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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).

 May 23, 2023
 #1
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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)​​.

 May 23, 2023
 #2
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Try: sqrt(32 + sqrt(2)) ==sqrt(33.14142) ==5.7805

 May 23, 2023

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