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# Bouncing Point

Best Answer Anza Power, 03 August 2013 - 07:34 PM

I have a proof:

Spoiler for
Go to the full post

12 replies to this topic

### #1 gavinksong

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Posted 02 August 2013 - 05:47 PM

This may not be very good, since it is my first problem - but here it goes...

There is a two-dimensional box and a point.

The point starts somewhere in the box and moves straight in some direction. When it hits a wall, it bounces off of it.

The point leaves behind a black line as it travels.

Is there ever a case where the entire box becomes colored if the point travels forever? If not, what is the minimum area of the "white space"?

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### #2 bonanova

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Posted 02 August 2013 - 09:24 PM

Spoiler for for starters

Edited by bonanova, 02 August 2013 - 11:25 PM.
Equation for bounces

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Vidi vici veni.

### #3 DeGe

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Posted 03 August 2013 - 12:29 AM

Spoiler for I think

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### #4 gavinksong

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Posted 03 August 2013 - 05:11 AM

Spoiler for I think

I was hoping somebody would bring this up.

Of course, since a line has zero area, the area to my minimum white space question is almost certainly zero or the area of the entire rectangle - so the way I phrased that question is actually rather unnecessary. It was mostly to get people to think about how it is possible for an infinite series of lines to completely color a plane, in contrast to the solution for the fractal problem. If you look at bonanova's post on this thread, you can see how he defines "completely coloring the plane," and he is correct.  If it is possible to calculate, for any point in the rectangle, the time at which the line crosses that point, then the rectangle is completely colored.

Your explanation is not the one I was looking for - although I have to admit that since this is an original problem, I am not quite sure of my own solution. Could you explain how you came upon this conclusion: "Then for any point within this area, there is a line that passes through this point at each and every angle from 0 to 360 degrees"? Isn't it sufficient for every point within the area to have just one line that passes through it?

Edited by gavinksong, 03 August 2013 - 05:12 AM.

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### #5 gavinksong

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Posted 03 August 2013 - 05:20 AM

Spoiler for for starters

These are all good observations.

In your original post, you predicted that it is possible to color the entire rectangle if the slope is irrational, but in your edit, it looks like you concluded that your equations might not have a solution. So it looks like it could go either way at this point.

Edited by gavinksong, 03 August 2013 - 05:20 AM.

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### #6 DeGe

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Posted 03 August 2013 - 09:11 AM

Your explanation is not the one I was looking for - although I have to admit that since this is an original problem, I am not quite sure of my own solution. Could you explain how you came upon this conclusion: "Then for any point within this area, there is a line that passes through this point at each and every angle from 0 to 360 degrees"? Isn't it sufficient for every point within the area to have just one line that passes through it?

Within an "area", if you consider any random point, then all the points around it must also be colored, meaning that there must be lines through this point in all possible directions, hence the 0 to 360 degree comment. The problem is similar to the triangle problem where the naked eye tells you that there is a "black area" but infact it is so because you choose the thickness of a point, or a line in this case. If instead of point bouncing off, you had a ball bouncing, I would definitely agree with Bonanova that the complete rectangle can be colored and a 0 white area left off as minimum

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### #7 Rainman

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Posted 03 August 2013 - 11:08 AM

Spoiler for

Edited by Rainman, 03 August 2013 - 11:11 AM.

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### #8 gavinksong

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Posted 03 August 2013 - 12:31 PM

Your explanation is not the one I was looking for - although I have to admit that since this is an original problem, I am not quite sure of my own solution. Could you explain how you came upon this conclusion: "Then for any point within this area, there is a line that passes through this point at each and every angle from 0 to 360 degrees"? Isn't it sufficient for every point within the area to have just one line that passes through it?

Within an "area", if you consider any random point, then all the points around it must also be colored, meaning that there must be lines through this point in all possible directions, hence the 0 to 360 degree comment. The problem is similar to the triangle problem where the naked eye tells you that there is a "black area" but infact it is so because you choose the thickness of a point, or a line in this case. If instead of point bouncing off, you had a ball bouncing, I would definitely agree with Bonanova that the complete rectangle can be colored and a 0 white area left off as minimum

Just because all the points around a point are colored does not mean that there must be lines going through the point at all angles. Just consider the scenario where there are an infinite number of parallel lines going through all of the points.

In the triangle problem, it was because the "black areas" for a given initial point were in fact points-sized and were sparsely packed. In other cases, you can consider any solid area to be made up of an infinite number of points. The infinitesimal thickness of a line or a point is not sufficient to conclude that an area composed of an infinite number of these is also zero.

bonanova was getting very close to the answer.

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### #9 gavinksong

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Posted 03 August 2013 - 12:56 PM

Spoiler for

This is honestly a much more complicated proof than I had in mind. To be honest, this is going to take me a while to read and understand. So give me a moment. Maybe several.

But first, I'll go ahead and say that you are correct in saying that the rectangle can never be colored completely, and that there IS a simpler explanation out there. And also, I meant for the box to be rectangular.

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### #10 Rainman

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Posted 03 August 2013 - 03:00 PM

It would be nice to see a proof that doesn't use the concept of countability
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