A random chord longer than the inscribed square's side

Question

A chord is drawn at random in a circle of radius $r$ .

What is the probability that the chord is longer than the side of the inscribed square?

As in Bertrand's paradox, the answer depends on the sampling method, so give it under all three and note anything surprising.

Accepted Answer

Take r = 1; the square's side is \sqrt2 \approx 1.414, so we want P(	ext{length} \sqrt2). Method 1, random endpoints. Length 2\sin\alpha with \alpha uniform on [0,\pi). Require 2\sin\alpha \sqrt2 \iff \sin\alpha 	frac{\sqrt2}{2} \iff \alpha \in \left(	frac{\pi}{4}, 	frac{3\pi}{4}ight), an interval of length 	frac{\pi}{2} out of \pi, so P1 = 	frac12. Method 2, random radial midpoint. d uniform on [0,1], length 2\sqrt{1-d^2} \sqrt2 requires 1 - d^2 	frac12 \iff d < 	frac{1}{\sqrt2} \approx 0.707. So P2 = 	frac{1}{\sqrt2} \approx 0.707. Method 3, random midpoint in the disk. With P(d …

A random chord longer than the inscribed square's side

Your answer

More Probability questions