AMC12 2001 A

AMC12 2001 A · Q15

AMC12 2001 A · Q15. It mainly tests 3D geometry (volume), Geometry misc.

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.)

一只昆虫生活在棱长为 $1$ 的正四面体表面上。它希望沿四面体表面从一条棱的中点走到与之相对的棱的中点。这样的最短路径长度是多少？（注：四面体的两条棱若没有公共端点，则称为相对棱。）

(A) $\frac{1}{2}\sqrt{3}$ $\frac{\sqrt{3}}{2}$

(B) 1 1

(D) $\frac{3}{2}$ $\frac{3}{2}$

(E) 2 2

Answer

Correct choice: (B)

正确答案：（B）

Solution

Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. By symmetry (as the tetrahedron is regular, so all of its faces are equilateral triangles), its length is the same as the length of the tetrahedron's edge, i.e. $\boxed{\text{(B) }1}$.

对于表面上的任意路径，我们都可以将表面展开到平面上，得到一条在平面中长度相同的路径。考虑下图所示的四面体展开图，一对相对的点用点标出。显然在平面中最短路径就是连接这两点的线段。由对称性（正四面体的所有面都是等边三角形），该线段长度与四面体的棱长相同，即 $\boxed{\text{(B) }1}$。

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