The vertices of a pentagram are five uniformly random points on a circle.
Is the following conjecture true: The probability that the pentagram contains the circle's centre is $\frac38$.
(The pentagram is said to contain the circle's centre if the central pentagon, or any of the five triangles adjecent to the central pentagon, contains the circle's centre.)
A simulation with $10^7$ such random pentagrams yielded a proportion of $0.3750079\approx1.00002\times\frac38$ containing the circle's centre. Thus, my conjecture.
How to set up a simulation
Let the circle be $x^2+y^2=1$ with centre $O\space(0,0)$.
Assume the first random point is $(1,0)$ and let the other four random points be, going anticlockwise from $(1,0)$: $(\cos\theta_1,\sin\theta_1)$, $(\cos\theta_2,\sin\theta_2)$, $(\cos\theta_3,\sin\theta_3)$, $(\cos\theta_4,\sin\theta_4)$.
Let $A,B,C,D,E$ be the five regions in the circle and outside the pentagram, going anticlockwise starting with the region between $(1,0)$ and $(\cos\theta_1,\sin\theta_1)$.
- $O$ lies in $A$ if and only if $\theta_2>\pi,$ and $\theta_4-\theta_1<\pi$.
- $O$ lies in $B$ if and only if $\theta_2>\pi,$ and $\theta_3-\theta_1>\pi$.
- $O$ lies in $C$ if and only if $\theta_3-\theta_1>\pi$, and $\theta_4-\theta_2>\pi$.
- $O$ lies in $D$ if and only if $\theta_3<\pi,$ and $\theta_4-\theta_2>\pi$.
- $O$ lies in $E$ if and only if $\theta_3<\pi,$ and $\theta_4-\theta_1<\pi$.
The pentagram contains $O$ just if $O$ lies in none of $A,B,C,D,E$.
If my conjecture is true, then, given the simplicity of the probability, there might be a proof based on some kind of symmetry.
Best Answer
Fix some circle (e.g. unit circle as a 1D variety), and a probability measure on it. The probability space we start with is the one of all $P=(P_0,P_1,P_2,P_3,P_4)$ tuples of five points on the circle. Denote this configuration space by $C(5)$. Measure on spaces (derived from probability spaces) will always have total mass zero. Let $\Pi=\Pi(P)=\Pi(P_0,P_1,P_2,P_3,P_4)$ be the "solid" area / set of points obtained in the following manner.
We will write below some integrals on configuration spaces, the mass will be omitted, (so we simply write $\int f$ as a "functional in $f$" instead of the measure theoretical notation $\int f\; d\mu$). The passage from one configuration space to the other is done by the transport formula.
Then we can compute the probability $p$ that $O$ is inside $\Pi(P)$.
The complement, opposite event is the one where $O$ is in one of the regions denote by OP as "region $A,B,C,D,E$". Each such region has the same probability to contain $O$, and exactly one contains it. By rotational symmetry we compute the probability $q$ that $O$ is in the "wrong half-space" w.r.t. both lines $A_0A_2$ and $A_1A_3$, i.e. that the corresponding chords/segments have lengths $\ge 1$. The probability $q$ is thus: $$ \begin{aligned} q & =\int_{C(5)}1_{O\text{ in region $A$ w.r.t. }\Pi(P)} \\ &=\int_{C_0(5)}1_{O\text{ in region $A$ w.r.t. }\Pi(A)} \\ &=\int_{\Delta(5)}1_{O\text{ in region $A$ w.r.t. }x} \\ &=\int_{\Delta(5)}1_{\substack{ 0=x_0\le x_1\le x_2\le x_3\le x_4\le 2\\ x_2-x_0 \ge 1\\ x_3-x_1 \ge 1 }} \\ &= \frac {\operatorname{Volume} \left( \begin{split} x\ :\ 0=x_0\le x_1\le x_2\le x_3\le x_4\le 2 \\ 1\le x_2\\ 0\le x_1\le 1\\ 1+x_1, x_2\le x_3\le 2\\ x_3\le x_4\le 2 \end{split} \right)} {\operatorname{Volume}(x\ :\ 0=x_0\le x_1\le x_2\le x_3\le x_4\le 2)} \\ &= \frac 1{2^4/4!} \int_1^2 dx_2 \int_0^1dx_1 \int_{\max(1+x_1, x_2)}^2dx_3 \int_{x_3}^2dx_4 \\ &= \frac 1{2^4/4!} \int_1^2 dx_2 \int_0^{x_2-1}dx_1 \int_{x_2}^2dx_3 \int_{x_3}^2dx_4 + \frac 1{2^4/4!} \int_1^2 dx_2 \int_{x_2-1}^1dx_1 \int_{1+x_1}^2dx_3 \int_{x_3}^2dx_4 \\ &= \frac 1{2^4/4!} \int_1^2 dx_2 \int_0^{x_2-1}dx_1 \int_{x_2}^2(2-x_3)\;dx_3 + \frac 1{2^4/4!} \int_1^2 dx_2 \int_{x_2-1}^1dx_1 \int_{1+x_1}^2(2-x_3)\;dx_3 \\ &= \frac 1{2^4/4!} \int_1^2 dx_2 \int_0^{x_2-1}\frac 12(2-x_2)^2\;dx_1 + \frac 1{2^4/4!} \int_1^2 dx_2 \int_{x_2-1}^1\frac 12(1-x_1)^2\;dx_1 \\ &= \frac 1{2^4/4!} \int_1^2 \frac 12(2-x_2)^2(x_2-1)\;dx_2 + \frac 1{2^4/4!} \int_1^2 \frac 16(2-x_2)^3\;dx_2 \\ &= \frac 1{2^4/4!} \left(\frac 1{24}+\frac 1{24}\right) = \frac{24}{16}\cdot \frac 2{24} =\frac 2{16}=\frac 18\ . \end{aligned} $$ This leads to the value of the wanted probability $p$: $$ \bbox[lightyellow]{\qquad p=1-5q=1-\frac 58=\frac 38\ .\qquad} $$
Bonus: Let us also compute the probability $p^*$ that $O$ is in the inner pentagon $\Pi^*$. $$ \begin{aligned} p^* & =\int_{C(5)}1_{O\in \Pi(P)} =\int_{C_0(5)}1_{O\in \Pi(A)} =\int_{\Delta(5)}1_{\substack{ 0=x_0\le x_1\le x_2\le x_3\le x_4\le 2\\ x_2-x_0 \le 1\\ x_3-x_1 \le 1\\ x_4-x_2 \le 1\\ 2 - x_3 \le 1\\ 2+x_1-x_4\le 1 }} \\ &= \frac {\operatorname{Volume}\left( \begin{split} x\ :\ 0=x_0\le x_1\le x_2\le x_3\le x_4\le 2 \\ x_2\le 1\\ x_3-x_1 \le 1\\ x_4-x_2 \le 1\\ 2 - x_3 \le 1\\ 2+x_1-x_4\le 1 \end{split} \right)} {\operatorname{Volume}(x\ :\ 0=x_0\le x_1\le x_2\le x_3\le x_4\le 2)} \\ &= \frac 1{2^4/4!} \int_{x_2\in[0,1]}dx_2 \int_{x_1\in[0,x_2]}dx_1 \int_{x_3\in[1,1+x_1]}dx_3 \int_{x_4\in[1+x_1,1+x_2]}dx_4 \\ &= \frac 1{2^4/4!} \int_{x_2\in[0,1]}dx_2 \int_{x_1\in[0,x_2]}dx_1 \cdot x_1(x_2-x_1) = \frac 1{2^4/4!} \int_{x_2\in[0,1]}\frac 16x2^3\;dx_2 \\ &=\frac 1{2^4/4!}\cdot\frac 1{4!}=\frac 1{2^4}=\frac 1{16}\ . \end{aligned} $$
Simulation: We let $x$ be a random configuration with five components, one of them equal to zero, in the interval $[0,2]$, and check that no two consecutive cords have length bigger one. The code runs through the sage interpreter.
(The
pattern
is something like the sample[False, False, False, True, True]
, and showsTrue
in the bad case of a chord bigger one. The decision to not count is when two cyclically consecutiveTrue
values occur.And this time i've got:
We can modify the do-not-count condition to also simulate $p^*$, by just testing that there is no
True
entry in thepattern
, sook = True not in pattern
, a one liner.And this time i've got:
Which supports the computed result in the bonus part...
The moral is that "similar probabilities" (using possibly more points, and special patterns of the position of the center $O$ w.r.t. the sides and diagonals of the cyclically ordered polygon) are volumes of polyhedra (in more dimensions), and can be computed via Fubini as above.