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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 diadic, 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 diadic 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 $\E A^n$.

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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 $\E A^n$.

$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=} \newcommand{\tsi}{\tilde\si}$

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 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 natural $n$) are rational numbers, which are actually diadic, 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 diadic (complex) 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 $\E A^n$.

$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=} \newcommand{\tsi}{\tilde\si}$

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 diadic, 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 diadic 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 $\E A^n$.

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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 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 natural $n$) are rational numbers, which are actually diadic, with denominators being natural powers of $2$. This is so because the integrand, equal \begin{equation} (e^{-2 u_1} - e^{2 u_1} - e^{-2 u_1 - 2 u_2} + e^{-2 u_2} - e^{2 u_2} + e^{ 2 u_1 + 2 u_2})^n/4^n, \end{equation} is a polynomial in $e^{2iu_1},e^{2iu_2},e^{-2iu_1},e^{-2iu_2}$ with diadic 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 for any integers $k,\ell$.

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

$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=} \newcommand{\tsi}{\tilde\si}$

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 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 natural $n$) are rational numbers, which are actually diadic, 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 diadic (complex) 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 $\E A^n$.