A Dynamic View
Here's an interesting and unexpectedly recent problem (Am Math Monthly, v. 109, n. 4, April 2002, pp 396-397.)
This is an amazing theorem (proposed by Floor van Lamoen, Goes, The Netherlands) that only employs the most basic notions of triangle geometry (medians, circumcenters), and yet the problem appears to be quite recent. van Lamoen refers to Clark Kimberling's TCCT. (This is a full pledged book, not a paper, as claimed by the Problems' editors.) Kimberling has pioneered the use of computers as a research tool in geometry. He used QBASIC programs to systematically search for interesting points, lines and curves related to the triangle. The results of one such program, SIXTRI, suggested that the incenters of the small triangles formed by three intersecting cevians through special points, lie on an ellipse, while their circumcenters lie on a conic. In van Lamoen's problem, the cevians - the medians - pass through the centroid of the triangle. It was noted by Antreas P. Hatzipolakis that in this particular case the circumcenters trivially lie on a conic as a consequence of Pascal's theorem.
As of now, in addition to the solution that appeared in the Monthly, I am aware of two entirely synthetic proofs of the statement. One was submitted by the proposer along with the problem and on which the editors' solution is partially based. The other is by K. Y. Li, one of the cited solvers, that has appeared in the Excalibur, a magazine of the Hong Kong University of Science and Technology. (I am grateful to Floor van Lamoen for the link.) The latter proof stands somewhat apart. The other two utilize van Lamoen's argument that I summarize in the following
Let each side of the hexagon O1O2O3O4O5O6 be parallel to its opposite mate. Then the hexagon is inscribable in a circle if and only if the three pairs of opposite sides define isosceles trapezoids.
The proof below, partly analytic and partly synthetic, also resorts to Lemma at its final stage. The analytic portion, credible after experimentation with the applet below, might have been demonstrated with Maple or Mathematica. It's simple enough to be carried out by hand with the help of a diagram.
First, I embed the problem in a one parameter family of problems. For every value of the parameter there will be present 6 triangles whose circumcenters form a hexagon with opposite sides parallel. I then show that if for one value of the parameter the opposite sides form isosceles trapezoids, the same holds true for all values of the parameter. Together with an observation that for a certain parameter value, the trapezoids degenerate into rectangles which, ipso facto, are isosceles, we only have to invoke the lemma once to complete the proof.
Originally, let MA, MB and MC denote the midpoints of sides BC, CA and AB, respectively. Let G be the centroid of ΔABC. We consider 6 triangles: AMBG, MBCG, CMAG, MABG, BMCG, MCAG and their circumcenters. Now, let's make points MA, MB and MC vary so as to keep ΔMAMBMC homothetic to its original position with the center of homothety at G. Notationally, we are interested in the "same" 6 triangles: AMBG, MBCG, CMAG, MABG, BMCG, MCAG.
The circumcenters of the triangles lie at the intersections of the stationary perpendicular bisectors of AG, BG, CG and the dynamic bisectors of GMA, GMB and GMC. The former three form a triangle (blue in the applet, call it S), a half-sized replica of the antipedal triangle of G and ABC. (The antipedal triangle is formed by the lines orthogonal to the cevians at the vertices of the triangle.) The latter three form a homothetic triangle (red, call it D) with the center of homothety at G.
As the parameter changes (drag any of the M points), the circumcenters appear to glide along the sides of S, two per side.
|What if applet does not run?|
The Exceptional Configuration
Wherever AG = GMA we also have BG = GMB and CG = GMC. The six triangles all become equal. S becomes equal to D. And, because of the symmetry of the configuration, the trapezoids prove to be rectangles. Their common center coincides with G.
(To see the trapezoids and, when the time comes, the rectangles, keep clicking on the "++" button.)
All For One, One For All
This is the crucial part of the proof. If for one parameter value the trapezoids are isosceles, they remain isosceles for all other values of the parameter. Go figure. The diagram below, although far from being a proof without words, helps visualize the configuration. It's useful, although ugly. I proceed with apologies.
The diagram depicts the static triangle S (blue) and two positions of the dynamic triangle D (red).
Let's consider just one pair of parallel sides OBCOBA and OABOCB. If a, b, c denote the lengths of the medians from A, B, and C, respectively, variations in two positions of the points M, differ proportionally to a, b and c along each of the medians. These variations are shown as at, bt and ct. At G, the medians form angles a, b and g, which are inherited by the antipedal triangle and, importantly, by the triangle formed by the medians.
In the direction of OBCOBA, the midpoint of OBCOBA shifts by
(ct/sin(a) - at/sin(g))/2,
whereas the midpoint of OABOCB moves in the same direction by
(bt/tan(a) - bt/tan(g))/2.
Therefore, in order to establish the result we have to prove that these two differences are equal. The goal is thus to prove the identity
c/sin(a) - a/sin(g) = b/tan(a) - b/tan(g).
By the Law of Sines applied to the triangle formed by the medians, a = b·sin(a)/sin(b), while c = b·sin(g)/sin(b), which reduces the above to
(sin(g)/sin(a) - sin(a)/sin(g))/sin(b) = cos(a)/sin(a) - cos(g)/sin(g),
sin2(g) - sin2(a) = (sin(g)cos(a) - cos(g)sin(a))·sin(b).
The left-hand side here equals sin(g - a)·sin(g + a)! (For a derivation, see [Barbeau, pp 32-33].) In turn, the right-hand side equals sin(g - a)·sin(b). Since
Now, let's try to learn more about the problem. First of all, there's another peculiar configuration that occurs when the moving points M fall on top of the vertices A, B, C. The hexagon of the circumcenters appears then as a doubly traversed triangle that coincides, on the occasion, with both S and D. Thus the circle in question serves as the circumcircle of triangle S.
It's obvious that as the points M trace straight lines, so do the circumcenters of the six triangles and so does the center of the circle on which all of them lie. That center, therefore, traces the straight line through the centroid of ΔABC and the circumcenter of triangle S. Since S is homothetic to the antipedal triangle of G with G as the center of homothety, the above line passes through the circumcenter of the antipedal triangle.
Now, why is the antipedal triangle so called? The reason is that, with regard to the antipedal triangle of a point, the original reference triangle ABC serves as the pedal triangle of that point. As is well known, the symmedian, or Lemoine, point of a triangle serves as the centroid of its pedal triangle. That point is also uniquely defined [Honsberger, p. 107] as the point whose distances to the side lines of a triangle are proportional to the triangle's side lengths. Either of these two facts can be used to prove that the centroid G of ΔABC coincides with the Lemoine point of its antipedal triangle. The same of course holds for the half-sized replica S of the latter. The Lemoine point is usually denoted by K.
To sum up, the center of the circle that passes through the six circumcenters lies on the line that joins the Lemoine point K with the circumcenter O of the antipedal triangle of G. This line does not have a name. But the segment OK does. It's known as the Brocard diameter of the (antipedal) triangle. The Brocard diameter of triangle S is half as long. The segment owes its fame to the fact that both Brocard points lie on the circle (the Brocard circle) with diameter KO. How does all that relate to the problem of 6 circumcenters? Simple. 6 circumcenters form 2 equal triangles that split between them the equal sides of the isosceles trapezoids we encountered above. (To see the triangles, one in magenta, the other in dark pink, check the "Triangle" box.)
|What if applet does not run?|
The triangles are similar to S and spirally rotate in opposite directions as their vertices glide over the side lines of S. The vertices of each triangle, taken 2 at a time, together with the vertex of S that is the intersection of the sides the 2 points belong to, define a circle. By the Pivot theorem, the 3 circles due to each of the triangles concur at a point. Thus, for each configuration in our one parameter family, there are two points of note. A well deserved note, indeed. For, those points stay fixed for all configurations in the family! (See the remark between problems 64 and 65 in [Yaglom, p. 78] and also a page on spiral similarity.) The points are exactly the Brocard points of triangle S [Yaglom, problem 65]. The line OK is the perpendicular bisector of the segment joining the two Brocard points!
- E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000.
- R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995.
- C. Kimberling, Triangle Centers and Central Triangles, Congressus Numerantium 129, Utilitas Mathematica Publishing, 1998.
- I. M. Yaglom, Geometric Transformations II, MAA, 1968
Copyright © 1996-2018 Alexander Bogomolny