# Four Equal Incident Circles

What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander BogomolnyThe applet is supposed to suggest the folloing proposition:

Let there be four incident circles of equal radius. Four of the common tangents (as shown in the applet) form a quadrilateral. This quadrilateral is always cyclic.

Denote the centers of the circles A_{1}, ..., A_{4} and the corresponding vertices of the quadrilateral in question B_{1}, ..., B_{4}. Since the four circles have the same radius, the four points A_{1}, A_{2}, A_{3}, A_{4} are concyclic and lie on a circle with center at the common point of the four given circles and radius equal to their common radius. Thus the quadrilateral A_{1}A_{2}A_{3}A_{4} is cyclic. This in particular means that the opposite angles in A_{1}A_{2}A_{3}A_{4} add up to 180°.

Turning to the quadrilateral B_{1}B_{2}B_{3}B_{4}, note that its sides are parallel to the corresponding sides of A_{1}A_{2}A_{3}A_{4}. For example, A_{1}A_{2}||B_{1}B_{2}. This is because the points A_{1} and A_{2} lie at the same distance from the common tangent B_{1}B_{2} to the two circles. The corresponding angles of the two quadrilaterals are, therefore, equal. In particular, this implies that the opposite angles in B_{1}B_{2}B_{3}B_{4} add up to 180°. The quadrilateral B_{1}B_{2}B_{3}B_{4} is therefore cyclic.

### References

- R. Honsberger,
*More Mathematical Morsels*, MAA, 1991, pp. 36-37

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Copyright © 1996-2018 Alexander Bogomolny