关于是否存在独立随机弦选取方式使其交点在圆内均匀分布的证明或证伪问询

Hey there! Great question—this is a fun twist on the classic random chord paradox problems. Let's break this down clearly so it makes sense:

Yes, such a method of choosing independent random chords does exist, and we can construct it explicitly to make the intersection points uniformly distributed across the unit circle.

First, why the "random two points per chord" method fails #

As you noticed with your Excel test, when you generate each chord by picking two uniform random points on the circle, the intersection points aren't uniform. Your result that intersections lie inside the radius-1/2 circle only 1/6 of the time (instead of the 1/4 we'd expect for uniform distribution) confirms this—this method produces intersection points biased toward the circle's center.

The working construction #

Here's a step-by-step way to generate independent random chords such that their intersection points (when they exist inside the circle) are uniformly distributed:

For each independent random chord:

1. Pick a random direction: Generate an angle ( U ) uniformly in the interval ([0, 2\pi)). This defines the orientation of the chord's perpendicular bisector (i.e., the direction from the circle's center to the chord's midpoint).
2. Pick a corrected distance from the center: Generate a uniform random number ( W ) in ([0,1]), then set ( V = \sqrt{W} ). This ( V ) is the distance from the circle's center to the chord—using the square root of a uniform number gives ( V ) a probability density of ( 2v ) over ([0,1]), which fixes the center bias from the classic method.
3. Define the chord: The chord is the line perpendicular to direction ( U ), lying at distance ( V ) from the center.

Why this works #

The key fix here is the ( \sqrt{W} ) step. For any point ( (r, \theta) ) in polar coordinates inside the circle:

• A chord passing through this point at an angle ( \psi ) relative to the radius will have a center-to-chord distance of ( V = r \cos \psi ).
• The ( 2v ) density of ( V ) ensures that the number of chord pairs intersecting at ( (r, \theta) ) scales with the area element ( r , dr , d\theta )—this is exactly the measure we need for uniform distribution across the circle's area.

In practical terms, this adjustment spreads out the possible chord positions enough to counteract the center bias, making intersections uniformly distributed instead of clustering near the middle.

A quick check of your radius-1/2 test case #

For this construction, the probability that an intersection lies inside the radius-1/2 circle would be exactly ( \frac{\text{Area of small circle}}{\text{Area of unit circle}} = \frac{1}{4} ), which matches the uniform distribution expectation you'd hope for.

备注：内容来源于stack exchange，提问作者Dan