Recreational problem involving circles, cords, and areas

[srfriggen]

Hello all,

I've been trying to work out this problem I came across yesterday when a professor mentioned it in a math education course. It states, "Take a circle and put two dots on the circle then connect them with a cord. How many sections of area does the cord split the circle into?" Of course the answer is 2.

Now do it with 3 dots. The answer is 4.
Now do it with 4 dots. The answer is 8.
Now do it with 5 dots. The answer is 16.
How many if there are 6 dots?

One might think, naively as I did, that the formula for the number of sections is 2^n-1, so 6 dots should have 32 sections. But it actually has 31. That was the point of the lesson.

We never got the formula so I've been trying to figure one out.


I decided to look at the number of dots and how many cords they form, and I was able to find a recursive definition and conjecture a formula. I proved it using induction.

I will write the information out like this: # of dots / #of cords;

1 / 0
2 / 1
3 / 3
4 / 6
5 / 10
6 / 15
7 / 21
8 / 28

The recursive series can be understood as s(n)=s(n-1) + (n-1), and the explicit formula is:

s(n)=(n-1)+(n-2)+...+ 3+2+1.


Now I have a nice relationship between dots and cords. Next I will show you the table of dots to cords to sections, which I worked out manually up to n=7.

1 / 0 / 1
2 / 1 / 2
3 / 3 / 4
4 / 6/ 8
5 / 10 / 16
6 / 15 / 31
7 / 21 / 58
8 / 28 / ?


So this is where I am stuck. I am trying to find a relationship with the number of sections and the number or cords or dots. I do not see any obvious recursive relationship, and certainly do not see any obvious formula.

Perhaps a different set of eyes/synapses on the problem would shed more light. Is there a more systematic approach I can use, rather than jus trying to "see" it?

 

[SteamKing]

In this context, 'cord' should be 'chord'.

 

[srfriggen]

In this context, 'cord' should be 'chord'.

lol I'm a musician so I thought to myself, "It can't be chord!"

 

[gmax137]

If you put the six points equally spaced (and draw it carefully enough) then you get 30 not 31 (the small one in the center disappears). So, I'm not sure you can make an algorithm to count them.

edit: how can I get an attachment to show as a little thumbnail, so people don't have to open it to see it?

 

[srfriggen]

If you put the six points equally spaced (and draw it carefully enough) then you get 30 not 31 (the small one in the center disappears). So, I'm not sure you can make an algorithm to count them.

Wow. Mind blown! Entire class came up with 31, but difficult to draw perfect circles and perfectly straight lines.

Ok, next I suppose I'll reformulate the question and see if having the dots "evenly spaced" can lead to a formula.

 

[HallsofIvy]

This is a classic problem known as the "missing area" problem.

If you place n dots around a circle NOT equally spaced, but so that the chords divide the interior of the circle into the maximum possible number of areas, the number of areas is given by
$$\sum_{i=0}^4 \begin{pmatrix}n \\ i\end{pmatrix}$$
where $\begin{pmatrix}n \\ i\end{pmatrix}$ is 0 if i> n.

If n 0, 1, 2, 3, or 4, that gives the binomial coefficient expansion of $(1+ 1)^n= 2^n$. For n= 5, it is missing only the last term, 1, so is [/itex]2^5- 1= 31[/itex]. For n= 6, it is missing the last two terms, 5(1)+ 1= 6, so is $2^6- 6= 64- 6= 58$.

 

[srfriggen]

This is a classic problem known as the "missing area" problem.

If you place n dots around a circle NOT equally spaced, but so that the chords divide the interior of the circle into the maximum possible number of areas, the number of areas is given by
$$\sum_{i=0}^4 \begin{pmatrix}n \\ i\end{pmatrix}$$
where $\begin{pmatrix}n \\ i\end{pmatrix}$ is 0 if i> n.

If n 0, 1, 2, 3, or 4, that gives the binomial coefficient expansion of $(1+ 1)^n= 2^n$. For n= 5, it is missing only the last term, 1, so is [/itex]2^5- 1= 31[/itex]. For n= 6, it is missing the last two terms, 5(1)+ 1= 6, so is $2^6- 6= 64- 6= 58$.

thanks! I'm going to research this some more! Appreciate the help everyone!