>>14868166

Reiterating that the answer is 1/4. I may give the analytic proof soon if no one else does it.

Starting from my parametric form {R cos(t) + x0, R sin(t) + y0} you get

{R cos(t1) + x0, R sin(t1) + y0} = {x1,y1}

{R cos(t2) + x0, R sin(t2) + y0} = {x2,y2}

{R cos(t3) + x0, R sin(t3) + y0} = {x3,y3}

This is 6 equations (3 vector equations with 2 components each) for 6 unknowns (R, x0,y0,t1,t2,t3)

Use the x-component of the first equation to solve for t1, then plug that t1 into the y component of the first equation e.g.

t1 = arccos((x1 -x0)/R)

sin(t1) = sqrt{1 - (x1-x0)^2/R^2}

So the y component of the first equation reads

sqrt{R^2 - (x1-x0^2)} + y0 = y1

In this way you can eliminate the three times t1,t2,t3 and obtain 3 polynomial equations that determine x0,y0, and R

then you have to compute a lot of Jacobians and integrate over dx1,dx2,dx3,dy1,dy2,dy3 over the unit square