Question
Look at this convex polygon.
How many triangles are formed by drawing all the diagonals from a single vertex?
How many triangles are formed by drawing all the diagonals from a single vertex?
Answers
GPT-4o mini
To determine the number of triangles formed by drawing all the diagonals from a single vertex of a convex polygon, we can use the following reasoning:
Let's denote the number of sides (vertices) of the convex polygon as \(n\). A convex polygon with \(n\) vertices has \(n\) sides.
When we choose one vertex of the polygon, we can draw diagonals to all other non-adjacent vertices. The key points are:
- From the chosen vertex, you cannot draw a diagonal to itself.
- You cannot draw a diagonal to the two vertices that are adjacent to it, as they form sides of the polygon, not diagonals.
Thus, the total number of vertices that the chosen vertex can connect to with diagonals is:
\[
n - 1 \quad \text{(total vertices excluding the chosen vertex)} - 2 \quad \text{(excluding the two adjacent vertices)}.
\]
This leads to the formula:
\[
\text{Diagonals from one vertex} = n - 3.
\]
Once we draw diagonals from the chosen vertex to the \(n - 3\) non-adjacent vertices, we form triangles. Each triangle consists of the chosen vertex and two of these \(n - 3\) vertices.
The number of ways to choose 2 vertices from \(n - 3\) vertices is given by the binomial coefficient:
\[
\binom{n - 3}{2} = \frac{(n - 3)(n - 4)}{2}.
\]
Therefore, the number of triangles formed by drawing all the diagonals from a single vertex in a convex polygon with \(n\) vertices is:
\[
\boxed{\frac{(n - 3)(n - 4)}{2}}.
\]
Let's denote the number of sides (vertices) of the convex polygon as \(n\). A convex polygon with \(n\) vertices has \(n\) sides.
When we choose one vertex of the polygon, we can draw diagonals to all other non-adjacent vertices. The key points are:
- From the chosen vertex, you cannot draw a diagonal to itself.
- You cannot draw a diagonal to the two vertices that are adjacent to it, as they form sides of the polygon, not diagonals.
Thus, the total number of vertices that the chosen vertex can connect to with diagonals is:
\[
n - 1 \quad \text{(total vertices excluding the chosen vertex)} - 2 \quad \text{(excluding the two adjacent vertices)}.
\]
This leads to the formula:
\[
\text{Diagonals from one vertex} = n - 3.
\]
Once we draw diagonals from the chosen vertex to the \(n - 3\) non-adjacent vertices, we form triangles. Each triangle consists of the chosen vertex and two of these \(n - 3\) vertices.
The number of ways to choose 2 vertices from \(n - 3\) vertices is given by the binomial coefficient:
\[
\binom{n - 3}{2} = \frac{(n - 3)(n - 4)}{2}.
\]
Therefore, the number of triangles formed by drawing all the diagonals from a single vertex in a convex polygon with \(n\) vertices is:
\[
\boxed{\frac{(n - 3)(n - 4)}{2}}.
\]