# Moments of area of random triangle inscribed in a circle

The $2m$th moment of the (random) area of the triangle whose vertices are three independent, uniformly distributed random points on the unit circle appears to be $((3m)!/(m!)^3)/16^m$. Can anyone prove this? Better yet, can anyone give a conceptual explanation for why this moment should be rational? (If this observation is not new, references would be appreciated.)

This question was inspired by John Baez's posts https://johncarlosbaez.wordpress.com/2018/07/10/random-points-on-a-sphere-part-1/ and https://johncarlosbaez.wordpress.com/2018/07/12/random-points-on-a-sphere-part-2/.

• Curious: based on what evidence does the answer "appear" to be $((3m!)/(m!)^3/16^m)$? A few symbolic calculations for low $m$, numerical calculations for lots of $m$, or something else? Jul 13 '18 at 22:10
• Symbolic calculations for small $m$, up through $m=7$. Jul 14 '18 at 1:55
• I'm hoping someone looks at moments of the area of a triangle determined by three random points on the $d$-sphere (for $d>1$). Jul 14 '18 at 2:01

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Let $z_0,z_1,z_2$ be the random points on the circle. We may assume that $z_0=1$, $z_1=e^{i\thh_1}$, $z_2=e^{i\thh_2}$, where $\thh_1,\thh_2$ are iid uniformly distributed on $[0,2\pi]$, and so, by the formula $A_\bigtriangleup=\frac12\,ab\sin C$, the area of the random triangle is \begin{equation} A=2\sin U_1\,\sin U_2\,|\sin(U_1+U_2)|, \end{equation} where the $U_i:=\thh_i/2$'s are iid uniformly distributed on $[0,\pi]$. So, in view of the Euler formula $\sin\thh=(e^{i\thh}-e^{-i\thh})/(2i)$, all the even moments \begin{equation} \E A^n=\frac1{\pi^2}\iint_{[0,\pi]^2}(2\sin u_1\,\sin u_2\,\sin(u_1+u_2))^n\,du_1\,du_2\, \end{equation} (with even natural $n$) are rational numbers, which are actually dyadic, with denominators being natural powers of $2$. This is so because the integrand, equal \begin{equation} (e^{-2i u_1} - e^{2i u_1} - e^{-2i u_1 - 2i u_2} + e^{-2i u_2} - e^{2i u_2} + e^{ 2 iu_1 + 2 iu_2})^n/(-4i)^n, \end{equation} is a polynomial in $e^{2iu_1},e^{2iu_2},e^{-2iu_1},e^{-2iu_2}$ with dyadic rational coefficients, whereas \begin{equation} \frac1{\pi^2}\iint_{[0,\pi]^2}e^{2iku_1}\,e^{2i\ell u_2}\,du_1\,du_2 \end{equation} is an integer (actually, $0$ or $1$) for any integers $k,\ell$.

In particular, for $n=2,4,6$ we have the respective values $3/8,45/128,105/256$ of $\E A^n$.

• ... and 45/126 isn't in reduced form anyway... Jul 13 '18 at 22:36
• @j.c. : Thank you for spotting the typo. It should be $128$ instead of $126$, which is now fixed, and then, according to my calculations, the values for $n=2,4,6$ are the same as in the conjectured formula. Jul 13 '18 at 22:42

If $q$ is uniformly distributed on the unit circle, $E[q^n]=\delta_{n0}$. So if $f$ is meromorphic, $E[f(q)]$ is the constant term in $f$.

The area of triangle $pqr$ is $\frac12\Im[(q-p)(\bar{r}-\bar{p})]$. By symmetry we can say $p=1$. For complex numbers on the unit circle, $\bar{z}=z^{-1}$. Also use $\Im(z)=(z-z^{-1})/(2i)$. So if $q$ and $r$ are both uniformly distributed we want the constant term in $$\frac{1}{(4i)^{2n}}\left((q-1)(\frac 1r-1)-(\frac1q-1)(r-1)\right)^{2n}$$ Rearranging gives $$\frac{1}{(4i)^{2n}}(r-1)^{2n}(q-1)^{2n}\left(\frac1r-\frac1q\right)^{2n}$$ The constant terms can only arise by picking an $j$th power of $r$ from $(r-1)^{2n}$ and a $(2n-j)$th power of $q$ from $(q-1)^{2n}$. So the final result is $$\frac{1}{(-16)^{n}}\sum_{j=0}^{2n}(-1)^j{2n\choose j}^3$$ As pointed out in the comments, the equivalence to the original formula follows from Dixon's formula. I like the proof at Wikipedia using the MacMahon-Master theorem.

• According to Mathworld this sum is treated in de Bruijn, N. G. Asymptotic Methods in Analysis. New York: Dover, 1981. See also OEIS A006480
– j.c.
Jul 13 '18 at 23:41
• There’s a factor of 1/2 in the triangle area formula you don’t mention initially, though you do apply it later. And though I believe the expression after “Rearranging gives” is correct, the expression before seems wrong to me. $(q-1)(1/r-1) = 1-q-1/r+q/r$, and then mapping $z-z^{-1}$ on each term gives $1/q-q-1/r+r+q/r-r/q$. Jul 13 '18 at 23:54
• @GregEgan Thanks. I think I was missing a $q$ in that expression. Jul 13 '18 at 23:58
• The evaluation of this sum was treated by Dixon, "On the sum of the cubes of the coefficients in a certain expansion by the binomial theorem" in Messenger of Math. 20, 79 (1891)! See also en.wikipedia.org/wiki/Dixon%27s_identity. Jul 14 '18 at 17:33

The result follows from Dyson's Conjecture (A) from Part I, p.151 of Dyson, Freeman J., Statistical theory of the energy levels of complex systems. I-III, J. Math. Phys. 3, 140-156, 157-165, 166-175 (1962). ZBL0105.41604..

This conjecture follows from a famous integral of Selberg, and there's an overview with more references in Section 1.2 of Cacciatori, S. L.; Dalla Piazza, F.; Scotti, A., Compact Lie groups: Euler constructions and generalized Dyson conjecture, Trans. Am. Math. Soc. 369, No. 7, 4709-4724 (2017). ZBL1360.22009.

The relevant case $N=3$ of Dyson's conjecture, and hence the proof of the desired identity, is discussed by Dyson on p.152 of his paper above. Dyson traces this case back to Ramanujan's first letter to Hardy (1913) and an even earlier identity proven by Dixon in 1891 -- the series mentioned in the answer by Dan Piponi.

Here's the connection between the area-moment question and Dyson's conjecture and modern reformulations.

Consider three points $z_1, z_2, z_3 \in U(1)$, with $z_j = e^{i \theta_j}$ for $j = 1,2,3$. Let $A = A(z_1, z_2, z_3)$ be the signed area of the oriented triangle they bound. Partitioning the triangle into three triangles with vertices $0$, $z_j$, $z_k$, for $j \neq k$, one finds $$2 A = \sin(\theta_2 - \theta_1) + \sin(\theta_3 - \theta_2) + \sin(\theta_1 - \theta_3).$$ Note that, for all $j,k$, $$\sin(\theta_j - \theta_k) = \Im(z_j / z_k) = \frac{z_j / z_k - z_k/z_j}{ 2i}.$$ This gives a nice area formula that's a polynomial in the $z_1, z_2, z_3$ and their inverses. $$A = \frac{1}{4i} \left(z_1 z_2^{-1} - z_2 z_1^{-1} + z_2 z_3^{-1} - z_3 z_2^{-1} + z_3 z_1^{-1} - z_1 z_3^{-1} \right).$$

Let $T = U(1)^3$, with Haar measure normalized so that $Vol(T) = 1$. Then the question asks about the integral $$E(A^{2m}) = \int_T A(z_1, z_2, z_3)^{2m} d z_1 dz_2 dz_3.$$

View $T$ as the standard maximal torus in the compact Lie group $G = U(3)$. Let $W = S_3$, the permutation group on $\{ 1,2,3 \}$, and the Weyl group of $G$ with respect to $T$. Choose the usual system of positive roots $\{ \alpha, \beta, \alpha + \beta \}$, with simple roots $$\alpha(\vec z) = z_1 / z_2, \quad \beta(\vec z) = z_2 / z_3.$$ The half-sum of positive roots is $\rho(\vec z) = z_1 z_3^{-1}$ and the Weyl denominator is (miraculously), $$\Delta(\vec z) = - 4i A(\vec z).$$

So we have $$E(A^{2m}) = \frac{1}{16^m} \int_T \lvert \Delta(t) \rvert^{2m} dt.$$ This is the integral considered by Dyson (or very close to it). Or in terms of MacDonald's conjecture for $U(3)$ (proven for more general root systems by Opdam), $$E(A^{2m}) = \frac{1}{16^m} \left( {3m} \atop m \right) \left( {2m} \atop m \right) = \frac{(3m)!}{16^m \cdot m! m! m!}.$$

• Combining your and Dan Piponi's answers gives a proof of the formula for $\sum_{j=0}^{2n} (-1)^j {2n \choose j}^3$; is this a known proof? Jul 16 '18 at 16:01
• I'm not sure -- I'd guess that people like Opdam and Dyson are aware of the fact that the modern MacDonald-conjecture style formulas specialize to Selberg's integrals, which are equivalent in a special case to Dixon's sum. Maybe ask Opdam directly? Jul 16 '18 at 21:17

Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.

I claim that for $d\ge 2$ and $m\ge 1$:

$$E(A_d^{2m})=\frac{3}{4^m} \prod _{q=1}^{m-1} \frac{3 d+6 m-2 q-6}{d+2 m-2 q-2}\prod _{q=1}^m \frac{d+2 m-2 q-1}{d+2 m-2 q}\\ = \frac{3\ \Gamma \left(\frac{d}{2}\right)^2 \Gamma \left(\frac{d-1}{2}+m\right) \Gamma \left(\frac{3 d}{2}+3 m-3\right)}{4^m\ \Gamma \left(\frac{d-1}{2}\right) \Gamma \left(\frac{d}{2}+m-1\right) \Gamma \left(\frac{d}{2}+m\right) \Gamma \left(\frac{3 d}{2}+2 m-2\right)}$$

For $d=2$ (and any $m\ge 1$) this simplifies to the established formula:

$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$

For $m=1$ (and any $d\ge 2$) it simplifies to:

$$E(A_d^2)=\frac{3(d-1)}{4d}$$

To prove the general formula, first note that the squared area of a triangle can be described in terms of a Grammian determinant:

$$A_d^2 = \frac{1}{4} \det{\left(s_i \cdot s_j\right)}$$

where the triangle has vertices $v_0, v_1, v_2$ and:

$$s_i = v_i - v_0, \: i=1,2$$

For $d\ge3$, we can always rotate a triangle with vertices on the sphere into the configuration:

$$\begin{array}{rcl} v_0 & = & e_0 \\ v_1 & = & \cos(\theta_1)\, e_0 + \sin(\theta_1)\, e_1 \\ v_2 & = & \cos(\theta_2)\, e_0 + \sin(\theta_2)\cos(\phi_2)\, e_1 + \sin(\theta_2)\sin(\phi_2)\, e_2 \end{array}$$

The expectation values of the even moments can then be expressed as an integral over three coordinates of a suitably weighted version of the squared area raised to a power:

$$E(A_d^{2m}) = \frac{(d-2)\ \Gamma \left(\frac{d}{2}\right)}{2 \pi ^{3/2}\ \Gamma \left(\frac{d-1}{2}\right)} \int_{0}^\pi \int_{0}^\pi \int_{0}^\pi (A_d^2)^m \sin(\theta_1)^{d-2} \sin(\theta_2)^{d-2} \sin(\phi_2)^{d-3} \,d\theta_1\,d\theta_2\,d\phi_2$$

where the vertex $v_1$ is a representative of a $(d-2)$-sphere of radius $\sin(\theta_1)$ over which it can be rotated while keeping $v_0$ fixed, and $v_2$ is a representative of a $(d-3)$-sphere of radius $\sin(\theta_2)\sin(\phi_2)$ over which it can be rotated while keeping $v_0, v_1$ fixed, and the weights incorporate the measures of these spheres with respect to the whole $(d-1)$-sphere.

For $m=1$, the integrand expands as a sum of products of non-negative integer powers of sines, and the integral can be carried out explicitly to obtain:

$$E(A_d^2)=\frac{3(d-1)}{4d}$$

For $m\ge 2$, we can integrate by parts to obtain the recursion relation:

$$E(A_d^{2(m+1)}) = \frac{(d-1) (3 d+4 m)}{4 d^2} E(A_{d+2}^{2m})$$

The general formula then follows by induction.

Although we derived this formula for even moments, it also gives correct values for odd moments using half-integer values for $m$, including the average area if we set $m=1/2$:

$$E(A_d) = \frac{3\ \Gamma \left(\frac{3 (d-1)}{2}\right) \Gamma \left(\frac{d}{2}\right)^3}{2\ \Gamma \left(\frac{d-1}{2}\right)^2 \Gamma \left(\frac{d+1}{2}\right) \Gamma \left(\frac{3 d}{2}-1\right)}$$

For example:

$$\begin{array}{rcl} E(A_2) & = & \frac{3}{2\pi} \\ E(A_3) & = & \frac{\pi}{5} \end{array}$$

Though this does not directly address the moments question, I was curious as to whether the full probability distribution for the triangle area could be written in closed form. It turns out that this is possible, and I derive the result in the answer to the question here:

To summarise:

$$P(A) = \frac{4\sqrt{2}\,(a+1)\,K'\left(\frac{2 k(a) i}{g(a) + k(a) i}\right)}{\pi^2 \sqrt{a(a+2)}\,\sqrt{g(a) + k(a) i}}$$

where:

$$\begin{array}{rcl} A&=&\frac{\sqrt{a^3 (a+2)^3}}{2 (a+1)}\\ g(a)&=&a^2(a+2)^2-3\\ k(a)&=&(a+1)^3\sqrt{(1-a)(a+3)}\\ K'(m)&=&\frac{1}{\sqrt{m}}\left(K\left(\frac{1}{m}\right)+i K\left(1-\frac{1}{m}\right)\right) \end{array}$$

Here $K$ is the complete elliptic integral of the first kind, and $K'$ is its analytic continuation across the branch cut on the real axis from 1 to $\infty$ that would mess up the result if we used $K$ rather than $K'$.

The parameter $a$ is found by solving the equation above that connects it to $A$, and choosing the real-valued solution that monotonically increases from $0$ to $1$ as $A$ ranges from $0$ to the maximum possible triangle area, $\frac{3\sqrt{3}}{4}$.

The distribution looks like this: 