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A red line segment's endpoints are random (independent and uniform) points on two tangent circles, with one endpoint on each circle. A green line segment is tangent to both circles and is a shared side of two isosceles triangles with central angles $\alpha$ and $\beta$, where $0\le\alpha+\beta\le\pi$.

Two circles, green and red line segments

Show that the probability that the red line segment crosses the green line segment is $\dfrac{\alpha+\beta}{\pi}$.

I will post my answer, which involves a complicated integral that I calculated numerically. I am looking for a more intuitive answer, or an answer that does not rely on a computer.

Context: I use this result to answer a probability question about a random triangle on three mutually tangent circles.

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  • $\begingroup$ The whole thing looks a bit fishy to me: ok, you state that $\alpha + \beta \leq \pi$, but nevertheless, let's imagine that both $\alpha$ and $\beta$ are equal to $\frac{\pi}{2}$ (both being right angles). In that case it would still be possible for both lines segments not to cross while the probability would be equal to $1$, indicating that the not-crossing of the line segments is not possible. Do you see what I mean? $\endgroup$
    – Dominique
    Commented 17 hours ago
  • $\begingroup$ @Dominique If $\alpha$ and $\beta$ are both right angles, then the two circles have equal radii and the green line segment's length equals the diameter of the circles, so the probability that the red line segment crosses the green line segment is $1$. $\endgroup$
    – Dan
    Commented 17 hours ago
  • $\begingroup$ Why would both circles have equal radii in case of right angles? I'll show you an example in an answer, you'll see what I mean. $\endgroup$
    – Dominique
    Commented 17 hours ago
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    $\begingroup$ I guess the "whole point" or "unexpected thing" about this is that the probability $\dfrac{\alpha+\beta}{\pi}$ is independent of the (ratio of the) two circles' radii. $\endgroup$
    – Adam Rubinson
    Commented 17 hours ago

2 Answers 2

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Self-answering. This answer involves a complicated integral that I calculated numerically. I am looking for a more intuitive answer, or an answer that does not rely on a computer.

Two circles, green and red line segments, angles x and y

In the diagram, the red line segment goes through the top endpoint of the green line segment. Angle chasing, and the sine rule, give the (complicated) relationship between angles $x$ and $y$ (these angles are spanned by dashed lines):

$$y= \begin{cases} 2\arccos\left(\dfrac{\sin\left(\dfrac{\alpha+\beta}{2}-v\right)}{\cos\left(\dfrac{\beta}{2}\right)}\right) &0\le x\le\pi+2u\\ 0&\pi+2u<x<2\pi\\ \end{cases}$$

where

$$u=\arcsin\left(\frac{a+b-b\sec\beta}{a}\cos\beta\right)$$

$$v=\arcsin\left(\frac{\cos\left(\frac{\alpha}{2}+\beta+u-x\right)}{\sqrt{\sec^2\left(\frac{\alpha}{2}\right)-2\sec\left(\frac{\alpha}{2}\right)\sin\left(\frac{\alpha}{2}+\beta+u-x\right)+1}}\right)$$

and $a$ and $b$ are the radii of the left and right circles, respectively.

The probability that a random red line segment lies above the green line segment is the average value of $\dfrac{y}{2\pi}$ for $0\le x<2\pi$, that is,

$$\frac{1}{4\pi^2}\int_0^{2\pi}y\space dx$$

Numerical calculation strongly suggests that this probability is $\dfrac12\left(1-\dfrac{\alpha+\beta}{\pi}\right)$, which implies that the probability that the red line segment crosses the green line segment is $\dfrac{\alpha+\beta}{\pi}$.

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Hereby an example of a situation that, according to your question, should not be possible:

Screenshot

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    $\begingroup$ According to the OP, the green line segment must be tangent to the circles. $\endgroup$
    – Dan
    Commented 17 hours ago
  • $\begingroup$ Indeed: when the green line must be tangent to both circles, in case of both $\alpha$ and $\beta$ being $\frac{}{}$, this indeed means that the red line not crossing the green one seems impossible. $\endgroup$
    – Dominique
    Commented 17 hours ago

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