Three Tangent Circles
Three circles (Oi), with centers at Oi,
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Copyright © 1996-2018 Alexander BogomolnyThree circles (Oi), with centers at Oi,
The center Ak of circle (Ak) lies at the intersection of lines CiQk and CjRk. One passes through Pj, the other through Pi. Note that, say Pj, is the center of similarity of (Oi) and (Ok). Therefore, triangles PjOkPi and PjOiQk are similar. Which makes lines OkPi and OiQk parallel. Thus, OkOj||OiAk. Similarly, OkOi||OjAk. The quadrilateral OiOkOjAk is a parallelogram. Which implies
AkOi = OjOk = rj + rk.
But the radius of circle (Ak) equals
AkQk = AkOi + OiQk = rj + rk + ri.
Which means that the radius of circle (Ak) is the sum of the radii of the given three circles. Since there is nothing special about the index k, all three circles A have the same radius and are, therefore, equal.
References
- J. Hadamard, Leçons de géométrie élémentaire, tome I, 13e édition, reprint 1988, #322
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Copyright © 1996-2018 Alexander Bogomolny72111987