What's the area of a round can in which a 1" cube thrown into it will land on its center 50% of the time?

Question

The area of the round bottom of the can, that is. Its center is just a point, over which the cube may or may not land.

A round can in which a 1" cube can just fit inside already has area of about 1.5708. The area of the can in this problem has to be larger.

Answer 1

(Corrected)

Hello,

I found that the radius R of the can must be the solution of:

sqrt(4*R^2-1)+1 =
= (pi-4*asin(1/2/R))*R^2

R~=1.4639

so the area A of the can is: A=6.73244

I just leave the cube immobile in the center of the axis, and move the circle in every possible direction having the cube inside.

Sorry for my English.

Answer 2

Note; I'm only writing this because no one else has bothered to answer.

Yet.

Let me think on it a day or so. I have a partial answer, I spent an hour writing, but now I really have to go to bed. I think there are a few errors in it anyway....