Concyclic Points in Equilateral Bumps: What is this about?
A Mathematical Droodle

What if applet does not run? 
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander BogomolnyThe applet illustrates a problem suggested by Bui Quang Tuan:
Point B is on segment AC. Construct two equilateral triangles XAB, YBC on one side of segment AC. Prove that the following ten points are concyclic:


What if applet does not run? 
Proof 1
For a straightforward proof, denote K, L, N, P, Q, S, T the midpoints of AY, CX, BX, CY, AC, AB, and BC, respectively, and let M be the intersection of AY and CX.
First of all observe that ΔABY = ΔXBC by SAS:
∠AMX = 60°. 
Depending on the position of B relative to the midpoint Q, ∠KML is either 60° or 120°.
Now, LQ is a midline in ΔACX and so LQAX. Similarly, KQCY. This means that ∠KQL = 60° so that the four points K, L, M, Q are concyclic (the angles at M and Q being subtended by the same segment KL.)
Referring again a property of midlines, NLBC and KPAB implying NLKP. Also LTXB (in ΔBCX) and PTCY (in Δ BCY), and since XBCY so, too, LTPT. Therefore L, P, and T are collinear. It follows that
∠LPK = ∠LTB = 60°. 
Similarly, ∠NKP = 60& making quadrilateral KNLP an isosceles trapezoid and hence cyclic. Furthermore, ∠KPL is subtended by the same segment KL so that points, K, L, P, Q are concyclic, which shows that six points L, N, K, Q, P, and M lie on the same circle. B also belongs to that circle because ∠NBP = 60° and is subtended by the same segment as ∠NKP.
Proof 2
Bui Quang Tuan offers a more elegant solution based on the following lemma:
Let two circles C(P) and C(Q) intersect in points C and D. A lines through C intersects the second time C(P) at A and C(Q) at B. Let O be the midpoint of PQ. Then the circle C(O) with center O through C and D meets AB at the midpoint T. 
To make use of the lemma draw the circumcircles of the two equilateral triangles. Then points L, K, and Q are concyclic with M and B as the midpoints of lines through one of the points of intersection of the circles  M or B. The same in fact holds for N and P as well as the midpoints of tangents BY and BX  the extreme case of the crossing line.
Further, S'QXC and SPAY therefore ∠PVQ = 60°. It means V, B, Q, P are concyclic. The same argument applies to W.
Finally, consider U, the centroid of ΔACD. The perpendicular to AC at A intersects the circumcircle ABX at X', the perpendicular to AC at C intersects the circumcircle BCY at Y'. Then, first of all, X', M, Y' are collinear. The itersection of X'Y' with AD is midpoint of X'Y', for CDXX'YY'. This point U is the centroid of ΔACD because it is collinear with both N, A and C, P. (It helps also to observe that
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny