angles of triangle XRU (we will do this later), and thus we have a good chance of finding another triangle similar to it. Now, the ratio $\frac{XB}{BU}$ fixes the position of the point B on the side XU of triangle XRU. If we find, in the triangle similar to it, a point which divides the corresponding side in the same ratio, then this point corresponds to B in this latter triangle, and we have an occasion for finding < ABR (actually, we will find < XBR, but that's more or less equivalent). Actually, the triangles XRU and BXC are inversely similar. The proof is straightforward: Since the circle k passes through the points X, R and U, while the line AC is the tangent to this circle k at the point X, the tangent-chordal angle theorem yields < (AC; XU) = < XRU. In other words, < CXB = < XRU. Consequently, < XRU = < CXB = - < BXC. Also, since $XP_1\parallel BC$, we have $\measuredangle\left(BX;\;XP_1\right)=\measuredangle\left(BX;\;BC\right)$, what rewrites as < UXR = - < CBX. Thus, we have shown that the triangles XRU and BXC are inversely similar. It remains to find the point corresponding to B in triangle BXC. This is easily done: Let the parallel to the line CT through the point X meet the line BC at a point S. Then, XS || CT yields by Thales $\frac{BS}{SC}=\frac{BX}{XT}$. Combining this with $\frac{XB}{BU}=\frac{BX}{XT}$, we see that $\frac{XB}{BU}=\frac{BS}{SC}$. Thus, the points B and S are corresponding points in the inversely similar triangles XRU and BXC (since they lie on the corresponding sides XU and BC and divide them in the same ratio). Corresponding points in inversely similar triangles form oppositely equal angles; thus, it follows that < XBR = - < BSX. But since XS || CT, we have < BSX = < BCT, so that < XBR = - < BCT. Finally, since the point T lies on the circumcircle of triangle ABC, we have < ABT = < ACT, so that < ABR = < ABX + < XBR = < ABT + < XBR = < ACT + (- < BCT) = < ACT - < BCT = < ACB. And the proof is complete. An unreasonably difficult problem, even for a 239MO. The classical geometry questions of 239MO 2002 were all very easy... Darij

Consider the points $Q'$ and $P'$ on sides $BC$ and $AB$ such $XQ'\parallel AB, XP'\parallel BC$. Let $K$ is the intersection point of lines $P'Q'$ and $AC$. Then $KA/KX=KP'/KQ'=KX/KC$, and $KX^2=KA\cdot KC$, hence radical axis of the circles passes through the point $K$. But the middle of $BX$ lies in that axis, hence $P'=P, Q'=Q$. Further, $AP/PB=AP/XQ=KP/KQ=PX/QC=BQ/QC$. Let $T$ is the second point of intersection of circles, which touching of $AB$ and $BC$ at the point $B$ and passes through the points $C$ and $A$. Then the triangles $ATB$ and $BTC$ is directly similar, then $\angle TPB=\angle TQC$, hence the points $B, P, T, Q$ is coincircle. Thus $T$ is this fixed point.

Let $MN\cap AC=Y$. Z is the reflection of X in Y. Then obviously Z,X,Y,A are harmonic conjugate. Apply Menelaus two times, we can find the length of PB and BQ, which is $\frac{BA}{AC}XC$ and $\frac{BC}{AC}XA$ respectively. A further short computation shows that $\frac{PP'}{QQ'}$ is constant, which implies the circumcircle of $\triangle BPQ$ passes through another fixed point as X varies along AC. btw, Darij, it is not really that hard you think. I think it is only because you follow grobber's idea for the weaker case.

Why you define $ Z$? To calculate lengths of$ BP$, $ BQ$, Y is enough. Isn't it?