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Difficulty: HardCombinations

A convex polygon has 4444 diagonals. What is the total number of distinct triangles that can be formed by joining any three vertices of this polygon?

Answer: 165

Answer

165
The number of diagonals of an nn-sided convex polygon is given by (n2)−n=n(n−3)2\binom{n}{2} - n = \frac{n(n-3)}{2}. Setting this equal to 4444 yields n(n−3)=88n(n-3) = 88, which simplifies to n2−3n−88=0n^2 - 3n - 88 = 0. Factoring gives (n−11)(n+8)=0(n-11)(n+8) = 0, so n=11n = 11. The total number of distinct triangles formed by choosing any 3 vertices from an 11-sided polygon is (113)=11×10×93×2×1=165\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165.

Step-by-Step Solution

1
Set up the equation for the number of diagonals in terms of the number of vertices nn
n(n−3)2=44\frac{n(n-3)}{2} = 44
Choosing any 2 vertices from nn vertices gives (n2)\binom{n}{2} total connecting line segments. Subtracting the nn boundary sides leaves the diagonals.
2
Solve the quadratic equation for nn
n^2 - 3n - 88 = 0 \implies (n-11)(n+8) = 0 \implies n = 11
A polygon must have a positive integer number of vertices, so n=11n = 11.
3
Compute the number of distinct triangles using combinations
\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165
In a convex polygon, no three vertices are collinear, so every unique combination of 3 vertices forms a distinct triangle.

Key Concept

Application of combinations to geometry (polygon diagonals and triangle selection)
Estimated Time:2m 0s
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Topics

Combinations

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