# Circles Tangent to CircumcircleWhat is this about? A Mathematical Droodle

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Explanation The applet illustrates the following statement:

 (*) In ΔABC, let K be a point on BC. Consider two circles tangent to AK, BC and the circumcircle of ΔABC and there points of tangency, say Kb and Lb of one and Kc and Lc of the other. Then lines KbLb and KcLc meet at the excenter E of ΔABC opposite vertex A. In addition, if K coincides with the point of tangency of the incircle of ΔABC, then the two circles have the same radius as the excircle (E).

### 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.

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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

 Let K' be a point on BC and circle γ' tangent to BK', AK' and the circumcircle of ΔABC. Let M and L be the points of tangency of the circle with BC and AK', respectively. Then ML passes through E.

We'll use the lemma to show that, for K' = K, the radius of γ equals ra, 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 = 2ra.

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 = ra. Thus the radius of γ equals ra.

(This solution is due to V. Protasov.)

### References

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