Circles Tangent to Circumcircle
What is this about?
A Mathematical Droodle
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|
Copyright © 1996-2018 Alexander BogomolnyThe 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 K_{b} and L_{b} of one and K_{c} and L_{c} of the other. Then lines K_{b}L_{b} and K_{c}L_{c} 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). |
What if applet does not run? |
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 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
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
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,
(This solution is due to V. Protasov.)
References
- Problem 1.4, Matematicheskoe Prosveshchenie, Ser 3, No 5 (2001), pp. 218-221
- Thébault's Problem I
- Thébault's Problem II
- Thébault's Problem III
- Circles Tangent to Circumcircle
|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|
Copyright © 1996-2018 Alexander Bogomolny65842266