How to Generate All Possible Triangles from Regular Polygon Vertices

Understanding Triangle Density from Regular Polygons

What We're Solving:

We need to determine whether triangles formed by choosing three vertices from regular n-gons (as n varies) can approximate ANY possible triangle shape. In other words, can we get arbitrarily close to any triangle configuration by picking the right regular polygon and the right three vertices?

The Approach:

This is a beautiful problem that connects discrete geometry with continuous topology! We're essentially asking: "If I have an infinite collection of regular polygons (triangle, square, pentagon, hexagon, etc.), and I can pick any three vertices from any of these polygons, can I create triangles that come arbitrarily close to matching ANY triangle I might draw?"

The strategy is to:

• 1. Understand what "dense" means in this context
• 2. Parametrize triangles by their angles
• 3. Show our polygon-generated triangles can approximate any target triangle

Step-by-Step Solution:

Step 1: Understanding the Triangle Space Every triangle is determined by its three angles (α, β, γ) where α + β + γ = π. So our "space of all triangles" is really the 2-dimensional set: {(α, β) : α, β > 0, α + β < π}.

Step 2: Triangles from Regular n-gons In a regular n-gon, vertices are evenly spaced around a circle. If we pick three vertices at positions i, j, k (where 0 ≤ i < j < k ≤ n-1), the triangle's angles depend on the "gaps" between these vertices.

Step 3: The Key Insight - Inscribed Angle Theorem Here's the beautiful part! When we form triangles from vertices of a regular n-gon, we're really forming triangles inscribed in a circle. The angles of these triangles are determined by the arcs between the chosen vertices.

Step 4: Showing Density As n gets larger, we get more vertices, which means more possible arc lengths. The crucial observation is that we can approximate any desired arc length arbitrarily well by choosing n large enough and selecting appropriate vertices.

For any target triangle with angles (α₀, β₀, γ₀), we can:

• Choose n large enough
• Find vertices that create arcs approximately proportional to 2α₀, 2β₀, 2γ₀
• The resulting inscribed triangle will have angles very close to (α₀, β₀, γ₀)

Step 5: Making it Rigorous The formal proof uses the fact that as n → ∞, the set of possible "normalized arc lengths" becomes dense in [0,1]. Since triangle angles are determined by these arc ratios, we can approximate any triangle configuration.

The Answer:

YES! The triangles formed by vertices of regular polygons ARE dense in the set of all possible triangles.

This means: Given any triangle shape you can imagine, and any tiny tolerance ε > 0, there exists some regular n-gon and three of its vertices that form a triangle within ε of your target triangle (in terms of angle measurements).

Memory Tip:

Think of it this way: "More sides = more flexibility!" As regular polygons get more sides, their vertices become like a "fine mesh" around the circle, giving you incredible precision to approximate any triangle shape you want. It's like having a adjustable triangle-making machine that gets more precise with each upgrade!

The beauty here is that something purely discrete (picking 3 points from finite sets) can approximate something continuous (all possible triangles). That's the magic of density in topology! 🔺✨