Cut The Knot!

An interactive column using Java applets
by Alex Bogomolny

Concyclic Circumcenters:
A Sequel

June 2002

The discussion will then necessarily show some very bad features; points which are made can never be properly documented; and a certain over-all superficiality of the discussion becomes unavoidable.

John von Neumann, The Mathematician.
From The World of Mathematics,
James R. Newman
Simon and Schuster, 1956, p. 2053

The nice little problem discussed in the previous column deserves a few more words. The remark by the Monthly's editors that "... a [synthetic] solution may shed light on why the result is true" kept me weighing the known solutions against that criterion: does any of them clarify why the result is true? While the idiom of shedding light only refers to the source of enlightenment, it clearly assumes an interaction with an alleged receiver of the truth. The enlightenment, the moment of "aha!" is a very individual event that, to be engendered, may need more than one light source and a proper backdrop, but most of all it needs active involvement on the receiver's part. Sunbathing has little effect when the light is shed by a mathematical proof.

In my case, the first rays came through on an umpteen rereading of the editors' proof. Their first action was to show that the three trapezoids formed on the 6 circumcenters are isosceles by establishing that the three diagonals of the hexagon that connect the opposite vertices are equal. (The rest of the proof might have been replaced with a reference to a known problem, see for example #109 in [Shklyarsky].)

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

Once the diagonals have been shown to be equal, and the trapezoids isosceles, it takes just one remark to finish the proof. The common perpendicular bisectors of parallel sides (the axes of symmetry of the trapezoids) bisect the angles of the triangle formed by the extensions of the three equal diagonals. Therefore, the hexagon lies on a circle centered at the incenter of that triangle. (There are other characterizations of that point. The lines midway between parallel sides of the hexagon also form a triangle. As the Monthly's editors showed, the point in question, coincides with the circumcenter of that triangle. This point is also the midpoint of the segment joining the circumcenters of triangles S and D, see the applet above.)

A hexagon with three elements equal is a well known construct. The Tucker hexagons that we briefly encountered on another occasion, are formed by intermingling 3 parallels and 3 antiparallels in a given triangle. A most accessible account of Tucker hexagons and circles could be found in [Honsberger, Ch. 9]. Tucker hexagons are always cyclic. Antiparallel sides of Tucker hexagons are always equal. It also so happens that the Lemoine point K, that has popped up at the end of the previous column, plays an important role in Tucker's theory. In particular, both Lemoine circles are also Tucker's.

The point K we came across is the Lemoine point of both triangles S and D. (It was also the centroid of the original triangle.) The triangles S and D are homothetic with the center of homothety at K. The six circumcenters are found at the intersections of the side lines of these two triangles. The applet below sidesteps the original triangle, emphasizing instead the stationary triangle S (blue) and its homothetic image D (red). The same 6 points could be listed in different order. One ordering gives us the hexagon with three pairs of parallel sides and equal diagonals. Another yields a Tucker hexagon with intermittent parallels and antiparallels, the latter being equal.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

Aha! That's it! Since Tucker hexagons are cyclic, so are the hexagons of the six circumcenters at hand. How nice! Have you felt the light? I had certainly had a warm sensation when I got to that point. But things must be sorted out yet. Let's summarize:

  1. Given a triangle. Perpendicular bisectors of two pieces of its medians form a hexagon with opposite sides parallel (and some additional features). The bisectors also form two homothetic triangles with the center of homothety at G, the centroid of the base triangle. G serves as the Lemoine point K for both triangles.

  2. Consider a triangle and its homothetic image in the Lemoine point of the triangle. The six points of intersection of the sidelines of the two triangles form a Tucker hexagon. All Tucker hexagons can be obtained that way, by selecting a homothetic triangle. (The original problem from #1 is obtained with the coefficient of similarity equal to -1/2.)

  3. Tucker hexagons are cyclic. Therefore the hexagon in #1 is also cyclic.

  4. Given the pair of homothetic triangles in #2, is it possible to reconstruct the base triangle? In fact only one triangle is needed. Expand that triangle to twice its size with homothety in K. In the new triangle, the base triangle is then found as the pedal triangle of K.

  5. What about the Tucker circles corresponding to the coefficients other than -1/2? They correspond to the hexagons obtained in the dynamic framework where the medial triangle was allowed to expand homotheticly in G.

#2 bears some clarification.


Given ΔABC and its Lemoine point K. Let ΔA'B'C' be a homothetic image of ΔABC in K so that A' corresponds to A, and so on. Then the six points of intersection of the side lines of the two triangles define in ΔABC (and naturally also in ΔA'B'C') a Tucker hexagon. Conversely, any Tucker hexagon can be obtained that way.


Two points A and A' define a parallelogram in which AA' is a diagonal that passes through K, the Lemoine point of ΔABC. AK is thus a symmedian in ΔABC. It is known [Honsberger, Ch. 7] that, as cevians, the symmedians are characterized by the fact that they bisect the antiparallels. From which we infer that the second diagonal in the parallelogram defined by A and A' is an antiparallel. The hexagon is thus Tucker's.

Conversely, given a Tucker hexagon, extend, if necessary, its side lines to form triangle A'B'C'. Its antiparallel sides, each serves as a diagonal of a parallelogram. The other diagonals, that join A and A', B and B' and C and C' are naturally symmedians and thus cross at K. The triangles ABC and A'B'C' are therefore homothetic in K.

In the opening paragraph of this column I have suggested that the idiomatic "shedding light" does not tell the whole story. However, the metaphor of "shedding light" is very apt in the context of imparting eye opening ideas by means of a proof. It might have been, say, "throwing light" or "shedding skin", perhaps, in the sense of uncovering the core of the problem, or letting a revealing peek below the surface. It seems to me that both belittle the importance of the receiving end. Worse, they both conjure up the idea of a one time effort. As a metaphor for instructive communication, "shedding light" is unmatched. Shedding light means to diffuse and radiate, making available without enforcement, an expanding process reaching far and wide. The reach and the impact of the process could be enhanced by placing the source in a reflective setting of generalization. It's often the case that a generalization is at least as easy to prove as the result itself. It's almost always the case that a generalization is easier to comprehend than the original problem.


  1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995.
  2. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 2, Moscow, 1952.

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Copyright © 1996-2017 Alexander Bogomolny

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

This is supposed to be a proof without words, but just in case, here is a short clarification. We are dealing with two homothetic triangles ABC and A'B'C' and a Tucker hexagon. I claim that the circumcenter of the latter is the midpoint of the segment joining the circumcenters of the two triangles. Consider the midpoints A*, B*, and C* of the 3 antiparallel sides of the hexagon. These are in fact the centers of the three parallelograms with diagonals AA', BB' and CC'. Triangle A*B*C* is homothetic to both ABC and A'B'C' and, thinking "homotheticly", is half way between the two. In addition, since in the hexagon the antiparallel sides are all equal, their centers are equidistant from the center of the circle circumscribing the hexagon. The circumcircles of ΔA*B*C* and the hexagon are therefore concentric, and the assertion follows.


  1. All about Symmedians
  2. Symmedian and Antiparallel
  3. Symmedian and 2 Antiparallels
  4. Symmedian in a Right Triangle
  5. Nobbs' Points and Gergonne Line
  6. Three Tangents Theorem
  7. A Tangent in Concurrency
  8. Symmedian and the Tangents
  9. Ceva's Theorem
  10. Bride's Chair
  11. Star of David
  12. Concyclic Circumcenters: A Dynamic View
  13. Concyclic Circumcenters: A Sequel
  14. Steiner's Ratio Theorem
  15. Symmedian via Squares and a Circle
  16. Symmedian via Parallel Transversal and Two Circles
  17. Symmedian and the Simson
  18. Characterization of the Symmedian Point with Medians and Orthic Triangle
  19. A Special Triangle with a Line Through the Lemoine Point

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Copyright © 1996-2017 Alexander Bogomolny


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