# Algebraic expression for triangles within polygons

• elementbrdr
In summary: The Attempt at a Solution Treating vertices of polygons as possible outcomes from which 3 are chosen: Sample space S = nC3. I can't figure out how to express the number of outcomes within event E={number of triangles sharing at least one side with polygon}. My way of thinking about it is that values in E will be all combinations that contain at least a pair of consecutive numbers (which, since they represent vertices of the polgon, constitute a side of both the polygon and the triangle). Unfortunately, I can't come up with a way to express this algebraically. It may not even be the right approach.
elementbrdr

## Homework Statement

If a polygon has n>=4 sides, what is the probability, in terms of n, that a triangle made up of vertices of the polygon shares at least one side with the polygon.

## The Attempt at a Solution

Treating vertices of polygons as possible outcomes from which 3 are chosen: Sample space S = nC3. I can't figure out how to express the number of outcomes within event E={number of triangles sharing at least one side with polygon}. My way of thinking about it is that values in E will be all combinations that contain at least a pair of consecutive numbers (which, since they represent vertices of the polgon, constitute a side of both the polygon and the triangle). Unfortunately, I can't come up with a way to express this algebraically. It may not even be the right approach.

I think you are on the right track. nC3 appears to the number of unique triangles that can be constructed.

When it comes to determining those which share no common side ...

consider any particular side, and you can immediately rule out two of the remaining points for vertices of your triangle because a line drawn to those would be common to the polygon. Sketching a few polygons might help with generalising the formula.

Last edited:
Here are my thoughts using your suggested strategy:

The number of unique triangles sharing no common edge with polygon = n(n-3)(n-?). This is because there are n possible polygon vertices to choose for the first triangle vertex. The next triangle vertex cannot be the first vertex (-1) and cannot be an adjacent vertex (-2). The final vertex cannot be the first or second vertex (-2) and cannot be adjacent to either. At first glance I thought this would rule out 4 additional vertices, but that is wrong because vertices that are adjacent to two vertices selected for a triangle should only be counted once. So I can't see how to finish counting unique triangles using this strategy.

Another way of doing it is to take the union of the following events: (1) the triangle shares exactly 1 common side with the polygon and (2) the triangle shares exactly 2 common sides with the polygon. (1) can be expressed as n(n-4) because there are n sides that could be common, but 2 vertices have already been used for that side and the two polygon vertices adjacent to the selected side will be counted by event (2). Event (2) can be expressed as 2n because there are n sides of the polygon that can be used for the common side, and then there are 2 remaining vertices of the polygon that can complete the triangle. So you get 2n+n(n-4) = n(n-2). I guess you also get the same result by just taking n(n-2) to reflect using a side of the polygon for the first two vertices and then any vertex of the polygon except those contained in the selected side to complete the triangle. Either way, that results in n(n-2) / nC3

For everyone's information, the answer is 3!(n-3) / (n-2)(n-1). I just don't know how to get there.

Taking a closer look, by sketching 4,5,6,7 and 8-sided polygons I arrive at the total number of possible triangles = (n-1)(n-2)-2

A pattern for the number of triangles which don't share the polygons side is a little more taxing.

## 1. What is an algebraic expression for finding the number of triangles within a polygon?

An algebraic expression for finding the number of triangles within a polygon can be written as n-2, where n represents the number of sides in the polygon. This is derived from the fact that every polygon can be divided into n-2 triangles.

## 2. How is the expression for triangles within polygons related to the number of sides in a polygon?

The expression n-2 is directly related to the number of sides in a polygon because it represents the number of triangles that can be formed within a polygon with n number of sides. This formula is applicable to all regular and irregular polygons.

## 3. Can the algebraic expression for triangles within polygons be used for any type of polygon?

Yes, the algebraic expression n-2 can be used for any type of polygon, whether it is regular or irregular. This expression is a general formula that applies to all polygons, as long as they have a finite number of sides.

## 4. How can the algebraic expression for triangles within polygons be applied in real-life situations?

The algebraic expression for triangles within polygons can be applied in various real-life situations, such as in construction and architecture. It can be used to calculate the number of triangular panels needed for a roof or the number of triangular tiles needed for a floor layout. It can also be used in geometry problems and in calculating the area of irregular polygons.

## 5. Is there any other way to find the number of triangles within a polygon besides using the algebraic expression?

Yes, there are other ways to find the number of triangles within a polygon, such as drawing the polygon and dividing it into triangles, or using the formula n(n-3)/2, where n represents the number of sides in the polygon. However, the algebraic expression n-2 is a simple and efficient way to calculate the number of triangles within a polygon, especially for larger polygons.

• Precalculus Mathematics Homework Help
Replies
2
Views
2K
• Precalculus Mathematics Homework Help
Replies
2
Views
2K
• Precalculus Mathematics Homework Help
Replies
5
Views
2K
• Precalculus Mathematics Homework Help
Replies
6
Views
3K
• General Math
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
1
Views
2K
Replies
10
Views
1K
• Calculus
Replies
6
Views
3K
• Other Physics Topics
Replies
1
Views
8K
• Calculus and Beyond Homework Help
Replies
3
Views
3K