On a regular two dimensional coordinate plane, you have a square with side length 1 unit.

Pick a point within the square at random, and from there travel a random but straight direction .5 units.

What is the probability that you end up still within the square?

I defined the square in the range - 0.5 < x < 0.5 and -0.5
The points of type 1, haven two angle zones where the end point lies outside. The points of type 2 only have an angle zone where the end point lies outside.

To find the probability (already multiplied by four):

Type 1:

The integral from x=0 to x=0.5 of
the integral from y=0 to 0.5-sqrt(0.5*0.5-(x-0.5)*(x-0.5)) of

(
2*pi
- 2*arcsin( sqrt(0.5*0.5-(0.5-x)(0.5-x))/0.5)
- 2*arcsin( sqrt(0.5*0.5-(0.5-y)(0.5-y))/0.5)
) /2/pi * dx * dy /0.25

Type 2:

The integral of x=0 to x=0.5 of
the integral of y=0.5-sqrt(0.5*0.5-(x-0.5)(x-0.5)) of

(
2*pi
-pi/2
- arcsin( sqrt(0.5*0.5-(0.5-y)(0.5-y))/0.5)
- arcsin( sqrt(0.5*0.5-(0.5-x)(0.5-x))/0.5))
) /2/pi * dx * dy /0.25


These two integrals has three parts each that give the result

(4-pi)/4

(pi*pi-12)/16/pi

(pi*pi-12)/16/pi

3Pi/16

(-pi*pi-4)/32/pi

(-pi*pi-4)/32/pi

Adding these 6 parts gives the result of:

1-7/4/pi aproximately 0.4429576992



Pablo Meraz

Posted by Pablo on 2004-01-29 12:30:44