# Three Circles and Common Chords

Three circles intersect pairwise but let's assume there is no point shared by all three of them. There are three pairs of circles. Two circles in any pair share a chord. The problem is to prove that the three chords meet at a point.

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?

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 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.

### Note

As Rob Rumppe noted, the above reasoning only holds if the centers of the spheres and hence the circles are not collinear.

There is another, more general solution that uses the notion of the *radical axis* of two cricles.

Abstract as it may appear, the theorem is the basis for the GPS navigation [Charming Proofs, p. 198]. A GPS unit communicates with three satellites and evaluates the distance to the satellites. The unit is located on the three spheres with centers at the satellites and the caculated radii. It is therefore positioned at one of the intersections of the three spheres.

### References

- C. Alsina, R. B. Nelsen,
*Charming Proofs*, MAA, 2010

### 2D Problems That Benefit from a 3D Outlook

- Four Travellers, Solution
- Desargues' Theorem
- Soddy Circles and Eppstein's Points
- Symmetries in a Triangle
- Three Circles and Common Chords
- Three Circles and Common Tangents
- Three Equal Circles
- Menelaus from 3D
- Stereographic Projection and Inversion
- Stereographic Projection and Radical Axes
- Sum of Squares in Equilateral Triangle

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny

68969694