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In a circle, a chord will be drawn by connecting two uniformly random points on the circle. Where in the circle should you draw a line segment of fixed length, to maximize the probability that it will be intersected by the chord?

My attempts suggest (but do not prove), perhaps counter-intuitively, that the line segment should be drawn so that it is a chord of the circle.

ATTEMPT $1$

Let the radius of the circle be $1$, and let the length of the line segment be $2L$, where $L<1$.

In deciding where to draw the line segment, the variables are:

$p=$ distance between centre of circle and midpoint of line segment
$\theta=$ acute angle between line segment, and line through centre of circle and midpoint of line segment.

The diagram shows the line segment in red, and one of the randomly chosen points on the circle.

enter image description here

For each value of angle $x$, the probability that the line segment is intersected by the random chord, is $C/\pi$.

The overall probability is the average value of $C/\pi$, as angle $x$ goes from $0$ to $2\pi$.

It takes a bit of work to express $C$ in terms of $L, p, \theta, x$:

$a=\sqrt{p^2+L^2-2pL\cos{(\pi-\theta)}}$ (cosine rule)
$A=\arcsin{\left(\frac{L\sin{(\pi-\theta)}}{a}\right)}$ (sine rule)
$b=\sqrt{1^2+a^2-2a\cos{(A+x)}}$ (cosine rule)
$c=\sqrt{1^2+p^2-2p\cos{x}}$ (cosine rule)
$\cos{B}=\frac{b^2+L^2-c^2}{2bL}=\frac{b^2+(2L)^2-d^2}{4bL}$ (cosine rule) $\implies d=\sqrt{2L^2-b^2+2c^2}$
$C=\arccos{\left(\frac{b^2+d^2-(2L)^2}{2bd}\right)}$ (cosine rule)

$P=P(\text{line segment is intersected by chord})=\frac{1}{2\pi}\int_0^{2\pi}\frac{C}{\pi}dx$

Putting all of this into my computer, it seems that, for any value of $L<1$, $P$ is maximized when the line segment is drawn as a chord of the circle. But trying to prove this analytically seems daunting.

ATTEMPT $2$

Consider a disk of radius $r<1$ placed inside a unit circle. The distance between their centres is $t$, and $P(t)$ is the probability that a random chord (chosen by connecting two uniformly random points on the unit circle) will intersect the disk.

We will try to show that $P(t)$ is an increasing function in $t$. Then in the original question, regard the line segment as a rigid row of many small disks. When the line segment (i.e. row of disks) is a chord of the unit circle, the $t$-value of each small disk is maximized, so the probability that the line segment is intersected by a random chord is maximized.

The diagram shows a disk, and one of the randomly chosen points on the circle.

enter image description here

For each value of angle $x$, the probability that the disk is intersected by the random chord, is $\alpha/\pi$.

The overall probability is the average value of $\alpha/\pi$ as angle $x$ goes from $0$ to $2\pi$.

$\alpha=2\arcsin{\left(\dfrac{r}{\sqrt{1+t^2-2t\cos{x}}}\right)}$

$P(t)=\dfrac{1}{\pi^2}\int_0^{2\pi}\arcsin{\left(\dfrac{r}{\sqrt{1+t^2-2t\cos{x}}}\right)}dx=\dfrac{2}{\pi^2}\int_0^{\pi}\arcsin{\left(\dfrac{r}{\sqrt{1+t^2-2t\cos{x}}}\right)}dx$

We can express the integral as a riemann sum, then differentiate with respect to $t$, then express the new riemann sum as an integral. So we get:

$\dfrac{dP(t)}{dt}=\dfrac{2r}{\pi^2}\int_{0}^{\pi}\dfrac{\cos{x}-t}{(1+t^2-2t\cos{x})\sqrt{1+t^2-2t\cos{x}-r^2}}dx$

Since $r$ will approach $0$, I think we can ignore the $r^2$ in the denominator.

So we just have to show that $\int_{0}^{\pi}\dfrac{\cos{x}-t}{(1+t^2-2t\cos{x})^{1.5}}dx>0$ for $0<t<1$, but I do not know how to show this. I found a related question, but it has not helped me.

ATTEMPT $3$

I tried to use the method in the answer of @Gribouillis below to get an expression for the probability that the line segment will be intersected by a random chord, but it does not seem to make the maximization problem any easier than my "Attempt $1$".

EXTRA INFORMATION

$P$ seems to be minimized when the line segment's midpoint is at the centre of the circle. The ratio of the maximum to minimum probabilities seems to approach $\pi/2$ as $L\to0$.

Just for fun, I asked my high school students this question, and asked them to use their intuition, with the following choices:

enter image description here

24 students chose A.
18 students chose B.
10 students chose C.
6 students chose D.

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  • $\begingroup$ is there a question? $\endgroup$ Commented Dec 9, 2022 at 3:45
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    $\begingroup$ @RyRytheFlyGuy Yes, the question is in the title, and in the first paragraph. $\endgroup$
    – Dan
    Commented Dec 9, 2022 at 3:46
  • $\begingroup$ okay, you answer your question in the title, sooo... ? $\endgroup$ Commented Dec 9, 2022 at 4:02
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    $\begingroup$ @RyRytheFlyGuy Numerical investigation suggests an answer, but that is not a proof. The question has not been answered yet. $\endgroup$
    – Dan
    Commented Dec 9, 2022 at 4:06
  • $\begingroup$ oh okay i see now $\endgroup$ Commented Dec 9, 2022 at 4:17

3 Answers 3

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Hint Let us say that a chord is of type $d\in [0,1]$ if its distance to the origin of the circle is $d$. First determine the density of chords of type $d$ in the set of all the random chords. This should not be too difficult.

Then consider the following figureenter image description here

Consider 4 cases, $d\in [0, r_1]$ or $[r_1, r_2]$ or $[r_2, r_3]$ or $[r_3, 1]$ and in each case determine the proportion of chords of type $d$ intersecting the segment. For this one only needs to draw lines passing throw the ends of the segment that are tangent to the circle of radius $d$, these lines determine one or two portions of the unit circle corresponding to chords that intersect the segment. Thus one can obtain the probability that a chord of type $d$ intersects the segment. For example if $d\ge r_3$, the probability is $0$.

It remains to integrate these results over $d$ to get the probability that a chord intersects the segment.

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  • $\begingroup$ If I understand you correctly, you are showing how to find the probability that a line segment of specified length and position will be intersected by a random chord. I already know how to find that probability (see my "Attempt 1"). My question is, where should a line segment of fixed length be drawn, so that the probability of being intersected by a random chord is maximized? It seems that the line segment should be drawn as a chord, but it seems difficult to prove this. $\endgroup$
    – Dan
    Commented Dec 10, 2022 at 15:52
  • $\begingroup$ Well I don't see a usable formula for the probability in the "Attempt 1". My hope is that the above method will produce a better formula to study the maximization problem. $\endgroup$ Commented Dec 10, 2022 at 15:59
  • $\begingroup$ OK, I'll have a try, thanks. $\endgroup$
    – Dan
    Commented Dec 10, 2022 at 16:01
  • $\begingroup$ I tried to use your method to get an expression for the probability that the line segment will be intersected by a random chord, but it does not seem to make the maximization problem any easier than my "Attempt 1". $\endgroup$
    – Dan
    Commented Dec 11, 2022 at 1:56
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Not an answer to the main question, but a rough sketch for why the limit should be $\frac{\pi}{2}$ as $L\rightarrow 0$.

When the line segment is a chord of the circle, the segment splits the circumference into two arcs: a small arc of length $2\arcsin(L)$, and a big arc of length $2\pi - 2\arcsin(L)$. A random chord intersects this segment iff it has one endpoint on the small arc and one on the big arc, which happens with probability:

$$p_{\text{chord},L}= 2\frac{\arcsin(L)}{\pi}\left(1-\frac{\arcsin(L)}{\pi}\right)$$

When $L$ is small, we have $p_{\text{chord},L}\approx \frac{2}{\pi}L$.

Now suppose the line segment is in the center, with endpoints $A=(0,L)$ and $B=(0,-L)$. Conditioning on $P=(\cos(\theta),\sin(\theta))$ as the first endpoint of the random chord, the probability the random chord intersects $AB$ is $\frac{2\angle APB}{2\pi}$.

By symmetry, we only need to average over the first quadrant:

$$p_{\text{center},L}=\frac{2}{\pi}\int_0^{\frac{\pi}{2}} \frac{\angle APB}{\pi} d\theta=\frac{2}{\pi}\int_0^{\frac{\pi}{2}} \frac{\text{arccot}(\tan\theta - L\sec(\theta))-\text{arccot}(\tan\theta + L\sec(\theta))}{\pi} d\theta.$$

Then \begin{align*} \lim_{L\rightarrow 0+}\frac{p_{\text{center},L}}{L}&= -\frac{2}{\pi^2}\int_0^{\frac{\pi}{2}}\left[\lim_{L\rightarrow 0+}\frac{\text{arccot}(\tan\theta + L\sec(\theta))-\text{arccot}(\tan\theta - L\sec(\theta))}{L}\right]d\theta\\ &= \frac{4}{\pi^2}\int_0^{\frac{\pi}{2}}\cos(\theta)d\theta\\ &= \frac{4}{\pi^2},\end{align*} i.e. when $L$ is small, $p_{\text{center},L}\approx \frac{4}{\pi^2}L$.

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From the OP's "Attempt $2$":

$P(t)=\dfrac{2}{\pi^2}\int_0^{\pi}\arcsin{\left(\dfrac{r}{\sqrt{1+t^2-2t\cos{x}}}\right)}dx$

$\overset{r\to0}{\to}\dfrac{2}{\pi^2}\int_0^{\pi}\dfrac{r}{\sqrt{1+t^2-2t\cos{x}}}dx$

$\overset{x=2u}{=}\dfrac{4r}{\pi^2}\int_0^{\pi/2}\dfrac{1}{\sqrt{(1+t)^2-4t\sin^2{u}}}du$

which for $0<t<1$ is equivalent to $\dfrac{4r}{\pi^2}\int_0^{\pi/2}\dfrac{1}{\sqrt{1-t^2\sin^2{u}}}du$, which is increasing in $t$.

As the OP's "Attempt $2$" explains, this implies that the line segment should be drawn as a chord, to maximize the probability that the line segment will be intersected by the random chord.

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  • $\begingroup$ I don't think it is correct to say that the probability that the disk is intersected by the random chord is $\alpha/\pi$ in the Attempt 2. It means that you may start with the wrong formula. $\endgroup$ Commented Dec 17, 2022 at 15:13
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    $\begingroup$ @Gribouillis I still think $\alpha/\pi$ is correct. In Attempt 2, the diagram shows the first random point on the circle, and lines through this point and tangent to the disk. We can extend these lines so that they intersect the circle on the other side of the disk. If the second random point on the circle is between those two intersections, then the chord will intersect the disk. The arc length between those two intersections is $2\alpha$, and the circumference of the circle is $2\pi$, so the probability that the chord will intersect the disk is $\alpha/\pi$. $\endgroup$
    – Dan
    Commented Dec 18, 2022 at 0:39
  • $\begingroup$ All right, I agree that the probability is $\alpha/\pi$ now. I'm working on understanding the rest of the argument... $\endgroup$ Commented Dec 18, 2022 at 9:00

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