Question

The entries in the Pascal Triangle can, for \(n \geq 2\), be used to determine the number of \(k\) -sided figures that can be formed using a set of \(n\) points on a circle. In general, the first entry in a row indicates the number of \(n\) -sided figures that can be formed, the second entry indicates the number of \((n-1)\) -sided figures, and so on. For example, if a circle contains 4 points, the row for \(n=4\) in the Pascal Triangle shows the number of possible quadrilaterals (1), the number of triangles (4), and the number of line segments (6) that can be formed using the four points. (a) How many hexagons can be formed using 8 points lying on the circumference of a circle? (b) How many triangles can be formed using 10 points lying on the circumference of a circle? (c) How many dodecagons can be formed using 20 points lying on the circumference of a circle?

Short answer

28 hexagons, 120 triangles, 125970 dodecagons.

Step-by-step solution

Identify the row in Pascal's Triangle for each problem

For each problem, we need to determine the correct row in Pascal's Triangle. The row number corresponds to the number of points, n.

Express each sub-problem in terms of combinations

Pascal's Triangle entries can be represented by combinations. The k-th entry in the n-th row is given by \(\binom{n}{k}\). We will use this formula to determine the required entries.

Solution for part (a)

To find the number of hexagons (6-sided figures) that can be formed from 8 points:Use the combination formula \(\binom{8}{6}\) which is the same as \(\binom{8}{2}\). Calculate \(\binom{8}{2} = \frac{8!}{2!(8-2)!} = 28\).

Solution for part (b)

To find the number of triangles (3-sided figures) that can be formed from 10 points:Use the combination formula \(\binom{10}{3}\). Calculate \(\binom{10}{3} = \frac{10!}{3!(10-3)!} = 120\).

Solution for part (c)

To find the number of dodecagons (12-sided figures) that can be formed from 20 points:Use the combination formula \(\binom{20}{12}\), which is also \(\binom{20}{8}\). Calculate \(\binom{20}{8} = \frac{20!}{8!(20-8)!} = 125970\).

Key concepts

Combinations

Combinations are a key concept in mathematics often used to determine how many ways we can select a group of items from a larger set. When dealing with combinations, the order of selection does not matter. This concept is expressed mathematically using binomial coefficients, which are often represented with the notation \(\binom{n}{k}\).

In the context of Pascal's Triangle, combinations indicate how we can choose subsets of points from a set on a circle. For example:

• To find how many 6-sided figures (hexagons) can be made from 8 points on a circle, we use the combination \(\binom{8}{6}\). This is the same as \(\binom{8}{2}\) due to the symmetry of combinations.
• For forming 3-sided figures (triangles) from 10 points, we use \(\binom{10}{3}\).
• To form 12-sided figures (dodecagons) from 20 points, \(\binom{20}{12}\) or \(\binom{20}{8}\) is used.

Understanding combinations allows us to solve these kinds of problems by simply plugging values into the combination formula and performing the calculations.

Geometry

Geometry is the branch of mathematics dealing with shapes, sizes, and properties of space. In this exercise, geometry helps us understand the arrangement of points on a circle and how they form various polygons.

When we have a set number of points on a circle, each possible subset of these points can form a different geometric figure. For instance:

• Hexagons are formed by choosing 6 points from a larger set of points on the circle.
• Triangles are formed by selecting 3 points.
• Dodecagons are created by selecting 12 points.

This spatial understanding is essential because it helps translate the abstract problem of combinations into tangible geometrical figures.

Binomial Coefficients

Binomial coefficients are essential in the study of combinations. They represent the coefficients in the expansion of binomials and are written as \(\binom{n}{k}\), which means the number of ways to choose k elements from a set of n elements.

In Pascal's Triangle, each entry represents a binomial coefficient. The entire triangle is constructed using the binomial coefficient formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(n!\text{ (n factorial)}\text{ is the product of all positive integers up to n}.\). For example:

• \binom{\text{8}}{\text{6}} = 28
• \binom{\text{10}}{\text{3}} = 120
• \binom{\text{20}}{\text{12}} = 125970

This mathematical framework is powerful, allowing us to efficiently solve problems involving combinations and geometrical formations.

Circle

A circle is a simple, yet widely studied geometric shape. In the context of this problem, the circle has points located on its circumference.

Selecting different subsets of these points will result in forming various polygons:

• Points on the circumference can connect to form edges of shapes like triangles, hexagons, or dodecagons.
• The symmetry of the circle makes transitional changes between figures smooth and calculable using combinations.

This symmetry ensures that selecting points to form a specific polygon is equitably distributed, which ties back into our combination calculations. Understanding how points on a circle configure into different shapes provides insight into real-world applications such as network design, molecular structures, and more.