In the diagram, four squares of side length 2 are placed in the corners of a square of side length 6. Each of the points W, X, Y, and Z is a vertex of one of the small squares. Square ABCD can be constructed with sides passing through W, X, Y, and Z. What is the maximum possible distance from A to P?https://latex.artofproblemsolving.com/3/a/e/3ae9131cc46a1aa7c3ba6923b1dfde6b8cee4537.png
You get sqrt(34) by rotating square ABCD so that the vertices were in the center of the 2x2 squares. Then, calling the midpoint of the bottom line E, you get right triangle AEP. AE = 5, EP = 3, so AP = sqrt(34).
The previous person also gave this answer but it incorrect.