Singapore Mathematics Project Festival
Saint Joseph’s Institution

Singapore Mathematics Project Festival
Saint Joseph’s Institution
The aim of our project is to investigate the number of chords and points of intersection within a circle with a variable number of points on the circumference and dissected from each point to every other point on the outside without any more than two lines intersecting each other (Fig. 1.1).
Our secondary aim is to discover the interrelationship between these two.
For example, in Fig. 1.1, fourth in a series of circles starting with the first with one point on the circumference, the circle has four points on the circumference, six lines
Fig. 1.1 joining the points, and one point of intersection.
This investigation can have many practical uses in the world. For example, in the printed circuit boards industry, in the design and layout of roads, in the architectural industry, and many more.
In the following pages, we will describe and go into greater detail about our investigation. So, what are you waiting for? Read On!
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
In the course of our project, we have found the number of chords in each of the circles with an increasing number of points on the outside. For
Figures 2.1 through 2.6, there are 0, 1, 3, 6, 10 and 15 chords respectively.
We observed that the number of chords increased proportionately to the number of points on the circumference of the circle. For example, in the third figure (Fig. 2.3), there are 3 chords and in the fourth figure (Fig. 2.4), there are 6 chords, a difference of 3. We have also found that this is true for all the figures. Thereafter, we realised that the formula for the number of chords is n ( n

1 )
, where n is the number of the sequence in the figures.
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Fig. 2.1 (0 chords) Fig. 2.2 (1 chord) Fig. 2.3 (3 chords)
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
Fig. 2.4 (6 chords) Fig. 2.5 (10 chords) Fig. 2.6 (15 chords)
The normal formula for finding triangular numbers is n ( n

1 )
. As in the case
2 of this sequence, which starts at 0 chords in the first figure , n has to be decreased by 1 due to the first figure in the sequence. By subtracting n by 1, we can get the formula n ( n

1 )
2
.
Fig. 1 Fig. 2
As it can be seen, the sequence of chords begin at the second figure instead of the first.
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
The number of points of intersection can be easily determined as the number of combinations that four different points can join is the number of points of intersection. This can be proved in the case of the following:
Fig. 4.1 Fig. 4.2
With four points, the number of combinations in Fig. 4.1 is 4, and therefore it only has 1 point of intersection. This also serves as proof to show that a figure with less than 4 points is unable to create a point of intersection. In the case of Fig. 4.2, the number of combinations is increased to 5, resulting in an increase of points of intersection to 5.
The method of finding the number of combinations is very simple. Since we want to find the number of combinations of 4 from the total number of points, we have to take n(n-1)(n-2)(n-3) divided by 4!, where n is the sequence in the figures. In the case of figure 10, the number of points of intersection will be
10
4 x x 3
9 x x
8 x
2 x 1
7
giving an answer of 210.
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
By taking the results from both the number of chords and number of points of intersection, we are able to find the number parts in the each circle. To find this number, we have to add 1, which is taken from the first figure which has no points of intersection and chords, plus the total number of chords within the circle, plus the number of points of intersection within the figure.
Fig. 5.1 Fig. 5.2
In Fig. 5.1, which is the fourth in the sequence, 1 + 6 chords + 1 point of intersection results in the total number of 8 parts. And in the case of Fig. 5.2, which is the fifth figure in the sequence, 1 + 10 chords + 5 points of intersection resulting in 16 parts.
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
In a situation such as in the following figure (Fig. 6.1),
Fig. 6.1
More than three lines are joined together resulting in fewer parts. In such a case, the same method is used to find the number of parts. [1 + 21 chords +
7 x 6 x 5 x 4
4 x 3 x 2 x 1
- 2 points of intersection] which gives a result of 33. This method of finding the number of parts can be used in any such figure.
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
It can be used in a this figure as well,
Fig. 7.1
In Fig. 7.1, one line is not joined from point to point. The number of parts can be found by taking 1 + 9 chords + 3 points of intersection which gives a total of 13 parts.
If this method were to be made into a single formula to find the maximum number of parts in such a figure, it will be:
24

12 n ( n

1 )
 n ( n

2 )( n

3 )
24
This formula is made up of the formulae of the different sequences and the steps to arrive to this formula is as follows:
1 + n ( n

1 )
2
+ n ( n

1 )( n

2 )( n

3 )
4 x 3 x 2 x 1
=
24

12 n ( n

1 )
 n ( n

2 )( n

3 )
24
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
In conclusion, we have discovered the formulae for finding both the number of chords and the number of points of intersection within a circle with a progressively increasing number of external points.
This project also shows that a relatively simple project like this can be used effectively in the real world, the most obvious of which are the planning of road layouts and architecture.
Therefore, we can proudly say that we think this project is a success in all ways given the limited amount of time that we had.
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Singapore Mathematics Project Festival
Saint Joseph’s Institution
Having completed our project successfully, we wish to extend our heartfelt gratitude and thanks to all who have helped us in the course of our project.
These include:
Mr Heng O.K., for his coaching and aid given in the course of the December holidays,
Ms Nicole Richards and Mrs Deborah Tan, our mathematics teachers, for joining us in the group meetings and offering us their expert advice,
All of our parents, for their support in our frequent group meetings,
And last but not least, YOU, the reader, for making this project the success that it is.
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