# Two Circles and a Limit

Proof #1

### Barbeau, Klamkin, Moser

We are solving the Two Circles and a Limit problem:

A stationary circle of radius 3 is centered at

What if applet does not run? |

The notations in the proof are as follows: C is the lowest point of the shrinking circle, D is the intersection of BC with the x-axis, F is the rightmost point of the stationary circle.

Since AC is the diameter of the shrinking circle, ∠ABC is right, an so is ∠DBE. Subtracting from both ∠DBF we obtain that

∠OBC = ∠FBE.

Of course we also have

∠OBC = ∠OCB.

On the other hand, angle AOB is formed by a tangent to and a chord of the stationary circle. Therefore,

∠AOB = ∠BFO.

By the Exterior Angle Theorem,

∠AOB = ∠OBC + ∠OCB, whereas

∠BFO = ∠FBE + ∠FEB.

From which we conclude that

∠FEB = ∠FBE.

As a consequence,

FE = FB.

As r goes to zero, B approaches the origin O, FB approached the diameter OF. Thus E goes to the point twice as far from he origin as F. Note that, since DBE is right and

### References

- E. J. Barbeau, M. S. Klamkin, W. O. J. Moser,
*Five Hundred Mathematical Challenges*, MAA, 1995, #396 - J. Konhauser, D. Velleman, S. Wagon,
*Which Way Did the Bicycle Go?*, MAA, 1996, #5

### Limits in Geometry

- Two Circles and a Limit
- A Geometric Limit
- Iterations in Geometry, an example
- Iterated Function Systems

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

61198088 |