Let AB and BC represent two adjacent sides of square ABCD. If P is a randomly selected point inside of the square, and segments PA, PB, and PC are drawn, what is the probability that angle APB and angle BPC are both obtuse?
the area where angle APB will be obtuse will be inside of a semicircle with AB as diameter.
Similarly BPC will be obtuse inside of a semicircle with BC as diameter.
Both will be obtuse in their intersectional ares which will be intersection of the two quarter circles:
<img src="http://i15.tinypic.com/2w7hgg2.jpg">
the solution will be the difference of integrals of the two curves:
y=(1-x²)½ and y= 1+(1-(x-1)²)½ from 0 to 1.
that area divided by four will give the required probability!
comes to an integration i cannot do, maybe theres a standard formula for that fraction of area
Edited on February 2, 2007, 6:30 am
Solution
