# The Longest Segment in Intersecting Circles

Given two intersecting circles. Draw a line through one of the intersection points, say, A. Measure BC, the segment of the line enclosed by the two circles. The problem is to find the direction of the line such that the segment BC is the longest.

### Solution

**Note:** elsewhere the probem appears under a a different guise.

- M. E. Larsen,
*Misunderstanding My Mazy Mazes May Make Me Miserable*, in*The Lighter Side of Mathematics*, R.K.Guy and R.E.Woodrow, Eds, MAA, 1994

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

Copyright © 1996-2018 Alexander Bogomolny

### Solution

The drawing is somewhat empty. Experiment with adding meaningful elements. Perhaps you will come up with the diagram on the right. Let D be the other point of intersection of the two circles. Connect D with B and C. In the triangle BCD, the angles B and C do not depend on the position of the line BC. Therefore, all triangles BCD obtained when the direction of BC changes are similar. Therefore, the longest BC corresponds to the longest DC. But DC is just a chord in a circle. And the longest chord is a diameter.

### Questions

- If BC is the sought line, does the above mean that BD is also a diameter?
- If BC is the sought line, does it follow that BC is parallel to the center-line?
- Let D be a point of intersection of two circles, draw two diameters to obtain points B and C. Is it true that BC passes through A, the second point of intersection of the circles?

The following problem is discussed in Honsberger, *Mathematical Morsels*, p. 126 and referred to in Nelsen, *Proofs Without Words II*:

Suppose two circles intersect in A and B. A point P is selected on one of the circles on the outside arc. P is projected through A and B to determine chord CD on another circle. Prove that no matter where P is chosen on its arc, the length of the chord CD is always the same.

### Solution

In the previous problem we used the fact that the angle P (APB) does not depend on the position of the point P. The same is of course true for the angle CPD. But the latter is defined by the difference of two arcs (on the left circle) CD and AB. Since the latter is fixed, so is the former. (Another solution comes with a dynamic illustration.)

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

Copyright © 1996-2018 Alexander Bogomolny

63034459 |