Question

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 $P$

Answer 1

In this problem, you have a square with side length 6 and four smaller squares with side length 2 placed in the corners. The points W, X, Y, and Z are the vertices of the smaller squares. The goal is to find the maximum possible distance between points A and P.

[asy] size(200); pair A,B,C,D,P,W,X,Y,Z; A=(0,4); B=(4,4); C=(4,0); D=(0,0); P=(2,2); W=(0,6); X=(6,6); Y=(6,0); Z=(0,0); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); draw(A--P--C); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("P",P,S); label("W",W,N); label("X",X,NE); label("Y",Y,S); label("Z",Z,SW); [/asy]

The point A is located on the bottom-left corner of the square with side length 6, and the point P is the center of the square $ABCD$. To find the maximum distance from A to P, we need to find the longest diagonal of the square $ABCD$.

The longest diagonal of a square is the one that goes from one corner of the square to the opposite corner. In this case, the longest diagonal goes from point A to point C, which is a distance of 6 units. Therefore, the maximum possible distance from A to P is 6 units.