Problem 2: Bertrand's Problem.

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These web pages were modified from Kyle Siegrist's pages on Bertrand's Problem as can be found in his Virtual Laboratories in Probability and Statistics.

Joseph Bertrand

Bertrand's problem is to find the probability that a "random chord" on a circle will be longer than the length of a side of the inscribed equilateral triangle. The problem is named after the French mathematician Joseph Louis Bertrand, who studied the problem in 1889.

It turns out, as we will see, that there are (at least) two answers to Bertrand's problem, depending on how one interprets the phrase "random chord". The lack of a unique answer was considered a paradox at the time, because it was assumed (naively, in hindsight) that there should be a single natural answer.

To formulate the problem mathematically, let us take (0, 0) as the center of the circle and take the radius of the circle to be 1. These assumptions entail no loss of generality because they amount to measuring distances relative to the center of the circle, and taking the radius of the circle as the unit of length. Now consider a chord on the circle. By rotating the circle, we can assume that one point of the chord is (1, 0) and the other point is (X, Y) where Y > 0. Then we can completely specify the chord by giving any of the following quantities:

  • The (perpendicular) distance D from the center of the circle to the midpoint of the the chord.
  • The angle A between the x-axis and the line from the center of the circle to the midpoint of the chord.
  • The horizontal coordinate X.

Note that 0 < D < 1, 0 < A < / 2, -1 < X < 1.

A chord in the circle

PROBLEM 2

Part (a):

1. Show that D = cos(A).

2. Show that X = 2D2 - 1.

3. Show that Y = 2D (1 - D2)1/2.

4. Show that A is also the angle between the chord and the tangent line to the circle at (1, 0).

Now consider the equilateral triangle inscribed in the circle so that one of the vertices is (1, 0). Consider the chord defined by the upper side of the triangle.

The equilateral triangle

Part (b):

Show that for this chord, the angle, distance, and coordinate variables are given as follows:

  1. a = / 3
  2. d = 1/2
  3. x = -1/2
  4. y = (3/4)1/2.

Now suppose that a chord is chosen in probabilistic way.

T he length of the chord is greater than the length of a side of the inscribed equilateral triangle if and only if the following equivalent conditions occur:

  1. 0 < D < 1/2
  2. / 3 < A < / 2
  3. -1 < X < -1/2

Can you see why?

The crux of Bertrand's paradox is the fact that the distance D and the angle A both seem to be a natural variable that determine the chord, but different models are obtained, depending on which variable is picked at random.

Bertrand's Experiment:

The Model Where D is Picked at Random

Suppose that D is picked at random on the interval (0, 1).

Part (c):

Show that the solution of Bertrand's problem is

P(D < 1/2) = 1/2

In Bertrand's experiment, (View in Microsoft Internet Explorer). Select the uniform distance model. Run the experiment 1000 times, updating every 10 runs. Note the apparent convergence of the relative frequency function of the chord event to the true probability.

The Model Where A is Picked at Random

Suppose that A is picked at random on the interval (0, /2).

Part (d):

Show that the solution of Bertrand's problem is

P(A > / 3) = 1/3

Solutions

In Bertrand's experiment, select the uniform distance model. Run the experiment 1000 times, updating every 10 runs. Note the apparent convergence of the relative frequency function of the chord event to the true probability.