Mary picks any three non-collinear points inside a given circle, what is the probability that the circumcircle of these 3 points will be covered by the original circle?
This is from a test question a few months ago.I got the result is $\frac{1}{\pi}$,I don't know my result right.What approaches do you think I could take to solving the step?
The probablity $P$ we seek is $\frac{2}{5}$.
WOLOG, choose the coordinate system so that our circle $\mathcal{C}$ is the unit circle centered at origin.
Let
It is clear the probability $P$ we seek is equal to the conditional probability
$$P = {\bf Pr}\left[\;\mathcal{E}(A,B,C) \mid A,B,C \in \mathcal{C}\;\right] = \frac{{\bf Pr}\left[\; \mathcal{E}(A,B,C)\; \land A,B,C \in \mathcal{C} \;\right]}{{\bf Pr}\left[\; A, B, C \in \mathcal{C}\; \right]} $$
Notice if $\mathcal{E}(A,B,C)$ is true, then $A, B, C \in \mathcal{C}$. This leads to
$$P = \frac{{\bf Pr}\left[\;\mathcal{E}(A,B,C)\;\right]}{{\bf Pr}\left[\; A, B, C \in \mathcal{C}\; \right]} = \frac{L^6}{\pi^3}{\bf Pr}\left[\;\mathcal{E}(A,B,C)\;\right] $$ To compute the probability on RHS of last expression, consider following parametrization for $A, B, C$.
$$[ 0,\infty )^2 \times [0,2\pi)^4 \ni ( \rho, r, \theta, \alpha, \beta, \gamma ) \quad\mapsto\quad \begin{cases} A &= \rho (\cos\theta,\sin\theta) + r( \cos\alpha, \sin\alpha )\\ B &= \rho (\cos\theta,\sin\theta) + r( \cos\beta, \sin\beta )\\ C &= \rho (\cos\theta,\sin\theta) + r( \cos\gamma, \sin\gamma) \end{cases}$$ Geometrically, $\rho(\cos\theta,\sin\theta)$ is the circumcenter and $r$ is the circumradius for the triangle $ABC$. With a little bit of algebra, one can show that $$dAdBdC = |\sin(\alpha-\beta)+\sin(\beta-\gamma)+\sin(\gamma-\alpha)| \rho r^3 d\rho dr d\phi d\alpha d\beta d\gamma$$
In terms of this parametrization, the condition $\mathcal{E}(A,B,C)$ simply becomes $\rho + r \le 1$.
Let $u = \alpha - \beta, v = \beta - \gamma$ and $$\begin{align} \Phi(u,v) &= \sin(u) + \sin(v) - \sin(u+v)\\ &= \sin(\alpha-\beta)+\sin(\beta-\gamma)+\sin(\gamma-\alpha) \end{align}$$ Integrate over $\phi, \gamma$ first and then $\rho, r$, we obtain: $$ \begin{align} P &= \frac{L^6}{\pi^3} \int_{\mathcal{E}(A,B,C)} \frac{dA dB dC}{L^6}\\ &= \frac{(2\pi)^2}{\pi^3} \left[\int_0^1 \rho \left( \int_0^{1-\rho} r^3 dr \right) d\rho \right] \left[\int_0^{2\pi}\int_0^{2\pi} |\Phi(u,v)| du dv\right]\\ &= \frac{1}{30\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi} |\Phi(u,v)| du dv \end{align} $$ Notice
We can split the region of integration for last integral into 4 equal pieces and get:
$$\begin{align} P &= \frac{2}{15\pi}\int_0^{\pi} \int_{-u}^u |\Phi(u,v)| dv du\\ &= \frac{2}{15\pi}\int_0^{\pi} \int_0^u (\Phi(u,v) - \Phi(u,-v)) dv du\\ &= \frac{2}{15\pi}\int_0^{\pi} \int_0^u 2\sin v ( 1 - \cos u) dv du\\ &= \frac{4}{15\pi}\int_0^{\pi} (1-\cos u)^2 du\\ &= \frac{2}{5} \end{align} $$
It turns out this question has been asked before on the web and Giaolino has given an yahoo answer 4 years ago. The probability he get is also $\frac25$.
Following is my reformatting of his answer for easy reference. All reformatting errors will be mine. I have make this answer an community wiki. If anyone notice any mistake in this reformatting, please feel free to fix it.
Let $E$ is be set of triples $A,B,C$ such that the circumcircle lies in the unit disc $D = D(O,1)$.
We want to show that $\int_E dA dB dC = \frac25 \pi^3$.
Let $R$ the radius of the circumcircle of $ABC$ and $G$ its center.
We fix $A$ and we change variables. We suppose that $B,C$ have polar coordinates $(r_b, t_b)$, $(r_c, t_c)$ with $A$ at the origin. So $r_b = |AB|$ and $r_c = |AC|$, $t_b = \angle(Ax, AB)$, $t_c = \angle(Ax,AC)$.
We are going to change variables and use $G$, $s_b = \angle(Gx,GB)$, $s_c = \angle(Gx,GC)$ as the new variables, $G$ having polar coordinates $(r_g,t_g)$ still with $A$ at the origin.
Let's compute the Jacobian going from $(r_b,t_b,r_c,t_c)$ to $(R,t_g,s_b,s_c)$.
We have the following formulas, in which we don't need to worry about signs since we'll only use the absolute value of the determinant. They follow from the half-angle property in the circle.
$$\begin{align} r_b &= 2 R \sin\left(\frac{t_g + \pi - s_b}{2}\right) = 2 R \cos\left(\frac{t_g - s_b}{2}\right)\\ r_c &= 2 R \cos\left(\frac{t_g - s_c}{2}\right)\\ t_b &= \frac{t_g + \pi + s_b}{2} + \frac{\pi}{2} = \frac{t_g + s_b}{2} + \pi\\ t_c &= \frac{t_g + s_c}{2} + \pi.\\ \end{align}$$
The partial derivatives are
$$\begin{bmatrix} 2 \cos\left(\frac{t_g - s_b}{2}\right) & -R \sin\left(\frac{t_g - s_b}{2}\right) & R \sin\left(\frac{t_g - s_b}{2}\right) & 0 \\ 2 \cos\left(\frac{t_g - s_c}{2}\right) & -R \sin\left(\frac{t_g - s_c}{2}\right) & 0 & R \sin\left(\frac{t_g - s_c}{2}\right) \\ 0 & \frac12 & \frac12 & 0\\ 0 & \frac12 & 0 & \frac12\\ \end{bmatrix}$$
The Jacobian is $R \sin\left(\frac{s_b - s_c}{2}\right) = \frac12 |BC|$.
So $$\begin{align} dB dC &= |AB| |AC| dr_b\,dt_b\,dr_c\,dt_c = \frac12 R |AB| |AC| |BC| dR\,dt_g\,ds_b\,ds_c\\ &= \frac12 |AB| |AC| |BC| dG\,ds_b\,ds_c \end{align} $$ where in the final line we switch back to cartesian coordinates for $G$.
We now exchange the order of integration, fix G and take A in polar coordinates with G as the origin.
Then $\frac12 |AB| |AC| |BC| dA,dG,ds_b,ds_c$ becomes $\frac12 |AB| |AC| |BC| dR\,dG\,ds_a\,ds_b\,ds_c$.
We turn to polar coordinates in $G$. Set $r = |OG|$ and $g = \angle(Ox, OG)$
We simplify $s_a,s_b,s_c$ in $a,b,c$ and we are reduced to
$$\frac12 \left(8 R^3 \left| \sin\left(\frac{a-b}{2}\right) \sin\left(\frac{b-c}{2}\right) \sin\left(\frac{c-a}{2}\right) \right| \right) r\,dR\,dr\,dg\,da\,db\,dc$$
with the conditions $r>0$, $R>0$ and $r+R < 1$, and $a,b,c,g \in [0,2\pi]$.
$(R,r)$ and $(a,b,c,g)$ can be separated.
The integral in $(R,r)$ gives $1/120$.
We set $u = a - b$, and $v = b - c$. The angular integral can be written as
$$\left|\sin\left(\frac{u}{2}\right) \sin\left(\frac{v}{2}\right) \sin\left(\frac{u+v}{2}\right)\right| dg\,da\,du\,dv$$
The variables $a$ and $g$ can be pulled out giving a factor $4 \pi^2$.
So far we have $\frac{8}{240}(4 \pi^2) = \frac{2}{15}\pi^2$.
So we are left with showing that
$$\int_0^{2\pi}\int_0^{2\pi} \left|\sin\left(\frac{u}{2}\right) \sin\left(\frac{v}{2}\right) \sin\left(\frac{u+v}{2}\right)\right| du dv = 3 \pi$$ which is straightforward.