Given a triangle , a circle centered at some point meets the segments in the pairs of points and , and , and , respectively, labeled in circular order: . Let be the Miquel point of the triangle (i.e., the point of concurrence of the circles ), and let be that of the triangle . Prove that the segments and have equal lengths.
Solution
We begin by reviewing some basic facts on conics. For an ellipse with center , foci and , semiaxes and , it is known that the orthogonal projections and of and on any line tangent to lie on the major auxiliary circle of , so that . Application to triangle (respectively, ) of a rotation (respectively, ) about (respectively, ) and a homothety of ratio with center (respectively, ) yields triangle (respectively, ), where , , , and both lie on . If varies and is constant, the locus of and is then a circle centered at .
By Cartesian geometry it is readily checked that and are bitangent, and the line supporting their common chord is also the radical axis of the circles and , with this real geometrical significance even if the bitangency is not real. Since and the circle are mutually inverse in , the inverse points of and in both lie on . Finally, the distance between the parallel lines and is given by . Similar considerations hold for a hyperbola .
Consider now an isopair (two isogonally conjugate in the triangle , the foci of a conic touching its sides), and take points (respectively, ) on lines , respectively, so that the lines (respectively, ) make the same directed angle (respectively, ) with the perpendiculars to , respectively; then the isopedal triangles of angles for the isopair have their Miquel points at and are inscribed in a common isopedal circle bitangent to , centered at a point on the perpendicular bisector of the segment .
Conversely, for any pair of triangles inscribed in a triangle and in a circle (as in the statement of the problem), the Miquel points are an isopair and is an isopedal circle of . (If are the Brocard points, is the Brocard ellipse and is a Tucker circle.)