gavinksong

That would be very impressive. The explanation I have in mind does, but bonanova didn't really mention it in his post, so perhaps we will see one.

Um. So this is embarrassing. There is a slight typo in my spoiler for infinite overhang.

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.

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.

If this is a paradox, it should probably go in the paradox section. I'm kinda curious as to what this one is about, because I am pretty confused right now.

GinaW, she's looking for coordinates, and she thinks they may be spelled out. Rainman, that's probably where queensgeek got her letter sequence in the first place... What I'm more curious about is where you got those coordinates. I think DeGe is right. Look through the codes you found in the previous rounds that haven't already been used to solve another puzzle. One of them might be the key to solving the letter sequence. Maybe you can post them onto the thread, and we can help you figure it out.

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.

I was hoping somebody would bring this up. I purposefully asked about the minimum "white space" to recall BMAD and bonanova's triangle problem. 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?

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"?

Excellent solution. There is an even easier (and non trivial solution. The trivial being that the image present is the initial. Can anyone find it?

Question: Are we allowed to arrange them in 3 by 2 or 1 by 6 or 6 by 1 rather than 2 by 3? Also, do they have to be packed in a grid, or can they be staggered such as in the spoiler above?

Yes. This is exactly what I found. bona_gold_star.gif Three fourths of the triangle are self-similar copies and the rest is white. The fractional whiteness of the four parts must be equal. That's the simplest and most elegant of the three proofs. If you like, try dissecting the trapezoid and do the same proof. It has twelve triangles of three varieties and takes two equations to solve.

We traveled a kind of intuitive path in doing this analysis. So to be clear of the result, we have to make statements that are a little more precise. In fact, the statement of the puzzle uses the intuitive term "white space." This leads to the idea that the triangle is partitioned somehow into disjoint regions that are intrinsically white or black. Or, as you say "the 'white space' that the points could never occupy." -- regions where points are not allowed to land. Such regions do not exist. Not even a single point is off limits: it could be the initial point. This beautiful puzzle forces us to view the triangle not as two sets of tiny triangles, some white and some black, (more precisely some that have No Trespassing signs, others that do not) but as two sets of points: one set that is rationally related to the arbitrarily selected first point, and all the others. Consider a one-dimensional analog. Color the interval [0, 1] of the x-axis white. Now paint the point x=.5 black. Now move half-way to 0 or 1, randomly chosen, and paint that point black. Repeat ad infinitum. The rational numbers will be painted black The non-rational numbers like 1/pi and 1/e remain white. What fraction of the interval is white? What you will find is there is no interval of infinitesimally small length that is free of black points, yet we can say that the "measure" of the interval that is white is 1. Or, define f(x) to be 1 if x is rational and 0 otherwise. What is the integral of f(x) on the interval? Our normal measuring device for lengths and areas is the ordinary Riemann integral. But it only works for functions that are reasonably continuous. When you get down to point-size regions, Mr. Riemann throws up his hands and says, don't bother me, I have no idea. Go talk with Mr. Lebesgue over there. And Mr. Lebesgue says, don't talk to me about length and area, I don't know what they are. I do know about measure, however. The measure of the rationals on [0, 1] is zero. The measure of the reals on [0,1] is 1. The measure of non-rational reals on [0, 1] is 1 - 0 = 1. On the triangle, the measure of the points with rational coordinates is 0. The rest of the points have measure equal to the area of the triangle - 0. In the Lebesgue sense, (and no other "sense" applies to this question,) the "area" of the triangle that is white is the area of the triangle minus something that has the value zero. What I really liked about your analysis was talking about the decomposition into triangles of ever decreasing size. Call the sequence of areas A1, A2, A3, ..., each value 1/4 of the preceding. That decomposition does not depend on the initial point. What you said, and I think you are correct in this, that wherever the initial point is placed, say it's in a triangle of area A13, none of the following points will land in triangles with area A1, A2, ... A12, or the other regions with area A13. The following points will be in triangular regions respectively with area A14, A15 ..., and so on. That analysis shows that once the initial point is placed, there ARE point-free regions. There are regions that will never (for that initial point placement) have a point in them. But those regions (A1, A2, ... A12) are not intrinsically "white areas". The initial point could land in a triangle of area A7. Then the A1, A2, ... A6 regions, and the other A7 regions will be point-free. And the initial point (my musings...) could land in A1. The probability of that happening is 0.25, but it is not 0. One final observation. my one-dimensional case in blue above behaves differently from the triangle, and makes the triangle at once more beautiful and intriguing to analyze. In one dimension, the point can return to any region arbitrarily close to the initial point. If we simulated this case, the whole interval would eventually look black because we use dots of non-zero size. Yet the interval remains white in the Lebesgue sense. In the triangle that is not possible; the point cannot back-track. It never moves directly away from a vertex, so it can never move back in the same direction. That is why the A1 region looks white every time the simulation is run. And the A2 regions do, also, and the A3 regions. Each of all the triangles with a given area contain at most one black dot. Perhaps this is why the square shape showed no structure. I see. Let me see if I get this. You are saying that after the initial point is selected, all of the points that could be selected afterwards (or the "black space") are rationally related to the first (regardless of whether the first point was rational or irrational). And since the set of these points are discontinuous point-regions (a bit of an oxymoron), the sum of the total area of these regions should be zero. It seems that we have different definitions of "black" and "white" space. I understood "white space" to be regions that points would never be allowed to land on in any simulation (I later redefined it to be something that can only be calculated for the nth point. eg. the area of white space for the third point is 7/16). On the other hand, your defined"black space" to be regions which would eventually become continuously filled with points if one simulation ran for an infinitely long time. Both of our definitions are valid, I think, but addresses rather different problems. It is actually a pretty interesting observation that there are no continuous regions. One way to explain this phenomenon using the image of dots always traveling down a path of white triangles of decreasing size would be to note that once a dot is placed in a white triangle, that triangle effectively becomes "white space" according to my definition (or another way to put it is that it stops "decomposing"). Thus, every dot will have a triangular area around it that future dots can never land.

It just takes longer.

DeGe has an insightfully creative solution. bona_gold_star.gif 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. Ok. I feel a little scared to be saying this (especially since I barely understood any of that) but... I disagree.. with the explanation mentioning the fact that points have zero area. It's not that I disagree with the fact itself. Points do indeed have zero area. It's just that I don't feel like it is relevant. The question wasn't to add up the area that the points occupy. It was to calculate the area of the "white space" that the points could never occupy, which is also different from the area that the points do not occupy. Just because points do not have area does not mean that "black spaces" (or "not white spaces") also do not have area. It just so happens that in this case, the entire triangle is white space - though even that is not accurate, because points do occupy the triangle, and that would go against my definition of "white space". To be accurate, it would have to be said that the true white space grows with step number and eventually its area converges on that of the entire triangle but never actually attains it. If we were asked to determine the white space of a certain step, that could easily be done, and the answer would not be equal to the area of the triangle. You can easily see that in this version of the problem, it is irrelevant that points have zero area... as it also is in the original problem.

A geometrical proof?

Rainman is a sniper!