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

Explanation

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Copyright © 1996-2018 Alexander Bogomolny

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

Through vertex A of ΔABC a straight line AD is drawn with D on BC. Let circle C1 tangent to AD at F, CD at E, and the circumcircle C2 of ΔABC at K be centered at P.Then the chord EF passes through the incenter I of ΔABC.

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 the 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.
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|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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