Circles Tangent to Circumcircle: What is this about?
A Mathematical Droodle

Explanation

The problem highlights an apparently overlooked property of the configuration from Thébault's third problem which, as the present applet shows, holds for circles touching the circumcircle externally as well as internally: the line joining the centers of the two passes through an excenter of ΔABC. (In the case of the internal tangency, i.e. for Thébault's problem proper, the point on BC that gives rise to equal circles is the point of tangency of the excircle opposite vertex A with BC.)

The latter result also follows from the proof Y. Sawayama's Lemma.

More accurately we shall use the following fact derived in the proof of Sawayama's lemma:

Lemma

We'll use the lemma to show that, for K' = K, the radius of γ equals r_a, the radius of the excircle centered at E.

Draw perpendicular EV from E to BC and extend it beyond E to the point W so that EW = EV. VW ⊥ BC and VW = 2EV = 2r_a.

Pick a point H on BC so that HW||LM. Let O be the midpoint of HW. Since E is the midpoint of VW and HW||LE by the construction, L is the midpoint of VH implying that OL is the midline ΔWVH and OL ⊥ BC. This is true for any position K'. Next we consider K' = K.

A homothety with center A maps the incircle of ΔABC onto the excircle centered at E and, thus, K onto W. Further, ΔMKL is isosceles and so is ΔWKH. Therefore KO bisects ∠WKH implying that O is the center of γ (as the unique point of intersection of the perpendicular LO with the bisector of ∠WKH.) It follows that, for K' = K, OL = r_a. Thus the radius of γ equals r_a.

(This solution is due to V. Protasov.)

References

1. Problem 1.4, Matematicheskoe Prosveshchenie, Ser 3, No 5 (2001), pp. 218-221

1. Thébault's Problem I
2. Thébault's Problem II
3. Thébault's Problem III
   • Y. Sawayama's Lemma
   • Y. Sawayama's Theorem
   • Thébault's Problem III, Proof (J.-L. Ayme)
4. Circles Tangent to Circumcircle