# Mixtilinear Circles and Concurrence What is this about? A Mathematical Droodle

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Explanation

L. Bankoff has introduced the term mixtilinear for a curvilinear triangle with two sides straight line segments and the third an arc of a circle. He also called mixtilinear the incircles of the mixtilinear triangles. Given a circle and a triangle (with none of the sides touching the circle), there exist circles tangent two the given circle and some two sides of the triangle. These, too, are called mixtilinear and, depending on the manner of the tangencies there are mixtilinear incircles and excircles.

The applet displays a circle, a triangle ABC and three mixtilinear incircles tangent to a given circle with tangency point A' opposite A, B' opposite B, and C' opposite C. We have the following result:

 (1) The three lines AA', BB', CC' are concurrent in point M say. Moreover, if I is the incenter of ΔABC and O is the center of the given circle, then points M, O, I are collinear.

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The theorem has been discovered by Matthew Lee when yet in high school. (Lee's statement requires that the triangle be entirely inside the circle, but the proof does not use this property.) For the case where points A, B, C lie on the given circle it was also shown (even in two ways) by Paul Yiu. Lee's proof is an extension of one of Yiu's and is based on an observation that the product of two homotheties is either another homothety or a translation.

Let C(p, r) denote the circle with center p and radius r. Let C(I, r) be the incircle of ΔABC and C(O, R) the given circle.

First focus on vertex A. Let C(OA, RA) be the circle tangent to AB, AC and C(O, R), A' be the point of tangency of the two circles.

Circles C(I, r) and C(OA, RA) are homothetic with center A and coefficient AOA/AI. H(A, AOA/AI) is the homothety. Circles C(OA, RA) and C(O, R) are homothetic with center A' and coefficient A'O/A'OA. H(A', A'O/A'OA) is the homothety. The product of two homotheties is a homothety with center on the line AA' and coefficient AOA/AI·A'O/A'OA ≠ 1.

But the same homothety can be constructed starting with vertices B and C and found to be on the lines BB' and CC', wherefrom the three lines AA', BB', CC' are indeed concurrent. Since this homothety maps C(I, r) onto C(O, R) its center is collinear with I and O.

Note: Mixtilinear circles admit simple construction.

### References

1. P. Yiu, Mixtilinear Circles, Am Math Monthly, v. 106, n. 10 (Dec., 1999), 952-955
2. M. Lee, Four circles and a triangle, The Math Gazette, v. 90, n. 517 (Mar., 2006), 130-131