Random Chord Length

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Problem Statement

A chord is drawn by choosing two random points uniformly and independently on the circumference of a unit circle (radius = 1). What is the probability that the chord length is greater than the radius?

Solution

This problem involves understanding how chord length relates to the central angle and using uniform distribution properties on a circle.

Step 1: Understand the Setup

We have a unit circle (radius r = 1). Two points are chosen uniformly and independently on the circumference. A chord connects these two points.

Step 2: Relate Chord Length to Central Angle

For a circle of radius r, if two points on the circumference subtend a central angle \theta at the center, the chord length is given by:

L = 2r \sin\left(\frac{\theta}{2}\right)

For a unit circle (r = 1), this simplifies to:

L = 2\sin\left(\frac{\theta}{2}\right)

Step 3: Determine the Condition

We want the chord length to be greater than the radius:

L > 1

Substituting the expression for chord length:

2\sin\left(\frac{\theta}{2}\right) > 1
\sin\left(\frac{\theta}{2}\right) > \frac{1}{2}

Step 4: Find the Range of Angles

We need to find all values of \theta such that \sin(\theta/2) > 1/2.

Since \sin(\alpha) = 1/2 when \alpha = \pi/6 or \alpha = 5\pi/6 (in the range [0, \pi]), we have:

\frac{\theta}{2} \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right)

Therefore:

\theta \in \left(\frac{\pi}{3}, \frac{5\pi}{3}\right)

Step 5: Calculate the Probability

Since the two points are chosen uniformly on the circumference, we can fix one point without loss of generality. The second point is then uniformly distributed along the circumference.

The central angle \theta between the two points is uniformly distributed on [0, 2\pi] (or equivalently, the arc length is uniformly distributed).

The probability that the chord length is greater than the radius is the probability that \theta \in (\pi/3, 5\pi/3):

P(L > 1) = \frac{\text{Length of favorable interval}}{\text{Total interval length}}
P(L > 1) = \frac{5\pi/3 - \pi/3}{2\pi} = \frac{4\pi/3}{2\pi} = \frac{2}{3}

Alternative Approach: Using Arc Length

We can also think in terms of arc length. On a unit circle, the circumference is 2\pi. The chord length is greater than 1 when the arc length between the two points (taking the shorter arc) is in a certain range.

However, since we’re choosing points uniformly, we consider the full range [0, 2\pi] for the angle, and the favorable region has length 4\pi/3 out of 2\pi, giving the same result.

Final Answer

P(\text{chord length} > \text{radius}) = \frac{2}{3} \approx 0.667

Key Insights

  1. Chord Length Formula: The relationship L = 2r\sin(\theta/2) is fundamental for problems involving random chords.

  2. Uniform Distribution on Circle: When points are chosen uniformly on a circle, the central angle (or arc length) is uniformly distributed.

  3. Symmetry: We can fix one point without loss of generality due to the rotational symmetry of the circle.

  4. Geometric Interpretation: The condition \sin(\theta/2) > 1/2 creates a specific angular range that determines the probability.

  5. Bertrand’s Paradox Connection: This problem is related to Bertrand’s paradox, which shows that “random chord” can be defined in multiple ways, leading to different probabilities. Here we use the “two random endpoints” method.

  6. Applications: Random chord problems appear in geometric probability, stochastic geometry, and various applications in physics and engineering.