Y. Sawayama's Lemma: What Is It About?
A Mathematical Droodle

Explanation

Copyright © 1996-2009 Alexander Bogomolny

The applet purports to illustrate Y. Sawayama's lemma:

Sawayama's paper has been discovered by J.-L. Ayme who used the lemma to directly solve Thébault's Problem. The proof by Ayme is a slight modification of that by Sawayama. Along the way, Ayme corrects a logical gap in the original proof.

1. Let M and N be the intersections of KE and KF with the circumcircle C₂. Then MN||EF due to C₁ and C₂ being homothetic at K. M is the midpoint of the arc BC not including K. Therefore, AM is the bisector at A and, thus, contains the incenter I.

2. Let J be the intersection of AM and EF. Consider the configuration of two lines AM and KN and the coaxal system of circles through A and K. One of the circles, viz. C₂, cuts the chord MN. Another circle - that through F - cuts a parallel chord, bound to lie on EF. This implies that quadrilateral AFJK is cyclic.

3. Apply now Miquel's Pivot Theorem to ΔAFJ with F on AF, E on FJ, and J on AJ. K is the pivot point common to three circles. Circle EJK is tangent to AJ (same as AM) at J.

4. Circle C₃ centered at M

   and radius BM passes through I. This circle is also orthogonal to C₁. Indeed,

   BKE = MAC = MBE

   so that the circumcircle of ΔBKE is tangent to BM at B. C₃ is orthogonal to the latter circle and, since M lies on EK, to all circles through K and E, in particular, to the circle EJK. Therefore, MB = MJ (but also MB = MI) so that J = I

and we are finished.

Reference

1. J.-L. Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum, v 3 (2003), 225-229.
2. Y. Sawayama, A New Geometrical Proposition, Amer. Math. Monthly, 12 (1905) 222–224.

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

Copyright © 1996-2009 Alexander Bogomolny