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
The answer is NOT sqrt(34).
We know that AB is a diagonal of the larger square, so AB=6*sqrt(2). Also, WP is a diagonal of one of the small squares, so WP=2*sqrt(2). Therefore, the maximum possible distance from A to P is AB−WP=4*sqrt(2).