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Concentric Circles

The region between two Concentric circles of different Radii is called an Annulus.

Given two concentric circles with Radii $R$ and $2R$, what is the probability that a chord chosen at random from the outer circle will cut across the inner circle? Depending on how the ``random'' Chord is chosen, 1/2, 1/3, or 1/4 could all be correct answers.

1. Picking any two points on the outer circle and connecting them gives 1/3.

2. Picking any random point on a diagonal and then picking the Chord that perpendicularly bisects it gives 1/2.

3. Picking any point on the large circle, drawing a line to the center, and then drawing the perpendicularly bisected Chord gives 1/4.
So some care is obviously needed in specifying what is meant by ``random'' in this problem.

Given an arbitrary Chord $BB'$ to the larger of two concentric Circles centered on $O$, the distance between inner and outer intersections is equal on both sides $(AB = A'B')$. To prove this, take the Perpendicular to $BB'$ passing through $O$ and crossing at $P$. By symmetry, it must be true that $PA$ and $PA'$ are equal. Similarly, $PB$ and $PB'$ must be equal. Therefore, $PB-PA = AB$ equals $PB'-PA' = A'B'$. Incidentally, this is also true for Homeoids, but the proof is nontrivial.

\begin{figure}\begin{center}\BoxedEPSF{ChordCirclesTheorem.epsf scaled 1000}\end{center}\end{figure}

See also Annulus

© 1996-9 Eric W. Weisstein