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 (3, 0). Another circle of variable radius r is centered at the origin and meets the positive y-axis in point A. Let B be the common point of the two circles in the upper half-plane. Let E be the intersection of AB extended with the x-axis. What happens to E as r grows smaller and smaller?

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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 FB = FE, F serves as the circumcenter of the triangle. So that we also have FD = FB and, by transitivity, FD = FE. F is the midpoint of DE!

References

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