Dear Alex,

I have send one message before with more two concyclic points but I still don't see it. In any case I resend it here with some more detail information:
If S' = midpoint of XA, T' = midpont of YC
and as your notations in applet:
S = midpoint of AB
T = midpoint of BC
Q = midpoint of AC
Suppose
V = intersection of S'Q and SP
W = intersection of T'Q and TN
then S'Q//XC and SP//AY therefore angle(PVQ) = 60. It means V, B, Q, P are concyclic.
Similarly with W then we have more two concyclic points V and W.
Now we have total ten concyclic points.

We can also use my proof method for two points V, W: they are also two another special cases. In fact:
- VM is tangent line of circumcircle of YBC at M
- WM is tangent line of circumcircle of XAB at M

I hope that we can find one complete nice elemetary proof for generalized fact when XAB, YBC are two similar triangles. In this case we still have ten points on one ellipse. This ellipse is Steiner circumellipse (St) of triangle BKL, here:
K, L is midpoints of AY, CX respectively. Our two circumcircles of XAB, YBC now are two Steiner circumellipses (St1), (St2) of triangles XAB, YBC respectively.
The fact we use for proof is still true:
Any line passing through one intersection of (St1), (St2) will cut three ellipse (St), (St1), (St2) again at three points and point on (St) is midpoint of two other points.

Best regards,
Bui Quang Tuan