“THE RELATIONSHIP OF PASCAL’S TRIANGLE TO THE

REGIONS FORMED BY CHORDS ON A CIRCLE”Imee Noli Sequerra

Stephie Andielic MabagaJea Andre Juego

Aiko BajitaKennan Paul TapangAaron Jacob Valino

Chapter I: INTRODUCTIONA chord of a circle is a geometric line segment whose

endpoints both lie on the circle and a diagonal is a line segment that joins two nonconsecutive vertices of a polygon or a polyhedron. These segments will be connected to the different points that lie on the circumference of a circle. The topic includes the help of the famous Pascal’s Triangle. It is a triangular array of binomial coefficients and is named after Blaise Pascal.

For instance, six points were placed randomly on the circle; join all possible pair of points by using a chord. As a result, the circle is separated into thirty one regions. To be precise, no three chords should ever pass or intersect through a common point. For you to attain that, the points on the circle must be placed ‘at random’ or in other words, the distance between the points must not be equal.

This topic is quite a rich problem, and there are many ways to attack it, some sophisticated, and some are quite elementary.

OBJECTIVES OF THE STUDYThe overall aim of the study is to collect reliable and valid

information about the relationship of the famous Pascal’s Triangle to the regions formed by chords on a circle. Within this topic, we have a number of specific objectives:

The main objective is to be able to find out the connection or the use of the Pascal’s Triangle in knowing the number of regions formed by chords on a circle.

Drawing chords on the circle is the easiest way to know how many regions were formed, but what if the number of points is big enough? We would also want to know if there is or are formula/s in solving the number of regions formed by chords on a circle.

Chapter II: STATEMENT OF THE PROBLEMThe purpose of this study is to know whether the number of

regions and the numbers of Pascal’s Triangle are related. If we were to find out the relationship between them, we’re going to make formula so that we will be able to get the result without looking at the Pascal’s Triangle and it will be more convenient to all of us. Unfortunately, we have this conclusion that it is not applicable to all points and we are kind of having a hard time to know how the number of regions formed by chords on a circle and the numbers of Pascal’s Triangle are related. In response to this problem, the following questions are raised to help us investigate the topic:

1. What is the connection of Pascal’s Triangle in finding the regions formed by chords on a circle?

2. What is or are the formula/s in finding the regions formed by chord on a circle?

3. What could the Pascal’s Triangle help and contribute to our study?

Chapter III: FORMULATING CONJECTURESA. First Conjecture: The sum of the numbers in a row of

the Pascal’s Triangle is the number of regions formed by chords on the circle.

The number of digits in a row of the Pascal’s triangle represents the number of points on a circle. We have formulated a hypothesis that if you add the

numbers on the Pascal’s Triangle horizontally, you will get the number of regions formed by chords on a circle.

No. of Points on CirclePascal’s Triangle

Illustration

Region

Conclusion

1 1 1 There is only one digit on the first row. If you have only one point on a circle you will not be able to form a chord so the region will remain one.

2 2 2There are two digits on the second row. The sum of those digits will result to two regions.

3 4 4There are three digits on the third row. The sum of those digits will result to four regions.

4 8 8There are four digits on the fourth row. The sum of those digits will result to eight regions.

5 16 16There are five digits on the fifth row. The sum of those digits will result to sixteen regions.

6 32 31There are six points on the fifth row. The sum of those digits will result to thirty-two regions. But six points can only form thirty-one regions. So this states that the hypothesis we formulated is applicable up to five points only.

Conclusion:Through our hypothesis, we can conclude that you can get the

number of regions formed by chords on a circle by adding the digits on the Pascal’s Triangle horizontally, but is applicable up to six points only.

B. Second Conjecture: A formula for solving the number of regions formed by chords on a circle.

Formulas lessen the burden of every student. With the help of the exponential functions, we were able to create a formula in solving the number of regions formed by chord on a circle. The formula is: No. of regions = 2n−1 where n is the number of points. When you subtract 1 to the number of points, it will result to exponential functions.

Number of digits in a column of Pascal’s Triangle

Formula 2n−1

Exponential Function Number of RegionsConclusion

1 21−1 20 1 If you substitute the number of digits in a

column to the formula, which is 1, the number of regions will result to

1.

2 22−1 21 2 If you substitute the number of digits in a

column to the formula, which is 2, the number of regions will result to

2.3 23−1 22 4 If you substitute the

number of digits in a column to the formula, which is 3, the number of regions will result to

4.4 24−1 23 8 If you substitute the

number of digits in a column to the formula, which is 4, the number of regions will result to

8.

5 25−1 24 16 If you substitute the number of digits in a

column to the formula, which is 5, the number of regions will result to

16.6 26−1 25 32 If you substitute the

number of digits in a column to the formula, which is 6, the number of regions will result to 32. So we can conclude

that the formula is applicable to any

number of digits in a column of Pascal’s

Triangle

Conclusion:In this conjecture, we found out that Exponential function

is much related to the Pascal’s Triangle. In our first conjecture, it tackles about the adding of the Pascal’s Triangle horizontally to find the regions. We noticed that the sum of each column of the Pascal’s Triangle has an equivalent to the Exponential Function.

20 = 1

21 = 2

22 = 4

23 = 8

24 = 16

25 = 32

26 = 64

27 = 128