Suppose that $ (a,ka^2),(b,kb^2),(c,kc^2),(d,kd^2)$ are some concyclic points on the parabola $ y = kx^2$, where $ k\neq 0$. We will prove that \[ a + b + c + d = 0 \] Let $ \omega(O,R)$ be the circle circumscribed around the chosen points and assume that $ O = (m,n)$. So $ (a - m)^2 + (ka^2 - n)^2 = R^2$, and the same equation holds for $ b,c$ and $ d$. Therefore, the polynomial $ P(x) = k^2x^4 - 2knx^2 + x^2 - 2xm + m^2 + n^2 - R^2$ has 4 real roots, namely $ a,b,c,d$. Since the coefficient in front of $ x^3$ equals $ 0$, by Vieta's formula we get: \[ a + b + c + d = 0 \] Now apply this result for cases $ k = 1$ and $ k = 2009$ and we get the desired result. P.S: Actually this is mine solution, try to figure out the official one. It is a bit harder though.

yeah my solution is similar, I reach the same equation but after that I take e new point and get it's coordinates as functions of the coordinates of the center of the first circle in such a way, that the new equation for the Bi-s and the new point (if it's K then KBi^2=const) is const - it's enoug the coeficients in frotn of x^2 and x to be equal in the 2 equations

Erken, I think your solution contains a bug. Yes it's true that you have proved; if $ (a,ka^2), (b,kb^2), (c,kc^2), (d,kd^2)$ are concyclic, $ a + b + c + d = 0$. But you also must prove that if $ a + b + c + d = 0$, $ (a,ka^2), (b,kb^2), (c,kc^2), (d,kd^2)$ are concyclic; for proving that the points on $ 2009x^2$ parabola are concyclic. There is a little bit more analysis.

Let $ N_i(x_i,ax_i^2)$. I proved that whether $ N_1,N_2,N_3,N_4$ is concyclic or not does not depend on $ a$. This is easy.

Umut Varolgunes wrote: Erken, I think your solution contains a bug. Yes it's true that you have proved; if $ (a,ka^2), (b,kb^2), (c,kc^2), (d,kd^2)$ are concyclic, $ a + b + c + d = 0$. But you also must prove that if $ a + b + c + d = 0$, $ (a,ka^2), (b,kb^2), (c,kc^2), (d,kd^2)$ are concyclic; for proving that the points on $ 2009x^2$ parabola are concyclic. There is a little bit more analysis. hmm...I personally disagree that this might be considered as a bug. To be honest, I put it down for an obvious fact, that doesn't require any further explanations, though it might be a bit awkward. - If $ a,b,c,d$ are different coordinates of the points on $ y=x^2$, then we should prove that the circumcircle of a triangle with vertices at 3 points from $ \{a,b,c,d\}$ intersects the graph of $ y=2009x^2$ at precisely $ 4$ points - if that is proven, then the rest easily follows from the made observations - sorry for lite notations and explanations. Assume the contrary, i.e the circumcircle is tangent to the graph. Clearly, the graph of both equations can be imagined as two branches, and two basic cases are possible, two of the points lie on each of the branches. and 1 point on one brach and 3 points on another. Suppose that $ a,b$ and $ c,d$ are on different branches. So circumcircle of $ \{a,b,c\}$ is tangent to the graph, and so is the circumcircle of $ \{a,b,d\}$. It yields that $ c=d$, contradiction. Now if $ a$ and $ b,c,d$ are on different branches then the circumcircle of $ a,b,c$ is tangent to the graph at $ a$ and so is $ a,c,d$,so all these points are concyclic.

I actually think, this is exactly what a bug is. A minor mistake, which can be fixed easily; but if it isn't fixed, the solution can't be counted as completed. You may have already done it on your own, but there was nothing written in your first post about it. Now everything is more clear, thanks for paying attention.