Two Circles and a Limit
We are solving the Two Circles and a Limit problem:
A stationary circle of radius 3 is centered at
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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
- 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
Copyright © 1996-2017 Alexander Bogomolny