This condition does not change (either way) if the droppings move radially to the boundary circle, so let’s instead compute the probability that a random triangle with vertices on a circle is an obtuse triangle (i.e., does not contain the center of the circle).

Start with the circle and three random lines through its center. This gives us 6 points on the circle: 3 antipodal pairs.

We can form a random triangle by choosing one end from each antipodal pair. That would be 8 possible triangles.

Of those 8 choices, 6 of them are obtuse, and two are acute!

So the probability of an obtaining an obtuse triangle is 3/4.

Of course this argument omits the possibilities of right triangles, line segments, or singletons, but these “degenerate cases” all have probability zero.

This solution generalizes to higher dimensions. The probability that a random n-simplex with vertices on the (n-1)-dimensional sphere (boundary of the unit n-ball) avoids the origin is (2^n – 1)/(2^n). Set n = 2 for the pizza, and n=3 for the meatball.

]]>Problems (397, 382)

http://www.projecteureka.org/problem?category=new

http://www.projecteureka.org/problem?category=new&page=2

Regards,

R.