A point P is chosen at random in the interior of a unit square S. Let d(P) denote the distance from P to the closest side of S. The probability that 1/5

Question

A point P is chosen at random in the interior of a unit square S. Let d(P) denote the distance from P to the closest side of S. The probability that 1/5<d(P)<1/3 is equal to m/n, where m and n are relatively prime positive integers. Find m+n.

perl · Answer

hmmm

btaylor · Answer

|dw:1394539080231:dw|
By my reasoning, probability that d(P)=1/5 is the area enclosed by those boxes...4(.2)(.2)+4(.2)(.6)=.16+.48=.64
Using similar reasoning, probability that d(P)=1/3 would be the area enclosed if the lines were 1/3 away from each side, leaving a middle of 1/3 x 1/3, so probability that d(P)=1/3=1-1/9=8/9.

Is that correct?

perl · Answer

we know that d(p) is between 0 and 1/2

perl · Answer

we want d(p) between 1/5 and 1/3

btaylor · Answer

If so, we have:
$$P \left( \frac{1}{5}<P(d)<\frac{1}{3} \right)=\frac{8}{9}-\frac{16}{25}=\frac{200}{225}-\frac{144}{225}=\frac{56}{225}$$
Then, m=56 and n=225 (relatively prime b/c no common divisors), so m+n=281.

perl · Answer

|dw:1394539527330:dw|