Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 158
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Passed - accepted as correct ?
Yeah.
It really isn't difficult. Got it right the first time :)
Not necessarily. It's brand new, it appeared on December 6, 2012. There's not a lot of statistics on it, so the score is low.
But in terms of difficulty, it still clearly does not look like a super-simple (although I got it right on the first try).
In general I decided in this way that, briefly speaking, the ones with white dot cannot be less, because no matter how many polygons with only black dots there are, we can arbitrarily take from them any n-gon and corresponding n+1-gon (the same, but with white dot). But we can take any triangle with white point and by removing it we will not obtain any 2-gon, agree :) because such figure does not exist, it will be just a segment (in fact it will be a 2-gon, but it will not be considered a polygon, because it is just a segment). So the conclusion is that no matter how you spin it, there will still be more of them with a white dot.
Right?
In general I decided in this way that, briefly speaking, the ones with white dot cannot be less, because no matter how many polygons with only black dots there are, we can arbitrarily take from them any n-gon and corresponding n+1-gon (the same, but with white dot). But we can take any triangle with white point and by removing it we will not obtain any 2-gon, don't you agree :) because such figure does not exist, it will be just a segment (in fact it will be a 2-gon, but it will not be considered a polygon, because it is just a segment). So the conclusion is that no matter how you spin it, there will still be more of them with a white dot.
Right?
Well, if I were a moderator on brainghams.ru, I wouldn't make that decision. It's not strict.
Think it over. I'll post my decision a little later.
Well, if I were a moderator on brainghams.rue, I wouldn't make that decision. It's lax.
Think again. I'll post my solution a little later.
Pfft. What are you talking about? It's the strictest decision I've ever made. What else could it be? My first decision was not strict, I really screwed up there, and of course I didn't get credit for not being strict. But then I wrote it, and now it was all clear, immediately scored (and scored the same moderator, who offered himself this problem on the site, so that the correctness of the solution all the more reason not to doubt). However, perhaps you misunderstood me. There I described it in a slightly different way. It is here I gave a brief answer, although the meaning seems to be the same. And the decision, which I was immediately given credit for and considered quite clear, read it yourself, here it is (in fact it is the same decision):
"Well, look at this. I think it's strict. Take the whole set of all polygons that can be drawn without a white dot. Take absolutely any such polygon (of course, each of them must have at least 3 points), chosen in a completely arbitrary way. Let's say it will be an n-gon. In that case we can always draw a so-called n+1-gon with white point (we will assume that it corresponds to our n-gon). Hence we may conclude that there are at least as many of them with white point, not less. But with the white dot there can be polygons that do not correspond to any polygon without it. This is the case if we take a triangle with two black dots. In this case we will not get a figure without the white point, we will get a line, a segment. So, out of the set of all possible polygons, the ones with the white dot are still more.
P.S.
Fortunately, all the points are on a circle, so no 3 points lie on the same line and hence any 3 or more random points can make a polygon."
Think about it some more. I will post my solution a little later.
With the white dot there are more options, because there are more vertices to build polygons.
Right.
So we have 2013 points on the circle, right?
Suppose 2013 is white, and among the set of all polygons with vertices at those points there are more with a white point numbered 2013, right?
"Well, look here. I think it's strict. Take the whole set of all the polygons you can draw without a white dot. Take absolutely any such polygon (of course, each of them must have at least 3 points), chosen in a completely arbitrary way. Let's say it will be an n-gon. In that case we can always draw a so-called n+1-gon with white point (we will assume that it corresponds to our n-gon). Hence we may conclude that there are at least as many of them with white point, not less. But with the white dot there can be polygons that do not correspond to any polygon without it. This is the case if we take a triangle with two black dots. In this case we will not get a figure without the white point, but we will get a line, a segment. So, out of the set of all possible polygons, the ones with the white dot are still more.
P.S.
Fortunately, all the points are on a circle, so no 3 points lie on the same line and hence any 3 or more random points can make a polygon."
Well, now it's clearly better and stricter. What you have written to me from the start is not strict. It's different:
RATIONALE:
Let the number of random polygons with N vertices be equal to p(N).
The number of all polygons without white dot is obviously p(2012). Let the set of all polygons without white point be {No white}.
To compute p(2013), we have to include into this number at least all different polygons from {No white}, adding to them two sides with a white point each (by connecting the white point with the start and the end vertex of the original polygon included in {No white}). We might not get all the polygons in {2013}, but it doesn't matter.
On the other hand, the addition of white-point connections to a polygon from {No white} is possible in at least 3 ways - if the original has three vertices (and there are no less than 3 vertices in {No white}). More precisely, if the initial polygon has N vertices, then by sequentially removing one of its sides, we can obtain from the same initial at least N different (N+1)-angles (because the sets of two sides with a common white vertex will be unique).
Hence, p(2013) > 3*p(2012), and therefore there are more white-point polygons.