Circles On Cevians: What Is It About?
A Mathematical Droodle


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Circles On Cevians

The applet suggests the following theorem [Coxeter, Theorems 2.45, 2.46]:

  1. If circles are constructed on two cevians (via different vertices) as diameters, their radical axis passes through the orthocenter of the triangle.

  2. The radical center of any three non-coaxal circles having cevians (via different vertices) for diameters coincides with the orthocenter of the triangle.


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Let in ΔABC a circle is drawn on a cevian CPc as diameter. Call it Sc. Unless CPc is perpendicular to the side line AB, the circle intersects AB in a second point, say Hc. The angle CHcPc subtends the diameter CPc and is, therefore, right. We conclude that CHc is an altitude in ΔABC.

If Pa is located on the side line BC, a similar construction produces circle Sa and a point Ha which is the feet of the altitude AHa. AHa and CHc intersect at the orthocenter H of the triangle. (For a different perspective, see Collinearity with the Orthocenter.)

The configuration contains several similar triangles. First, since the right triangles AHaB and CHcB share angle C, they are similar. It follows that

(1) ∠HaCHc = ∠BCHc = ∠BAHa = ∠HcAHa,

which shows that the right triangles AHHc and CHHa are also similar. This implies

(2) CH/AH = HHa/HHc, or CH×HHc = AH×HHa.

Now, CH×HHc is the power of H with respect to the circle on CPc, while AH×HHa is the power of H with respect to the circle on APa. The identity (2) means that H lies on the radical axis of the two circles. In other words, the radical axis of the two circles passes through H.

For an additional proof of this result see [Coxeter, p. 37, or Prasolov, p. 64, problem 3.23(b)]. Nathan Bowler has pointed to the fact that (1) means that the quadrilateral AHcHaC is cyclic. Let S denote its circumcircle. Then CHc is the radical axis of S and Sc, while AHa is the radical axis of Sa. The two meet at the orthocenter H, which is then of necessity the radical center of S, Sc and Sa. Therefore, the radical axis of Sc and Sa passes through H.

A circle Sb constructed on a third cevian BPb could be paired with either Sa or Sc. We thus obtain three radical axes that all pass through H.

If the circles are coaxal, the three radical axes coincide. Otherwise, the three axes meet in H, which is the radical center of the three circles. We are led to the famous Gauss-Bodenmiller theorem:

Corollary

The three circles constructed on the diagonals of a complete quadrilateral as diameters are coaxal: their centers are collinear, and, when taken two by two, their radical axes coincide [Coolidge, p. 101].

Indeed, the side lines of a complete quadrilateral taken three by three form four triangles, in each of which its three diagonals serve as cevians. From the foregoing discussion, it appears that each of the four orthocenters of the four triangles should play the role of the radical center of the same three circles. The implied contradiction does not arise only if the radical axes of the pairs of the circles coincide while the four orthocenters just lie on that line, the common radical axis. The centers of the circles lie on another line, and the two lines are orthogonal.

(It was pointed to me by Bui Quang Tuan that the problem has been discussed at the Hyacinthos group in January-February, 2000.)

References

  1. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971
  2. H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967
  3. V. V. Prasolov, Problems in Plane Geometry, Part 1, Nauka, Moscow, 1986

Radical Axis and Radical Center

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Copyright © 1996-2017 Alexander Bogomolny

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