# Geometry problem.

Acolyte
Posers and Puzzles 30 Oct '02 19:36
1. Acolyte
30 Oct '02 19:36
For this you will need a pen and paper (a ruler helps as well).

Draw a circle. Draw n lines across it. How many regions can you divide the circle into? Try
it for small n, and see if you can spot the pattern.
2. belgianfreak
stitching you up
30 Oct '02 20:24
I've already expressed my lack of knowledge in maths, so this has
got to be wrong. But isn't it just n*2??
3. Acolyte
30 Oct '02 23:13
4. 30 Oct '02 20:48
just trying: 1 + nx(n-1)/2 ?
5. 30 Oct '02 20:49
sorry I meant: 1 + nx (n+1)/2
6. 30 Oct '02 21:07
Gil,

I'll call 1 + n(n+1)/2 = f(n) for simplicity

f(3) = 1 + 3(4/2) = 7
f(4) = 1 + 4(5/2) = 10

Looks good so far! Nicely done! What was your reasoning? Or did you
just try to fit an equation to your observations?

Rein
7. 30 Oct '02 21:17
sorry: f(4) = 11
and I can make 11 area's with 4 lines, I believe...
8. 31 Oct '02 17:32
9. 30 Oct '02 22:06
Gilbert, do you have a proof? Can you prove your formula to be true
for when n = 100 say?

Mark
The Squirrel Lover

PS. There is more than one way of doing it. There is a particularly
attractive way of doing it however that does in fact kind of tie in with
the other thread on arithmetic progressions!!!!
10. 30 Oct '02 23:21
Oh Mark, that has been years ago since I tried things like this. I
cannot do that just like that. At most try to 'plan' a proof on intuitive
basis. The formal proof would have to follow. Let me see:

If I can proof that when it is true for in-1, then it holds also for n, AND
knowing it is true for say n=1, then it is ok isn't it? That is another
way of expressing the progression, I think.

I think the point is to prove that with n-1 non-parallel lines, the n-th
line creates n NEW regions, because at any (of the n-1) intersection
(s) the line splitses a region in two, but except from one extreme, the
others are counted twice (keft and right).

That's as far as I get right now. Saved by the bedtime-bell. lol.

Gil.
11. 03 Nov '02 11:20
Mark,
my math is too rusty. Why don't you give the correct proof (or the
correct answer should mine be wrong)?

I decided to stick to my intuition. As Einstein said once: &quot;as soon as
the mathematicians got involved I didn't understand relativity
anymore&quot;.
Sin.
12. 07 Nov '02 12:30

f(n) denotes the number of regions in the circle with n lines in it.

As you say, by drawing one more line there will be n points on that
line. In other words, of the f(n) regions the line will traverse n+1 of
them. Each region gets split into two so n+1 regions are added as a
result of this extra line.

So before we had f(n), now we have f(n+1) and this extra line has
given us n+1 extra regions.

Therefore, f(n+1) = f(n) + n+1

Now the cunning bit...

Rearranging the above line will give f(n+1) - f(n) = n+1

Consider what this is saying (sorry for all the dots, they are necessary
to line up the columns, if I leave spaces the forum software scrunches
everything up).

f(1) - f(0) = 1 (since here n = 0 so n+1 = 0+1 = 1)
f(2) - f(1) = 2 (since here n = 1 so n+1 = 1+1 = 2)
f(3) - f(2) = 3 (since...etc)
f(4) - f(5) = 4

and so on up to:-

f(n-1) - f(n-2) = n-1
f(n) - f(n-1) = n

Add up the columns (note how the top left to bottom right diagonal on
the left hand side of the equals sign cancels, this is what Acolyte
mentioned in one of his posts where all the terms cancel leaving just
two, (usually) just the first one and the last one).

We get f(n) - f(0) = 1+2+3+4+...+(n-1)+n

The right hand side is simply the sum of the first n integers!!! This is
where it ties in with the other thread involving Gauss and arithmetic
progressions!! I love maths! We can use the formula Hotpawn and
Gilbert gave in that thread (the proof of which follows the same
pattern as the explanation I gave Jon with n replacing 100).

So, we have:-

f(n) - f(0) = n(n+1)/2

We know f(0) to be 1 so...

f(n) - 1 = n(n+1)/2

Which, adding one to both sides, is the formula that Gilbert arrived at
by intuition and testing the first few cases.

f(n) = n(n+1)/2 + 1
13. bbarr
Chief Justice
03 Nov '02 11:26
sintubin's formula can be proven via mathematical induction, if you
are looking for a proof in the rigorous sense of the term.
14. 30 Oct '02 20:53
How are we drawing the lines? Through a central point? If so, then the
figure would have:

1 lines = 2 regions
2 lines = 4 regions
3 lines = 6 regions
4 lines = 8 regions
Which would wake the pattern regions = 2n

Or more simply, just noticing that every line divides 2 regions into 4
regions (double), we get the same result.

However, if you can make the like anywhere, the circle with three lines
in it can have 7 regions, the circle with 4 can have 10, the circle with 5
can have 14 regions (am I correct) and I fail to see a pattern.

Rein