Draw the semicircle with diameter AB that lies inside of the square. Angle APB is obtuse if and only if P lies in the region bounded by side AB and the semicircle. Similarly, by drawing the semicircle with diameter BC that lies inside of the square, angle BPC is obtuse if and only if P lies in the region bounded by side BC and this second semicircle. So both angles will be obtuse if and only if P lies in the closed region R bounded by both semicircles.
Without loss of generality, let the midpoint of side AB have coordinates (0,0) and point B have coordinates (r,0). So the square has area 4r^2 and the probability p of two obtuse angles equals the area of R divided by 4r^2. Let I(w) denote the integral from 0 to r of w dx. Since R is bounded by y1=sqrt(r^2-x^2) above and y2=r-sqrt(2rx-x^2) below,
p=(1/(4r^2))I(y1-y2)=(1/(4r^2))(I(y1)-I(y2))=
(1/(4r^2))(.25(pi)r^2 +.25(pi)r^2 - r^2)=(pi-2)/8. |
