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 $XY^2$.

Answer 1

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*}