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Explanation

The 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: B, Intersection of XC and YA, The midpoints of AC, XC, YA, XB, YB. Center U of ΔACD, where D is the intersection of AX and CY, V, the intersection of S'Q and SP, where Sp is the midpoint of AX, W, the intersection of T'Q and NT, where Tp is the midpoint of CY.

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### 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: AB = BX, BY = BC, ∠ABY = ∠XBC = 120° implying that a rotation around B through 60° maps one of the triangles on the other; in particular mapping AY onto XC. This means that the angle between the two is 60°:

 ∠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 LQ||AX. Similarly, KQ||CY. 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, NL||BC and KP||AB implying NL||KP. Also LT||XB (in ΔBCX) and PT||CY (in Δ BCY), and since XB||CY so, too, LT||PT. 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'Q||XC and SP||AY 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 CD||XX'||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 ∠ANP = ∠CPB = 90° meaning B, N, U, P are concyclic, the circle of concyclicity having BU as a diameter.)