|
|
|
|
|
|
|
|
Three Circles and Common Chords
 This question is akin to several other questions where an apparently difficult problem was solved by breaking implicit assumptions that the solver imposes at the outset. As in another problem concerning three circles, an almost immediate solution lies in embedding the circles into a 3-dimensional configuration. What might it be?
Copyright © 1996-2009 Alexander Bogomolny
Consider each circle as the equator of a sphere. Let the plane of the circles be horizontal. It cuts the spheres in two (of course symmetric) halves so that a circle is the projection of the corresponding sphere onto the horizontal plane. Any two spheres intersect at a circle. The chords at hand are the projections of three such (vertical) circles. However, the three spheres share two points which are symmetric with respect to the plane. Those points belong to all three pairwise intersections of the spheres - vertical circles. Therefore, projections of the three vertical circles - our chords - share a point which is the projection of the common points of the spheres. NoteAs Rob Rumppe noted, the above reasoning only holds if the centers of the spheres and hence the circles are not colinear. There is another, more general solution that uses the notion of the radical axis of two cricles. 2D Problems That Benefit from a 3D Outlook
Copyright © 1996-2009 Alexander Bogomolny |
| 33061852 |  |
