In convex quadrilateral $ABCD$, $X$ is the midpoint of $\overline{BC}$, and $Y$ is the midpoint of $\overline{DA}$. If $AB=BC=CD=DA=24$, and $\angle D=60^\circ$, then compute $XY^2$.
Let $\mathbf{a}$ be the displacement vector from the origin to $A,$ etc.
Note that
\[\mathbf{x} = \frac{\mathbf{b} + \mathbf{c}}{2} = \frac{\mathbf{b} + \mathbf{d}}{2} = \frac{\mathbf{c} + \mathbf{d}}{2}.\][asy]
unitsize (0.4 cm);
pair A, B, C, D, X, Y;
A = 3*dir(90);
B = 3*dir(90 + 8);
C = 3*dir(90 + 16);
D = 3*dir(90 + 24);
X = (B + C)/2;
Y = (D + A)/2;
draw(A--B--C--D--cycle);
draw(A--X);
draw(C--Y);
label("$A$", A, N);
label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, D);
label("$X$", X, S);
label("$Y$", Y, N);
[/asy]
By the Law of Cosines on triangle $ABX,$ with $\cos 60^\circ = \frac{1}{2},$ $AX^2 = (AB)^2 + (BX)^2 - AB \cdot BX = 648,$ so
\[\mathbf{x} = \frac{\mathbf{a} + \mathbf{d} + \sqrt{2} \sqrt{\mathbf{a} \cdot \mathbf{d} + 324}}{2}.\]Squaring both sides, and using the fact that $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2,$ $\mathbf{a} \cdot \mathbf{x} = \frac{\mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{d} + \sqrt{2} \mathbf{a} \cdot \mathbf{d} + 162}{2},$ etc. Furthermore,
\begin{align*}
\mathbf{x} \cdot \mathbf{y} &= \frac{\mathbf{a} \cdot \mathbf{d} + \sqrt{2} \sqrt{\mathbf{a} \cdot \mathbf{d} + 324}}{4} + \frac{\mathbf{b} \cdot \mathbf{c} + \sqrt{2} \sqrt{\mathbf{b} \cdot \mathbf{c} + 324}}{4} \\
&= \frac{\mathbf{a} \cdot \mathbf{d} + \mathbf{b} \cdot \mathbf{c} + \sqrt{2} |\mathbf{a} + \mathbf{d}||\mathbf{b} + \mathbf{c}|}{4} \\
&= \frac{\mathbf{a} \cdot \mathbf{d} + \mathbf{b} \cdot \mathbf{c} + \sqrt{2} |(\mathbf{a} + \mathbf{d})(\mathbf{b} + \mathbf{c})|}{4} \\
&= \frac{\mathbf{a} \cdot \mathbf{d} + \mathbf{b} \cdot \mathbf{c} + \sqrt{2} \sqrt{(\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c}) |\mathbf{a} + \mathbf{b}||\mathbf{a} + \mathbf{c}|}}{4}.
\end{align*}By symmetry, $\mathbf{a} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{d},$ so
\begin{align*}
\mathbf{x} \cdot \mathbf{y} &= \frac{2 \mathbf{a} \cdot \mathbf{d} + 2 \mathbf{a} \cdot \mathbf{c} + \sqrt{2} \sqrt{\left( 2 \mathbf{a} \cdot \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{d} + 2 \mathbf{b} \cdot \mathbf{c} + \mathbf{d} \cdot \mathbf{d} \right) (|\mathbf{a} + \mathbf{b}||\mathbf{b} + \mathbf{d}| + |\mathbf{a} + \mathbf{d}||\mathbf{b} + \mathbf{d}|)}}{4} \\
&= \frac{2 \mathbf{a} \cdot \mathbf{d} + 3 \mathbf{a} \cdot \mathbf{c} + \sqrt{2} \sqrt{\left( 2 \mathbf{a} \cdot \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{d} + 2 \mathbf{b} \cdot \mathbf{c} + \mathbf{d} \cdot \mathbf{d} \right) 2 \sqrt{21}}}{4} \\
&= \frac{4 \mathbf{a} \cdot \mathbf{d} + 6 \sqrt{21}}{4} \\
&= \mathbf{a} \cdot \mathbf{d} + \frac{3 \sqrt{21}}{2}.
\end{align*}We have that
\begin{align*}
|\mathbf{x} - \mathbf{y}|^2 &= |\mathbf{a} + \mathbf{d}|^2 - 2 \mathbf{a} \cdot \mathbf{d} \\
&= |\mathbf{a}|^2 + |\mathbf{d}|^2 + 2 \mathbf{a} \cdot \mathbf{d} - 2 \mathbf{a} \cdot \mathbf{d} \\
&= |\mathbf{a}|^2 + |\mathbf{d}|^2 \\
&= \frac{3}{2} \cdot 24^2 \\
&= \boxed{1296}.
\end{align*}
To find $XY^2$, we can use the Pythagorean Theorem.
Step 1: Draw a diagram. We have quadrilateral $ABCD$.
Step 2: Label the given information. We know that $AB=BC=CD=DA=24$ and $\angle D=60^\circ$.
Step 3: Draw segment $\overline{XY}$ connecting the midpoints $X$ and $Y$.
Step 4: Use the fact that $X$ is the midpoint of $\overline{BC}$ and $Y$ is the midpoint of $\overline{DA}$ to find the lengths of $\overline{XY}$.
Since $X$ is the midpoint of $\overline{BC}$, we have $BX=CX=\frac{BC}{2}=12$.
Similarly, since $Y$ is the midpoint of $\overline{DA}$, we have $AY=DY=\frac{DA}{2}=12$.
Therefore, $XY=BX+BY=12+12=24$.
Step 5: Apply the Pythagorean Theorem to find $XY^2$. The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two other sides.
In this case, we have a right triangle $BXY$ with a hypotenuse of length $XY$ and side lengths of $BX$ and $BY$.
Applying the Pythagorean Theorem, we have $XY^2 = BX^2 + BY^2 = 12^2 + 12^2 = 144 + 144 = 288$.
Therefore, $XY^2 = 288$.