Problem 1269

$$\begin{array}{l} \text { First, consider a special case of the problem. } \\ \text { There are two special cases for any five points in a plane. } \\ \text { The first case: These five points lie on the same line (Figure 1-13). } \\ \text { Suppose these five points lie on line } l \text { in the order } A, B, C, D, E \text {, and } \\ B C=\min \{A B, B C, C D, D E, A C, A D, A E, B D, B E, D E\} . \\ \text { In this case, it is clear that } \\ \quad \lambda=\frac{B C}{A E} \geqslant 4>2 \sin 54^{\circ} . \\ \text { The second case: These five points form the vertices of a regular pentagon (Figure 1-14). } \end{array}$$
$$\begin{array}{l} \text { The maximum distance between any two of the five points is the diagonal } A C \text { of the regular pentagon } A B C D E \text {, and the minimum distance is the side } A B \text {. } \\ \text { In this case, we only need to consider the isosceles triangle } A B C \text {. } \\ \text { Since } \angle A B C=108^{\circ} \text {, draw } B P \perp A C \text { at } P \text {, } \\ \text { then } \angle A B P=54^{\circ} \text {. } \\ \lambda=\frac{A C}{A B}=\frac{2 A P}{A B}=2 \sin 54^{\circ} . \end{array}$$

What inspiration can we draw from these special cases?
This means that the problem of five points can be reduced to a problem of three points, thus simplifying the situation.
Thus, the problem is reduced to:
What three points $A, B, C$ in the plane can satisfy
$$\lambda \geqslant 2 \sin 54^{\circ} \text {. }$$

Let $\angle A$ be the largest angle in $\triangle A B C$, and $\angle C$ be the smallest angle. That is,
$$A \geqslant B \geqslant C \text {. }$$

Thus, the length of $B C$ is the maximum distance between any two of the points $A, B, C$, and the length of $A B$ is the minimum distance.
$$\lambda=\frac{B C}{A B} .$$

By the Law of Sines, we have
$$\lambda=\frac{B C}{A B}=\frac{\sin A}{\sin C} .$$

Since
$$180^{\circ} \doteq(A+C)=B \geqslant C, \\ C \leqslant 90^{\circ}-\frac{A}{2}108^{\circ} \text{ (since } A \geqslant 108^{\circ} \text{) }.$$

How do we handle this problem? We can break down the scenario of any five points $A, B, C, D, E$ in the plane into several simpler cases.

The first case: The convex hull of these five points is a line segment. In this case, since three points lie on the same line, we must have
$$\lambda \geqslant 2>2 \sin 54^{\circ} .$$

The second case: The convex hull of these five points is a triangle $D B C$, and the other two points $A, E$ must be inside or on the boundary of $\triangle A B C$.

If one point is on the boundary, it reduces to the first case.
If one point is inside $\triangle D B C$, let this point be $A$. Then by
$$\angle D A B+\angle B A C+\angle C A D=360^{\circ} \text {, }$$

one of the angles $\angle D A B, \angle B A C, \angle C A D$ must be at least $120^{\circ}$. Without loss of generality, let $\angle B A C \geqslant 120^{\circ}>108^{\circ}$.

The third case: The convex hull is a quadrilateral $B C D E$, and $A$ is inside or on the boundary of the convex quadrilateral $B C D E$.
If $A$ is on one of the sides, it reduces to the first case.
If $A$ is inside, connect the diagonal $B D$, then $A$ must be inside or on the boundary of either $\triangle B D E$ or $\triangle B D C$.

Suppose $A$ is inside or on the boundary of $\triangle B D C$, this reduces to the second case.

The fourth case: The convex hull is a pentagon $A B C D E$. Since the sum of the interior angles of a convex pentagon is $540^{\circ}$, there must be one angle, say $\angle B A C \geqslant 108^{\circ}$. In this case, consider $\triangle A B C$.

Combining the above analysis, the problem is solved.
In solving this problem, our approach is to first consider special cases (collinear and regular pentagon), draw inspiration from them, then reduce the pentagon to a triangle, and finally reduce the problem to the fact that among the five points, there must be a triangle with one interior angle not less than $108^{\circ}$. This proposition is further broken down into four simpler cases, and reduction and decomposition, specialization and simplification are the keys to solving this problem.

For solid geometry problems, the strategy for simplification is to convert the relationships between spatial elements into relationships between elements in a plane. Therefore, reducing spatial problems to simpler cases means planarization, and using sections, projections, and unfolding are common methods for planarizing spatial problems.
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