Gianlino's Disc

@Michael Mendrin I've been waiting for you to post this question, since you were the one to originally ask it on Y!A. You may recall that I posted the same question there years after you did, not knowing its "history", so although I always wanted to post the question myself I felt that you had the honors.

If this problem doesn't end up at a 400 point level I will be surprised, and if someone comes up with an alternative solution I will be impressed, to say the least.

THIS IS NOT MY SOLUTION, this is copy pasted from yahoo answers. I'm in the process of latexing it to make it easier to read.

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 = (2/5) 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 $r_c = AC, t_b = (Ax, AB), t_c = (Ax,AC)$ .

We are going to change variables and use G, $s_b =(Gx,GB)$ , $s_c = (Gx,GC)$ 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

\(r_b = 2 R \sin{\dfrac{t_g + pi - s_b}{2}} = 2 R \cos{\dfrac{t_g - s_b}{2}}\)

$r_c = 2 R \cos{\dfrac{t_g - s_c}{2}}$

$t_b = \dfrac{t_g + pi + s_b}{2} + \dfrac{\pi}{2} = \dfrac{t_g + s_b}{2} + \pi$

$t_c = \dfrac{t_g + s_c}{2} + \pi$ .

The partial derivatives are

\color\green{2 \cos{\dfrac{t_g - s_b}{2}}, - R \sin{\dfrac{t_g - s_b}{2}}, R \sin{\dfrac{t_g - s_b}{2}},,,,,,,,0}

$2 \cos{\dfrac{t_g - s_c}{2}}, - R \sin{\dfrac{t_g - s_c}{2}},,,,,,,,, 0,,,,,,,, R \sin{\dfrac{t_g - s_c}{2}}$

,,,,,,,,,0,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,1/2,,,,,,,,,,,,,,,,,,,,,,,,,,,1/2,,,,,,,,,,,,,,,,0,,,,,,,,,,,,,,,

,,,,,,,,,,,,0,,,,,,,,,,,,,,,,,,,,,,,,,,,,,1/2,,,,,,,,,,,,,,,,,,,,,,,,,,,,0,,,,,,,,,,,,,,,,,1/2,,,,,,,,,,,,,,,

The Jacobian is $R \sin{\dfrac{s_b - s_c}{2}} = \dfrac{BC}{2}$

So $dB dC = AB*AC dr_b,dt_b,dr_c,dt_c$

$\Rightarrow R*AB*AC*BC/2 dR,dt_g,ds_b,ds_c$

$\Rightarrow AB*AC*BC/2 dG,ds_b,ds_c$ 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 $AB\times AC \times \dfrac{BC}{2} dA,dG,ds_b,ds_c$

becomes $AB\times AC\times \dfrac{BC}{2} dR,dG,ds_a,ds_b,ds_c$ .

We turn to polar coordinates in G. Set $r = OG$ and $g = (Ox, OG)$

We simplify $s_a,s_b,s_c$ in $(a,b,c)$ and we are reduced to

$\dfrac{8 |\sin{\dfrac{a-b}{2}} \sin{\dfrac{b-c}{2}| \sin{\dfrac{c-a}{2}}}R^3}{2} \times r \quad dR, dr, dg, da, db, dc$

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

$|\sin{\dfrac{u}{2}} \sin{\dfrac{v}{2}} \sin{\dfrac{u+v}{2}}| dg, da, du, dv$ .

The variables a and g can be pulled out giving a factor $4 \pi^2$ .

So far we have $(\dfrac{8}{240})\times 4 \times \pi^2 = \dfrac{2pi^2}{ 15}$ .

So we are left with showing that

$\displaystyle \int_0^{2pi} \int_0^{2pi} |\sin(u/2) \sin(v/2) \sin((u+v)/2)| du, dv = 3 pi$

which is straightforward.

edit The 2 bottom lines of the Jacobian got cut. They are (0,1/2,1/2,0) and (0,1/2,0,1/2)