Your non-mathy friends might enjoy this (especially the ones who like the missing dollar "riddle").

One possible answer here.

Your non-mathy friends might enjoy this (especially the ones who like the missing dollar "riddle").

One possible answer here.

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And for the non-non-mathy friends: http://oeis.org/A000127

Alternately, Euler formula for planar graphs (genus 0, Euler characteristic 2) holds for all of the cases if we count the 'interior' vertices and the 'external' face:

v + f -e = 2

1 2 1

2 3 3

3 5 6

5 9 12

10 17 25

21 32 51

No reason to assume doubling were valid, induction without the induction step, etc...

I don't see how Euler's formula applies here? A clique can never be a planar graph if it has more than 5 vertices.

It's not a clique - it just looks like it. All the intersections are nodes. So Euler does apply. Incidentally, the value for 10 points is a again a power of 2 - the wrong power!

It's not planar. it has K5 as a subgraph.

And k3,3

What's referred to as "planar" here is the graph with nodes at all intersections. That one doesn't have K5 or K3,3 in it.

One possible soln: The pattern doesn't hold for points greater than 5 points?

Ah poor english! - The pattern doesn't hold for greater than 5 points? so there is no missing region.. as its not meant to be there?

Either that, or the missing region is hidden in the 7th dimension (insert string theory explanation here).

The real question is how did you decide that the "riddle" was impossible for 96.2% of the population?

It's because 86,7% of statistics are wrong

No--102% of statistics are wrong. 86.7% of statistics are made up on the spot.

Today's smbc comic is related to this in a way!

Reference for inhabitants of the future: http://www.smbc-comics.com/index.php?db=comics&id=2386

Future me thanks you very much.

Hello from 2012! Thank you for the link.

The next time a student writes some variant of "it's obvious" where the inductive step of a proof should have been, I'll point them to this comic. Great teaching example.

It went right here. Waiting for you to come.

I think the pattern is to sum the first five numbers of the (n-1)th row of Pascal's Triangle, so it looks like 2^(n-1) for a while.

(n^4-6n^3+23n^2-18n+24)/24

A mathematician, a physicist, a statistician, and an engineer enter a mathematics contest, the first task of which is to prove that all odd number are prime.

The mathematician has an elegant argument: `1's a prime, 3's a prime, 5's a prime, 7's a prime. Therefore, by mathematical induction, all odd numbers are prime.

It's the physicist's turn: `1's a prime, 3's a prime, 5's a prime, 7's a prime, 9's not a prime, 11's a prime, 13's a prime, so, to within experimental error, all odd numbers are prime.'

Statistician: Let's try several randomly chosen numbers: 17 is a prime, 23 is a prime, 11 is a prime... Looks good to me.

The most straightforward proof is provided by the engineer: `1's a prime, 3's a prime, 5's a prime, 7's a prime, 9's a prime, 11's a prime....'

1 isn't prime.

Pbtpbtpbtpbe!

"Proof by counterexample." The very best form of proof.

(By the physicist, since 1 has one too few divisors, and 9 has one too many, it evens out.)

But, induction worked the last two problems I tried to solve!

On an unrelated note, if you do this same problem without the circle, and lines that continue on infinitely, the pattern will continue (in order to double the number of regions, you must cut every region in half, which is done by the new line passing through every single old line. A line that is not parallel to any of the old lines will pass through all of them and therefore cut every region in half and therefore double the number of regions.)

that is not ture. For 3 lines you only have 7 regions, and 4 lines with 11 regions. The pattern is n(n-1)/2+1

A line Intersecting each of the other lines doesn't necessarily pass through each region.

@bmonk: you forgot the economist: '1's a prime, 2's a prime, 3's a prime, 4's a prime... so, all odd numbers are prime.

Obviously, we have to invoke 'The Quantum' here... (it's very hip these days, you know...)

:P

>(especially the ones who like the missing dollar "riddle").

I liked Boob Girl better.

could it be hidden in the width of the already drawn lines?

A000127

Wrong formula mate :P

I found the missing region guys. It was in my jean pocket but it went through the wash. Do we still want it back? It's kinda all gross and torn up now.

I've watched Polya's lecture on this :D

Is it just me or is there a problem with the regions? Euler's Formula would apply if the graph was planar. This graph is not planar, because you can find a K5 in it. Also, If I am not mistaken (and I hope I am not sounding pretentious here...) a region is defined by the circuit, but since this graph is not planar, Euler's formula doesn't apply. A simple manipulation (staying true to the rules) of this graph would yield only 25 regions, while being isomorphic to the original graph.

Let f(n) be the number of regions created in this manner with n points. Prove that f(n)=1/24*(n-1)(n-2)(n^2-3n+12)+n

For those friends who still wonder why you can use Euler's formula: You do not apply Euler's formula (v-e+f=2) to the nonplanar graph on sight. Instead you apply it to the planar graph where every intersection that you see is a vertex. In this way you easily get that the number of regions inside the circle in the case of n points drawn is

"n choose 4"+"n choose 2"+"n choose 0".

I only count 30 !!! please see http://imageshack.us/photo/my-images/716/only30.png/ . So it depends on the way you locate the point around the circle, no formula then

You have multiple lines intersecting at the same point in the center.

message from future (now past when you read this) to the even more distant past - thanks!

Easy! # of regions = 1 + (# of lines) + (# of interior line intersections) = 1 + (n choose 2) + (n choose 4).

Answer: the 0 points case stole a region from the 6 point case.

The increase of regions isn't in respect to the number of points given but rather the number of intersections over previous lines. You can avoid dealing with the idea of a closed region by creating a function of the increase in number of angles. ∑4(n-1) where n=# of total intersections. Now that we understand how the areas are created then we can use the same n to see that the number of areas increasing linearly with the number of intersections(not simply the number of lines.)

Here:

http://en.wikipedia.org/wiki/Dividing_a_circle_into_areas

center field dos not multiply after five points