(I posted this problem earlier but was a little unclear, so I am adding a picture)

Larry chooses a unit square at random inside a 4x4 square, whose sides are parallel to the 4x4 square (shown in red). Laura also chooses a unit square at random, whose sides parallel to the 4x4 square (shown in blue). Find the probability that Larry's and Laura's square intersect.

Picture:https://latex.artofproblemsolving.com/miscpdf/suzrhsxt.pdf?t=1584828425846

MathCuber Mar 21, 2020

#1**+3 **

First, we start with placing the 4x4 square so that one corner is (0,0), and the opposite corner is (4,4). We can define the x and y "coordinates" of Larry's square as x_{1 }and y_{1}. We define for laura's square x_{2 }and y_{2 }for the coordinates. That means that Larry's square and Laura's square will only ever intersect when the intervals of X1 and X2 intersect, and likewise for Y1 and Y2.

Let X_{1} equal to [a, a+1] and X_{2} equal [b, b+1] with \(0 \leq a \leq3\) and \(0 \leq b \leq3\). This implies that the intervals given(x1 and x2) would only intersect in the case where |a - b| = 1(since that means the distance would be less than 1, which is the length of one of the unit squares. The region where the following holds true would be partitioned off into two 45-45 90 triangles, both with leg length of 2. The area of the desired region would be:

\(9-(2*2*1/2)*2 = 9-4= 5\)

Now, we have the probability that the intervals X_{1} and X_{2} intersect, which is \(5/9\). Likewise, the probability that Y_{1} and Y_{2} intersect is also \(5/9\), so our answer is just:

\((5/9)^2 = 25/81\)

I feel like I've done this problem somewhere before lol, can't seem to remember where

jfan17 Mar 22, 2020