Question

一只昆虫生活在边长为 $1$ 的正四面体表面。它想沿四面体表面，从一条棱的中点走到其对棱的中点。这样的最短路程是多少？（注：四面体的两条棱若没有公共端点，则称为对棱。）

An insect lives on the surface of a regular tetrahedron with edges of length $1.$ It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)

$\dfrac{1}{2}\sqrt{3}$

$1$

$\sqrt{2}$

$\dfrac{3}{2}$

$2$

答案：B

知识点：展开图（立体几何）立体几何菱形

难度评级：1660

小提示：

把路径经过的两个面展开到同一个平面图形中

Unfold the two faces the path crosses into a single flat figure

大提示：

这两个面形成一个边长为 $1$ 的菱形，而两个中点变成对边的中点

The two faces form a rhombus of side $1,$ and the midpoints become midpoints of opposite sides

解答：

将昆虫经过的两个面展开到平面上。它们形成一个由两个等边三角形组成、边长为 $1$ 的菱形。

这两个对棱中点会变成该菱形两条对边的中点，它们之间的直线距离正好是 $1$ 。折回四面体不会改变长度，因此最短路程是 $1$ 。

因此，正确答案是 B。

A shortest path leaves the starting edge through one of its two incident faces and reaches the opposite edge through one of its two incident faces. Any such pair of faces shares an edge. Unfolding that pair gives a rhombus of side $1$ made of two equilateral triangles.

The two opposite-edge midpoints become the midpoints of opposite sides of this rhombus, which are exactly $1$ unit apart along a straight segment. Folding back preserves the length, so the shortest trip is $1.$

Thus, the correct answer is B.

2001 AMC 12 第 15 题

答案：B

难度评级：1660

小提示：

大提示：

解答：

其他年份的第 15 题