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(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

 Mar 21, 2020
 #1
avatar+485 
+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 xand y1. We define for laura's square xand yfor 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 X1 equal to [a, a+1] and X2 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 X1 and X2 intersect, which is \(5/9\). Likewise, the probability that Y1 and Y2 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

 Mar 22, 2020
 #2
avatar+531 
+3

Thank you so much! That makes a lot of sense.

MathCuber  Mar 22, 2020
 #3
avatar+111329 
+1

Very nice, jfan !!!!

 

cool cool cool

CPhill  Mar 22, 2020

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