证明3×3分割的等边三角形中不存在特定顶点条件的等边三角形

Let’s work through this reasoning clearly, step by step, starting with concrete definitions to avoid confusion:

• We have a large equilateral triangle ( T ), split into 9 congruent smaller equilateral triangles (each with side length 1/3 that of ( T )). Let’s call these small triangles ( t_1, t_2, ..., t_9 ).
• "Orange regions" refer to the interiors of these small triangles (since the problem specifies vertices lie inside orange regions). The triangle we’re investigating would have all 3 vertices in orange regions, but not all in the same small triangle’s interior.

We’ll use proof by contradiction to show no such triangle exists.

Step 1: Assume the Opposite #

Suppose there exists an equilateral triangle ( \triangle ABC ) where:

1. Every vertex ( A, B, C ) lies strictly inside some orange region (i.e., inside the interior of a small triangle ( t_a, t_b, t_c ) respectively).
2. At least two of ( t_a, t_b, t_c ) are distinct (so vertices aren’t all confined to one orange region).

Step 2: Analyze Distances Between Small Triangle Interiors #

For the 3x3 grid of small triangles, we can categorize pairs of small triangles and their interior distance bounds:

• Adjacent small triangles: These share a full edge. The maximum distance between any two points in their interiors is less than ( \sqrt{3} ) (the length of the long diagonal of the rhombus formed by two adjacent small equilateral triangles with side length 1).
• Non-adjacent small triangles: These don’t share an edge. The minimum distance between any two points in their interiors is at least ( 1 ) (the side length of a small triangle).

Step 3: Derive a Contradiction #

Let’s break down the possible scenarios for ( \triangle ABC ):

Case 1: Two vertices in adjacent small triangles, third in a non-adjacent one #

Suppose ( A \in \text{int}(t_a) ), ( B \in \text{int}(t_b) ) (adjacent), and ( C \in \text{int}(t_c) ) (non-adjacent to both ( t_a ) and ( t_b )):

• ( AB < \sqrt{3} ) (from adjacent distance limits)
• ( AC \geq 1 ) and ( BC \geq 1 ) (from non-adjacent distance limits)
  For ( \triangle ABC ) to be equilateral, ( AB = AC = BC ). But any point in ( t_c ) will be too far from ( A ) and ( B ) to satisfy this equality—either the distance exceeds ( AB ), or ( C ) would have to lie on the boundary of ( t_c ) (not strictly inside the orange region).

Case 2: All three vertices in mutually non-adjacent small triangles #

Here, every pair of vertices is in non-adjacent small triangles, so ( AB \geq 1 ), ( AC \geq 1 ), ( BC \geq 1 ). The only equilateral triangles possible with these constraints would have vertices aligned with the large triangle’s grid boundaries—not strictly inside small triangle interiors. For example, the centers of three non-adjacent small triangles do not form an equilateral triangle, as their pairwise distances are not equal.

Case 3: Two vertices in non-adjacent triangles, third in an adjacent one #

This is symmetric to Case 1. The distance between the two non-adjacent vertices is at least 1, but the third vertex (adjacent to one) can’t match this equal side length requirement without falling on a small triangle’s boundary, violating the "inside orange region" rule.

Step 4: Conclusion #

In all possible scenarios, the requirement that all vertices are strictly inside orange regions and form an equilateral triangle spanning multiple orange regions leads to a contradiction. Therefore, no such equilateral triangle exists.

内容的提问来源于stack exchange，提问作者Tigerous215