$A C, A D, B C, B D, C D$ are six edges, only one of which is longer than 1. Find the maximum possible sum of the lengths of these six edges. (1935, 14th American High School Mathematics Olympiad)
Without loss of generality, assume the length of $AD$ is greater than 1, and the lengths of the other five sides are no more than 1. Suppose the lengths of the other five sides are fixed. To maximize the sum of the six side lengths, $AD$ should be as long as possible. Since the other five side lengths are fixed, i.e., $\triangle ABC$ and $\triangle DBC$ are fixed, $AD$ reaches its maximum length when and only when $ABDC$ is a planar convex quadrilateral, and $A, D$ are its opposite vertices.

Now, assume the positions of points $B, C$ are fixed. Since $|AB| \leqslant 1, |BD| \leqslant 1$, points $A, D$ should be within the circle centered at point $B$ with a radius of 1. Also, since $|CA| \leqslant 1, |CD| \leqslant 1$, points $A, D$ should be within the circle centered at point $C$ with a radius of 1. Thus, points $A, D$ should be in the intersection of these two circles. Noting that this region is centrally symmetric, the longest chord in this region coincides with the common chord of the two circles. If we take the endpoints of the common chord as $A, D$, then the lengths of the five sides $AD, AB, AC, BD$, and $DC$ are maximized, as shown in the figure. At this time, $AB = AC = BD = DC = 1$.

Finally, to determine the positions of $B, C$, the goal is to maximize $|BC| + |AD|$. Let $\angle ABO = \theta$, then $|BC| + |AD| = 2(\sin \theta + \cos \theta) = 2\sqrt{2} - \sin(\theta + 45^\circ)$. Noting that $|DC| \leqslant 1$, then $\angle BAC \leqslant 60^\circ$, $60^\circ \leqslant \theta < 90^\circ$, thus $105^\circ \leqslant \theta + 45^\circ < 135^\circ$. Clearly, the function $\sin(\theta + 45^\circ)$ is decreasing in this interval, so when $\theta = 60^\circ$, $|BC| + |AD|$ reaches its maximum value.
\[
\begin{aligned}
|BC| + |AD| & = 2(\sin 60^\circ + \cos 60^\circ) \\
& = \sqrt{3} + 1.
\end{aligned}
\]

Therefore, the maximum value of the sum of the six side lengths is
\[
4 + (\sqrt{3} + 1) = 5 + \sqrt{3}.
\]