A Median in Touching Circles: What is this about?
A Mathematical Droodle


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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

The applet attempts to illustrate a problem once suggested for but not used at an international math olympiad ([Honsberger, p. 29]):

With a reference to the applet, two circles with diameters AB and AE are tangent internally at A. E trisects AB. O is the center of the big circle. P is a point on the small circle and AP crosses the big one in C. BC and OP intersect in D. Prove that BC = CD.


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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

The two circles are homothetic at the point of tangency and the coefficient 3/2 (from the fact that E trisects AB.) Under this homothety, point C corresponds to P so that AC = 3/2·AP. In other words, P trisects AC.

On the other hand, O being the center of the big circle, AO = BO and so DO is a median in ΔABD. Further, ∠ACB = 90° as an inscribed angle subtended by a diameter. The same holds for ∠APE. Therefore, both BC and EP are perpendicular to APC. Hence, BC||EP. For two parallel transversals of the angle BOD we have a proportion: OP/OD = OE/BE = 1/2. Since OD is a median in ΔABD, P is just the center, read "the point of intersections of all three medians", of that triangle. AC that passes through P is thus one of them (and, as we seen before, is divided by P in the correct ratio). Thus BC = CD, as required.

(Nathan Bowler has observed that the result follows an application of Menelaus' theorem to ΔABC.)

References

  1. R. Honsberger, In Pólya's Footsteps, MAA, 1997
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|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny
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