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?
For angle APB to be obtuse, point P must lie
in the circle with diameter AB. For angle BPC
to be obtuse, point P must lie in the circle
with diameter BC. For both angles to be obtuse,
point P must lie in the intersection of the
circles. If ABCD is a unit square, then the
probability is just the area of the intersection.
2[(1/4)pi(1/2)^2 - (1/2)(1/2)(1/2)]
= (pi-2)/8 ~= 0.1427
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Posted by Bractals
on 2007-01-27 17:08:01 |
Solution
