Coaxal Circles on Perpendicular Bisector
The applet actually shows that a more general statement is true. In the applet, the vertices of the triangle and center O of circle (O) through B and C are draggable. (I) is a circle tangent to AB and AC and internally to (O). (P) is a circle tangent to AB and AC and externally to (O). You can observe that at least when (I) is close to the incircle of ΔABC one of the common internal tangents to (I) and (P) is parallel to BC. Points S and T are the points of tangency of pairs (O), (I) and (O), (P). (The line of tangency joins S and T and is seen to be different from either of the common intangents.)
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It is worth noting that when one of the common internal tangents of (I) and (P) is parallel to BC, the other one is antiparallel to it. The sides AB and AC cut off each of the the coaxal circles through B and C (and, hence, the center on the perpendicular bisector of BC) a chord which also antiparallel to BC.
- J. Marshall Unger, A new proof of a "hard but important" Sangaku problem, Forum Geometricorum, 10 (2010) 7--13.
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