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Problem 2010

National Olympiad second round; IMO P1/P4

Geometry Difficulty 7.3 Find the answer Middle European Mathematical Olympiad

We consider dissections of regular $n$ -gons into $n - 2$ triangles by $n - 3$ diagonals which do not intersect inside the $n$ -gon. A bicoloured triangulation is such a dissection of an $n$ -gon in which each triangle is coloured black or white and any two triangles which share an edge have different colours. We call a positive integer $n \ge 4$ triangulable if every regular $n$ -gon has a bicoloured triangulation such that for each vertex $A$ of the $n$ -gon the number of black triangles of which $A$ is a vertex is greater than the number of white triangles of which $A$ is a vertex.

Find all triangulable numbers.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

To solve the problem, we need to determine which positive integers $n \ge 4$ allow a regular $n$ -gon to be dissected into a bicoloured triangulation under the condition that, for each vertex $A$ , the number of black triangles having $A$ as a vertex is greater than the number of white triangles having $A$ as a vertex.

### Step-by-step analysis

1. Understanding the colours and conditions:
- In a bicoloured triangulation, each pair of triangles sharing an edge must be of different colours.
- For a vertex $A$ , the triangles sharing this vertex must fulfill the condition: more black triangles than white triangles.

2. Dissection characteristics:
- A regular $n$ -gon will be divided into $n-2$ triangles using $n-3$ diagonals.
- Since this is a bicoloured map, it implies a need for an alternating colour scheme.

3. Analyzing potential triangulable numbers:
- The colouring condition implies that for each vertex, the degree of connection, i.e., the number of triangles connected to it, should support this alternating pattern with more black triangles.
- This essentially translates to each vertex being part of a number of triangles that is odd, so as to favour a greater number of one colour.

4. Examining divisibility by 3:
- If $n$ is divisible by 3, we can construct an $n$ -gon such that each vertex is connected to a number of triangles conducive to having more black triangles, as follows:
- Divide the entire $n$ -gon into smaller sections or paths with exactly 3 connections or nodes, enabling cyclic colour breaking.

5. Proving the necessity:
- Suppose $n$ is not divisible by 3. Then attempting to uniformly distribute the triangles such that any vertex is part of more black than white becomes impossible without violating the bicolouring property.

6. Conclusion:
- The requirement translates to ensuring each vertex in the cyclic arrangement along the perimeter plays into alternating triangle counts.
- Therefore, only when $n$ is divisible by 3 can these conditions hold consistently for each vertex.

Thus, for a positive integer $n \geq 4$ to be triangulable, it must satisfy:

$3 \mid n$

Conclusively, the set of triangulable numbers are those that are multiples of 3, starting from 6. Hence, the triangulable numbers are:

$\boxed{3 \mid n}$

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.