What planar convex shape maximizes the probability that a random circle contains the centre? (Surprisingly, it's not a disk.)

Question

Given a planar convex shape (i.e. lamina) $S$, we define a "random circle" as the circle with diameter endpoints at two independent uniformly random points in $S$.

Let $P(S)$ be the probability that a random circle contains the centre of $S$. "Centre" can be defined however we want; if it is not explicitly defined, then assume it means the centre of mass.

What $S$ has the maximum value of $P(S)$, and what is that maximum value?

Surprisingly, the optimal shape is not a disk

I have found the following results. For the first three, I give proofs (which are supported by simulations); the last is only based on simulation.

• $P(\text{disk})=\dfrac12$
• $P(\text{square})=\dfrac12$
• $P(\text{equilateral triangle})=\dfrac13+\dfrac{\ln 3}{6}\approx 0.51643538$
• $P(\text{Reuleaux triangle})\approx0.50396$ from about $4\times 10^6$ trials

Disk

Suppose the first random point is the red point. The random circle will contain the centre just if the second random point is in the blue region. $\space\therefore P(\text{disk})=\dfrac12$.

Square

Essentially the same as the disk. Any line through the centre of the square divides the square into two regions of equal area. $\space\therefore P(\text{square})=\dfrac12$.

Equilateral triangle

Suppose the first random point is the red point. The random circle will contain the centre of the triangle just if the second random point is in the blue region.

Assume the side length of the triangle is $2$. Let $f(x)$ be the area of the blue region. Using angle chasing, the sine rule and the formula for the area of a triangle $\frac12ab\sin C$, we get

$f(x)= \begin{cases} \dfrac{4}{3\sqrt{3}\left(1-x^{2}\right)}&0\le x<\dfrac13\\ \dfrac{9x^{2}+12x-1}{6\sqrt{3}x\left(x+1\right)}&\dfrac13\le x\le 1 \end{cases}$

$\therefore P(\text{equilateral triangle})=\dfrac{1}{\sqrt3}\displaystyle\int_0^1f(x)dx=\dfrac12+\dfrac{\ln 3}{6}\approx 0.51643538$

(Note that, for uniform points in the equilateral triangle, $x$ is uniform in $[0,1]$. To see this, in the right triangle in the diagram, draw many cevians from the top vertex to the horiztonal base, such that their intersections along the base are uniform. The resulting thin sub-triangles have equal area, and thus equal probability of containing a random point in the equilateral triangle.)

Answer 1

Edit: leaving this (partial) answer to the original question since it led the OP to require convexity.

Point $P$ is in the disk with diameter $XY$ if and only if angle $XPY$ is at least a right angle. (Euclid Book III Proposition 31 - I've always like that proposition and am delighted to find a use for it).

Suppose $A$, $B$ and $C$ in this picture have equal area.

Let $S = A \cup B \cup C$. Then the probability that two randomly chosen points $X,Y$ in $S$ are in the same subregion is $1/3$ so the probability that $P$ is in the circle with diameter $XY$ is $2/3$.

You can choose $A$, $B$ and $C$ so that $P$ is the centroid of $S$, but that's not necessary.

The most symmetrical arrangement would place the regions containing $A$, $B$ and $C$ symmetrically with angle $\pi/6$.

If you shrink the black regions in this radiation warning logo to that angle and shrink the connecting central disk you can make a symmetrical connected example with probability arbitrarily close to $2/3$.

I conjecture that $2/3$ is the maximum for the original question and that the equilateral triangle is the only $S$ with the maximum for convex sets.

Thanks to the OP for an interesting question and to several commenters for hints at this general almost answer.

Answer 2

Too long for a comment.

A different way of framing this problem might make it easier to think about (or perhaps harder, haha). The convex shape can be parametrized as the curve $r=f(\theta)$ of a $2\pi$-periodic function in polar coordinates. The assumptions of the problem can be expressed as constraints on the function $f$. For instance, if we normalize by assuming the shape has unit area (in order to make probability calculations simpler), we get the constraint $$\int_0^{2\pi} \tfrac{1}{2}f^2(\theta) d\theta = 1 \tag{i}$$ and per this question the convexity constraint can be written as a differential inequality in $f$: $$f^2 + 2f'^2 - ff'' \ge 0 \tag{ii}$$ Further, the quantity to be optimized can be written as a double-integral in $f$, using the "blue region" trick from the OP's question: $$G(f) := \frac{1}{4}\int_{0}^{2\pi} \int_{\theta+\pi/2}^{\theta+3\pi/2} f^2(\theta)f^2(\phi)d\phi d\theta$$ So this becomes a functional optimization problem, namely optimizing $G$ subject to (i) and (ii). (Even if we don't want to assume that $f$ is smooth, smooth functions should at least get $G$ arbitrarily close to the true optimum.)

(Note that this formulation allows the "center" to be any point inside the region. This was left ambiguous in the OP's question, though some people are assuming it means the center of mass. This approach does not take that into account.)

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This is not relevant to the current formulation of the question, but to add to a previous answer, it appears that under the weaker condition of star convexity we can still get arbitrarily close to a probability of $2/3$ by using the following family of functions $f$ as $\epsilon\to 0$: $$f(\theta) = \frac{1}{1+\epsilon +\sin(3\theta)}$$ ...at least based on numerical experiments. I have not checked analytically that this gives $G(f)\to 2/3$ as $\epsilon\to 0$, but I suspect that this is so.

I suspect analyzing the Fourier coefficients of $f$ or $f^2$ might yield some insights here, though I'm not sure how (ii) can be expressed in terms of Fourier coefficients. They do allow us to prove some interesting related facts more easily though: for instance, from the Fourier series I believe we can show that if the convex region has 2-fold rotational symmetry about the chosen center point, then the probability is exactly $1/2$ (of which the square and circle are special cases).

Answer 3

Here's a method for regular polygons with odd numbers of sides. The figure below shows examples of regular $n$-gons for $n = 3,$ $n = 5,$ and $n = 7.$

In each figure the red point represents one end of the diameter of the circle, selected as a random point within the lamina. We construct a line segment between the centroid of the lamina and the red point; perpendicular to this segment we construct a line through the centroid. This line divides the lamina into two parts, one of which (the part containing the red point) is unshaded and the other of which is shaded.

The circle contains the centroid if and only if the other end of the diameter lies in the shaded portion of the lamina. Therefore, given the location of one end of the diameter, the conditional probability that the circle contains the centroid is equal to the ratio of the area of the shaded region to the area of the entire lamina.

We can construct a right triangle whose vertices are the centroid, the midpoint of an edge of the lamina, and a vertex on the same edge of the lamina. When the red point is within this triangle the conditional probability depends only on the angle $\theta$ between the segment connecting the red point to the centroid and the adjacent leg of the triangle. Moreover, by symmetry of the lamina the probability for the entire lamina is equal to the conditional probability given that the red point is in this triangle.

An intuition for why the overall probability is not $\frac12$ and a notion of how the probability differs from $\frac12$ depending on $n$ can be developed from the figure below showing a lamina in the shape of a regular pentagon.

Let $C$ be the centroid of the pentagon, $D$ the midpoint of a side, $F$ a vertex on the same side, and $CE$ the bisector of the angle $\angle DCF.$ Let $P$ be the first random point selected. Without loss of generality we can assume $P$ is in the triangle $\triangle CDF.$ We shade the portion of the lamina such that if a point $Q$ is placed in that part of the lamina the circle on diameter $PQ$ contains the centroid $C.$

If $P$ is on the line $CE,$ the shaded part of the lamina is bounded by a line $AB$ through $C$ perpendicular to $CE,$ where $A$ is the midpoint of a side and $B$ is a vertex. In that case the conditional probability that the circle contains the centroid is exactly $\frac12.$ Otherwise, if $P$ is in the triangle $\triangle CEF$ then a triangular region $\triangle ACK$ is added to the shaded region and a triangular region $\triangle BCL$ is removed. For the pentagon, $\triangle BCL$ is larger than $\triangle ACK$; in fact, $\triangle BCL$ can be subdivided into triangle $\triangle CMN$ (which is congruent to $\triangle ACK$) and the quadrilateral $BLMN.$ Therefore the conditional probability is less than $\frac12.$ But if $P$ is in triangle $\triangle CDE,$ the triangular region added to the shaded region is larger than the triangular region removed, producing a conditional probability greater than $\frac12.$

Let $\triangle CEG$ be the reflection of $\triangle CDE$ across $CE.$ By symmetry, the increases in probability for points $P$ in $\triangle CDE$ are exactly offset by the decreases in probability for points $P$ in $\triangle CEG.$ But $\triangle CDF$ can be partitioned into three triangles: $\triangle CDE,$ $\triangle CEG,$ and $\triangle EFG.$ There is a positive probability that $P$ will be in $\triangle EFG,$ and when it is there then the conditional probability is less than $\frac12.$ Therefore the probability that the circle will contain the centroid $C$ is less than $\frac12.$

When we do the corresponding construction on a lamina in the shape of an equilateral triangle, however, when $P$ is in $\triangle EFG$ the region that is added to the shaded part of the lamina is greater than the part that is removed.

In general, we can tell whether the probability for a regular $n$-gon will be greater or less than $\frac12$ by looking at the line through the centroid $C$ perpendicular to $CE.$ If that line passes through the midpoint of a side on the same side of $CE$ as $\triangle EFG$ and through a vertex on the opposite side of $CE,$ the probability will be greater than $\frac12.$ This is the case for all $n$ of the form $n=4k - 1.$ If the vertex is on the same side of $CE$ as $\triangle EFG$ and the midpoint is on the opposite side, the probability will be less than $\frac12.$ This is the case for all $n$ of the form $n = 4k + 1.$

Moreover, intuitively we should expect the probability to be farthest from $\frac12$ when the quadrilateral $BLMN$ for each angle $\theta$ is largest. This occurs for the equilateral triangle. The equilateral triangle therefore maximizes the probability over all regular polygons.

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The following is just a hunch, but it seems to me that the requirement of convexity reduces the upper bound on the probability. That is, convexity requires us to add material to the lamina that brings the probability closer to $\frac12.$

Consider the equilateral triangle, for example. Trisect each edge to construct the vertices of an equilateral hexagon. Divide the triangle into four pieces: that hexagon and three smaller equilateral triangles. The conditional probability that the circle contains the centroid given that both ends of the diagonal lie in the smaller triangles is some number $p > \frac12,$ but the probability given that both ends are in the hexagon is $\frac12$ and so is the probability given that one end is in a small triangle and one is in the hexagon. Since the probability for the entire triangle is a weighted average of $p$ and $\frac12,$ $p$ must be greater than the probability for the entire triangle. We could get as close to $p$ as we want while connecting the three triangles with thin strips, but we cannot get close to $p$ while taking the convex hull of the three triangles.

Taking this hunch to an extreme, we might hope to maximize the probability by removing the requirements of connectedness (which has trivial effect) and convexity (which appears to have a substantial effect). We might even maximize the probability by taking the lamina to be a set of small disks.

I cannot think of a better set of small disks than three disks placed at the vertices of an equilateral triangle. The probability that two random points chosen from these disks give the diameter of a circle containing the centroid is $\frac23.$ Taking the convex hull of these disks we get something like an equilateral triangle while lowering the probability somewhat. Extending the straight edges of that figure to make a true equilateral triangle just increases the probability again due to the chance that at least on end of the diameter is in the added pieces.

The disks could also be placed at vertices of a non-equilateral triangle and still achieve $\frac23$ probability, however, so the argument that the triangle should be equilateral doesn't seem as strong as the argument that the lamina should be a triangle.

Answer 4

Not a complete answer, but at least proves that for regular polygons the equilater triangle is the maximizer. It is probably a maximizer for all convex bodies, but even with polygons symmetric w.r.t. rotation for multiple of $2\pi/3$ it is difficult to prove

Let $G$ the center of the convex set $C$. Given two random points $X$ and $Y$ the centroid is inside the circle iif the angle between $X-G$ and $Y-G$ is obtuse, i.e. if $(X-G) \cdot (Y-G) \le 0$.

Let use a coordinate system with origin in the center. The probability is given by $$ \frac{1}{|C|^2}\iint_{C \times C} 1_{\le 0} (x\cdot y) dx dy $$

Let consider the function $$ r(\theta)= \max\{ \rho : \rho( \cos \theta,\sin\theta) \in C\} $$

In polar coordinates we have $$ \iint_{C \times C} 1_{\le 0}(x \cdot y ) dx dy= \int_0^{2\pi}d \theta \int_{0}^{2\pi} d \phi \int_0^{r(\theta)} d\rho \int_0^{r(\theta)} d\sigma 1_{\le 0}(\rho \sigma \cos(\theta-\phi) ) $$ The cosine is negative iif $\theta +\pi/2 \le \phi \le \theta +3/2 \pi$ and $\int_0^{r(\theta)} \rho d\rho=r(\theta)/2$, so the integral became $$ \frac{1}{4}\int_0^{2\pi} r(\theta)^2 \left(\int_{0}^{2\pi} 1_{(\pi/2,3/2 \pi)}\left( \theta-\phi\right) r(\phi)^2 d\phi\right) d \theta $$

Let call $f(\theta)=r(\theta)^2$, $K(\theta)=1_{(\pi/2,3/2 \pi)}(\theta)$ the function can be written as $\left< f,K *f\right>$, where $*$ is the convolution. By taking the Fourier series $$ \left< f,K *f\right>= 2\pi\sum_{n \in \mathbb{Z}} \hat{K}_n |\hat{f_n}|^2 $$ Is it possible to prove that $$ \hat{K}_0=\pi, \quad \hat{K}_n=(-1)^n \frac{2}{n} \sin \left(\frac{n\pi}{2}\right) $$

So $\hat{K}_n$ is $0$ if $n=2k \ne 0$, $\frac{2}{n}$ for $n=3 \mod 4$ and $-\frac{2}{n}$ for $n=1 \mod 4$. Let $\theta(n)$ such that $\hat{K}_n=\frac{1}{n} \theta(n)$. Using that $|C|= \left< f,1\right>= 2\pi \hat{f_0}$ we arrive to $$ \mathbb{P}(C)= \frac{1}{2} + \frac{1}{\pi^2 |C|^2}\sum_{ n > 0} \theta(n) \frac{|\hat{f_n}|^2}{ n } $$ where i have used $K_n=K_{-n}$

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For a regular polygon with fixed area $|C|$ and $m$ sides $r(\theta)= \sqrt{\frac{2|C|}{m \tan(\pi/m)}} \frac{1}{\cos(\theta-\phi(m))}=r_{in} \frac{1}{\cos(\theta-\phi(m))}$, For $\theta-\phi \in (-\pi/m,\pi/m)$ and extended by periodicity to $[-\pi,\pi]$ ($\phi$ is an angle that does not change the magnitude of the Fourier coefficients and depends only on the number of sides)

By periodicity we have that, for the regular $m$-gon $\hat{f_{n}} \ne 0$ iif $n=km$. This immediatly give us that the only $m$-gons that have probability bigger than $1/2$ are the one with $m=3 \mod 4$ and we have $$ |\hat{f}_{km}|= \left|\frac{2 r^2_{in}}{\pi} \int_0^{\pi/m} \frac{\cos(mk\theta)}{\cos^2(\theta)} d\theta \right| $$

Let $m \ge 5$, by integrating by parts $$ \left|\int_0^{\pi/m} \cos(mk\theta) \sec^2(\theta) d\theta\right| \le \frac{\tan^2(\pi/m)}{mk} $$ Thus $$ \sum_{ k > 0} \theta(km) \frac{|\hat{f_{km}}|^2}{ km } \le 8 |C|^2 \frac{\tan^2(\pi^2/m)}{\pi^2 m^3} \sum_{ \substack{k \ge 1\\ k \text{ odd}}} k^{-3} $$ As $\sum_{ \substack{k \ge 1\\ k \text{ odd}}} k^{-3}= 7/8 \zeta(3)$ and the maximum for $ m \in [5,\infty)$ is at $m=5$ we get $$ \sum_{ k > 0} \theta(km) \frac{|\hat{f_{km}}|^2}{ km } \le 7 |C|^2 \frac{\tan^2(\pi/5)}{ \pi^2 5^3} \zeta(3) $$

Now let $m=3$. The same bound gives us $$ \sum_{ k > 1} \theta(3k) \frac{|\hat{f_{3k}}|^2}{ km } \le \frac{8}{9\pi^2} |C|^2 \left(\frac{7}{8} \zeta(3)-1\right) $$

$$ |\hat{f}_3|= \left|\frac{12}{9 \pi} |C| \left(2 \sqrt{3} - 3 \operatorname{arctanh} (\sqrt{3}/2)\right)\right| $$ (see this) and so $$ \sum_{ k > 0} \theta(3k) \frac{|\hat{f_{3k}}|^2}{ km } \ge \frac{2}{3} |\hat{f}_3|^2-\frac{8}{9\pi^2} |C|^2 \left(\frac{7}{8} \zeta(3)-1\right) $$

But it is possible to check that the lower bound for $m=3$ is bigger than the upper bound for $m \ge 5$. So for regular polygons the equilater triangle is maximal.

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Note, numerically and looking at the Fourier series it seems that the biggest probability is achieved when $|\hat{f_3}|^2$ is maximized. Is it possible to prove that for the polygons of fixed area $|\hat{f}_3|$ is maximum when the polygon is the equilater triangle, but I wasn't able to prove that maximizing $|\hat{f_3}|$ actually maximize the sum