Circles On Cevians: What Is It About?
A Mathematical Droodle
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
Circles On Cevians
The applet suggests the following theorem [Coxeter, Theorems 2.45, 2.46]:
If circles are constructed on two cevians (via different vertices) as diameters, their radical axis passes through the orthocenter of the triangle.
The radical center of any three non-coaxal circles having cevians (via different vertices) for diameters coincides with the orthocenter of the triangle.
What if applet does not run? |
Let in ΔABC a circle is drawn on a cevian CP_{c} as diameter. Call it S_{c}. Unless CP_{c} is perpendicular to the side line AB, the circle intersects AB in a second point, say H_{c}. The angle CH_{c}P_{c} subtends the diameter CP_{c} and is, therefore, right. We conclude that CH_{c} is an altitude in ΔABC.
If P_{a} is located on the side line BC, a similar construction produces circle S_{a} and a point H_{a} which is the feet of the altitude AH_{a}. AH_{a} and CH_{c} 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 AH_{a}B and CH_{c}B share angle C, they are similar. It follows that
(1) | ∠H_{a}CH_{c} = ∠BCH_{c} = ∠BAH_{a} = ∠H_{c}AH_{a}, |
which shows that the right triangles AHH_{c} and CHH_{a} are also similar. This implies
(2) | CH/AH = HH_{a}/HH_{c}, or CH×HH_{c} = AH×HH_{a}. |
Now, CH×HH_{c} is the power of H with respect to the circle on CP_{c}, while AH×HH_{a} is the power of H with respect to the circle on AP_{a}. 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 AH_{c}H_{a}C is cyclic. Let S denote its circumcircle. Then CH_{c} is the radical axis of S and S_{c}, while AH_{a} is the radical axis of S_{a}. The two meet at the orthocenter H, which is then of necessity the radical center of S, S_{c} and S_{a}. Therefore, the radical axis of S_{c} and S_{a} passes through H.
A circle S_{b} constructed on a third cevian BP_{b} could be paired with either S_{a} or S_{c}. 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
- J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971
- H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967
- V. V. Prasolov, Problems in Plane Geometry, Part 1, Nauka, Moscow, 1986
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny