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

 

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Explanation

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Copyright © 1996-2017 Alexander Bogomolny

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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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: 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.)

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Copyright © 1996-2017 Alexander Bogomolny

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