In rectangle $ABCD$, $AB=1, BC=m, O$ is its center, $EO \perp$ plane $ABCD, EO=n$, and there exists a unique point $F$ on side $BC$ such that $EF \perp FD$. What conditions must $m, n$ satisfy for the angle between plane $DEF$ and plane $ABCD$ to be $60^{\circ}$?
Solution: As shown in Figure 4, let $FC = x$. Draw a perpendicular from point $O$ to $BC$, with the foot of the perpendicular being $G$. Then,
$$
\begin{array}{l}
GF = \frac{m}{2} - x, \\
OG = \frac{1}{2}, \\
EF^2 = EO^2 + OG^2 + GF^2 \\
= n^2 + \frac{1}{4} + \left(\frac{m}{2} - x\right)^2 \\
= x^2 - mx + \frac{m^2}{4} + n^2 + \frac{1}{4}, \\
DF^2 = CD^2 + CF^2 = 1 + x^2, \\
DE^2 = OD^2 + OE^2 = n^2 + \left(\frac{\sqrt{1 + m^2}}{2}\right)^2 \\
= \frac{m^2}{4} + n^2 + \frac{1}{4}.
\end{array}
$$

Since $EF \perp DF$, in $\triangle DEF$, by the Pythagorean theorem, we have
$$
\begin{array}{l}
\left(x^2 - mx + \frac{m^2}{4} + n^2 + \frac{1}{4}\right) + \left(1 + x^2\right) \\
= \frac{m^2}{4} + n^2 + \frac{1}{4},
\end{array}
$$

which simplifies to $2x^2 - mx + 1 = 0$.
Since point $F$ is unique, $\Delta = m^2 - 8 = 0$.
Solving this, we get $m = 2\sqrt{2}, x = \frac{\sqrt{2}}{2}$.
Since $EO \perp FD, EF \perp FD$, it follows that $FD \perp$ plane $OEF$. Therefore, $FD \perp OF$. Hence, $\angle EFO$ is the angle between plane $DEF$ and plane $ABCD$. Thus, we have
$$
\begin{array}{l}
= \sqrt{\frac{3}{4n^2} + 3} = \cos 60^\circ = \frac{1}{2}. \\
\end{array}
$$

Therefore, $n^2 = \frac{9}{4}, n = \frac{3}{2}$.
Thus, when $m = 2\sqrt{2}, n = \frac{3}{2}$, the angle between plane $DEF$ and plane $ABCD$ is $60^\circ$.
(Wang Xuesi, Anqing No. 3 Middle School, Anhui Province, 246003)