## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# Chaos calculation

Best Answer gavinksong, 30 July 2013 - 04:09 AM

I posted this in the original thread, but I think it belongs here now.

What fraction of the area is white (prohibited)

1. For the triangle? (straightforward)
2. For the trapezoid? (more thought-provoking)

Well, for a triangle, if we add up the white space, it is just the infinite sum of a geometric series with a factor of 3/4 (the big empty triangle in the middle is one-fourth of the total area, the three smaller triangles are one-fourth of the remaining area, and so forth). Strangely, it converges at the total area of the triangle. It we calculate the black space, it is three-fourths raised to the infinite power, which is zero. So apparently, the entire triangle is supposed to be prohibited space.

I think here bonanova's musings becomes relevant:

But I'm not clear what happens if your initial point were inside.

Instead of looking backwards, as we did to solve the problem initially, we should look forwards from the initial point. This is all intuition, but you will notice, that if the initial point is in the big empty triangle in the middle, then in the next step, it will always move to one of the three smaller white triangles. And from there, it will always move into one of its three smaller "shadows" (I think that is an appropriate term, and hopefully you understand what I mean), and so on. The important point here is that we always move into smaller and smaller "shadows".

When the initial point is thrown into the triangle at random, the probability that it lies into a white triangle is 100%. However, as the simulation progresses, the points are eventually forced into smaller and smaller white triangles. In other words, the amount of "prohibited space" is not actually an absolute value, but simply increases with step number (in a sense). As in, there is no prohibited space when you throw in the first point, but the second point cannot be in the large white triangle, the third point cannot be in the large white triangle or any of its shadows, the fourth cannot be in the large white triangle or its shadows or any of its shadows' shadows, and so on.

So to us, it seems as though there is some sort of absolute law because we see the larger empty white triangles, but in reality, it is only because we do not see the infinitesimally small white triangles with the dots in them. The entire process is just endless error-correction, moving toward an impossible figure, trailing behind an interesting asymptotic pattern in the process.

Go to the full post

19 replies to this topic

### #1 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 29 July 2013 - 02:51 AM

BMAD's chaos puzzle gave an algorithm for recursively placing points,

ad infinitum, within the interior of a triangle. The beautifully surprising

result was the appearance of well-defined areas, within which points

could not be placed. Call these areas white space.

The puzzle inspired Pickett to post images, and the code that generates

them, of several other shapes. The most interesting seemed to be

BMAD's original triangle, and the trapezoid, reproduced below.

My puzzle is a simple calculation:

What fraction of the triangle is white space?

What fraction of the trapezoid is white space?

Edit: Creativity will be rewarded: Got a different approach? Get another star.

Edited by bonanova, 29 July 2013 - 03:37 AM.
Creativity rewarded

• 0

Vidi vici veni.

Senior Member

• Members
• 1837 posts
• Gender:Female

Posted 29 July 2013 - 04:02 AM

Spoiler for triangle

Edited by BMAD, 29 July 2013 - 04:03 AM.

• 0

Senior Member

• Members
• 1837 posts
• Gender:Female

Posted 29 July 2013 - 04:13 AM

Spoiler for triangle

• 0

Senior Member

• Members
• 1837 posts
• Gender:Female

Posted 29 July 2013 - 04:20 AM

Spoiler for Trapezoid -hypothesis

• 0

### #5 DeGe

DeGe

• Members
• 132 posts
• Gender:Male
• Location:Paris

Posted 29 July 2013 - 11:10 AM

Spoiler for Triangle

• 0

### #6 DeGe

DeGe

• Members
• 132 posts
• Gender:Male
• Location:Paris

Posted 29 July 2013 - 12:28 PM

Spoiler for Triangle

Spoiler for Creativity point calculations reveal a different result!!

Edited by DeGe, 29 July 2013 - 12:28 PM.

• 1

### #7 gavinksong

gavinksong

• Members
• 403 posts
• Gender:Male
• Location:i'm a bird

Posted 30 July 2013 - 04:09 AM   Best Answer

I posted this in the original thread, but I think it belongs here now.

What fraction of the area is white (prohibited)

1. For the triangle? (straightforward)
2. For the trapezoid? (more thought-provoking)

Well, for a triangle, if we add up the white space, it is just the infinite sum of a geometric series with a factor of 3/4 (the big empty triangle in the middle is one-fourth of the total area, the three smaller triangles are one-fourth of the remaining area, and so forth). Strangely, it converges at the total area of the triangle. It we calculate the black space, it is three-fourths raised to the infinite power, which is zero. So apparently, the entire triangle is supposed to be prohibited space.

I think here bonanova's musings becomes relevant:

But I'm not clear what happens if your initial point were inside.

Instead of looking backwards, as we did to solve the problem initially, we should look forwards from the initial point. This is all intuition, but you will notice, that if the initial point is in the big empty triangle in the middle, then in the next step, it will always move to one of the three smaller white triangles. And from there, it will always move into one of its three smaller "shadows" (I think that is an appropriate term, and hopefully you understand what I mean), and so on. The important point here is that we always move into smaller and smaller "shadows".

When the initial point is thrown into the triangle at random, the probability that it lies into a white triangle is 100%. However, as the simulation progresses, the points are eventually forced into smaller and smaller white triangles. In other words, the amount of "prohibited space" is not actually an absolute value, but simply increases with step number (in a sense). As in, there is no prohibited space when you throw in the first point, but the second point cannot be in the large white triangle, the third point cannot be in the large white triangle or any of its shadows, the fourth cannot be in the large white triangle or its shadows or any of its shadows' shadows, and so on.

So to us, it seems as though there is some sort of absolute law because we see the larger empty white triangles, but in reality, it is only because we do not see the infinitesimally small white triangles with the dots in them. The entire process is just endless error-correction, moving toward an impossible figure, trailing behind an interesting asymptotic pattern in the process.

• 0

Senior Member

• Members
• 1837 posts
• Gender:Female

Posted 30 July 2013 - 04:29 AM

clearly i am better at writing questions than answering them

• 0

### #9 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 30 July 2013 - 04:53 AM

I posted this in the original thread, but I think it belongs here now.

What fraction of the area is white (prohibited)

• For the triangle? (straightforward)
• For the trapezoid? (more thought-provoking)

Well, for a triangle, if we add up the white space, it is just the infinite sum of a geometric series with a factor of 3/4 (the big empty triangle in the middle is one-fourth of the total area, the three smaller triangles are one-fourth of the remaining area, and so forth). Strangely, it converges at the total area of the triangle. It we calculate the black space, it is three-fourths raised to the infinite power, which is zero. So apparently, the entire triangle is supposed to be prohibited space.

I think here bonanova's musings becomes relevant:

But I'm not clear what happens if your initial point were inside.

Instead of looking backwards, as we did to solve the problem initially, we should look forwards from the initial point. This is all intuition, but you will notice, that if the initial point is in the big empty triangle in the middle, then in the next step, it will always move to one of the three smaller white triangles. And from there, it will always move into one of its three smaller "shadows" (I think that is an appropriate term, and hopefully you understand what I mean), and so on. The important point here is that we always move into smaller and smaller "shadows".

When the initial point is thrown into the triangle at random, the probability that it lies into a white triangle is 100%. However, as the simulation progresses, the points are eventually forced into smaller and smaller white triangles. In other words, the amount of "prohibited space" is not actually an absolute value, but simply increases with step number (in a sense). As in, there is no prohibited space when you throw in the first point, but the second point cannot be in the large white triangle, the third point cannot be in the large white triangle or any of its shadows, the fourth cannot be in the large white triangle or its shadows or any of its shadows' shadows, and so on.

So to us, it seems as though there is some sort of absolute law because we see the larger empty white triangles, but in reality, it is only because we do not see the infinitesimally small white triangles with the dots in them. The entire process is just endless error-correction, moving toward an impossible figure, trailing behind an interesting asymptotic pattern in the process.

His method was the infinite sum of triangular areas, and he gives an intuitive description of the point generation process and its implications. Kudos for explaining my wonderings about placement of the initial point. It's an amazing mental image to visualize an initial point at the center try to search for a "black area" to land on, but instead finding at each step it simply lands on a smaller white area. Every point travels a path of white areas; or putting it another way, there simply are no black areas!

[As an aside, therefore, my puzzle is a bit of a ruse. The figures clearly show black areas. A moment's reflection tells us why. I constructed the figures using dots that have non-zero area. Eventually they touch and overlap each other. It's OK that the construction used black dots: it's just that to avoid being misleading they should have had zero area. If I had done that, it is perfectly clear that the entire triangle would remain white forever, or until the computer reboots, whichever comes first.]

My first crack at calculating the white space used a geometrical approach, the answer to which surprised me enough to do the math. When that agreed, a third argument suggested itself to explain what seems an impossible result. Two more Gold Stars are available for proofs that don't require infinite sums.

As for the trapezoid, I think we can safely modify whatever is proved for the triangle, without considering it a different problem.
• 0

Vidi vici veni.

### #10 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 30 July 2013 - 05:15 AM

Spoiler for Triangle

Spoiler for Creativity point calculations reveal a different result!!

DeGe has an insightfully creative solution.

Without loss of generality the triangle can be scaled by a rational number times the inverse of the coordinates of the vertices and first random point. Or we could say without loss of generality the vertices and initial point can be assigned rational coordinates. Either way, all the computed points will have rational coordinates, and as such, although infinite in number, have the cardinality of the rational numbers, Aleph Null. Their Lebesgue measure is zero. Another way to say it, as DeGe points out, they are points. But so is the interior of the triangle. The distinction is that the rationals are not dense - they are isolated points.

The entire set of points inside the triangle has the cardinality of the continuum. Those points have real coordinates and are dense; thus, their Lebesgue measure equals the area of the triangle. The computed points can be removed without changing that measure. Therefore the area of the white space is the area of the triangle.

One Gold star remains to be achieved. A geometrical proof that does not use the word infinity or its implications.

• 0

Vidi vici veni.

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users