A lattice of 21 dots, some red, some blue, in 3 rows of 7. Show that some 4 dots of 1 colour form the vertices of a rectangle.

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A lattice of 21 dots, some red, some blue, in 3 rows of 7. Show that some 4 dots of 1 colour form the vertices of a rectangle.

Mathematics
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|dw:1350302056370:dw|
Yes, like that, thank u for drawing it:-)
each row must have either more red, or more blue dots

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therefore two of the rows has four or more of a certain colour
In total, there are 21 dots, so at least 11 are red or at least 11 are blue..
i get a minimum of 8
|dw:1350302644089:dw|
OK, maybe we can assume that 11 are blue and go from there....
|dw:1350302783459:dw|
|dw:1350302833654:dw|
Clearly, there are some different cases to consider.....
i dont know how to be general
There is some symmetry so we could assume that the top row has 7,6,5 or 4 blue dots.
1 blue 20 red is also possible is it, it wont work.. just trying to understand the question
Then you are just working with reds instead of blues but the problem is still the same.
is this some sort of probability question?
Um...no.
looks it is related to counting
Reasoning, I guess...
is it necessary that the dots along the edge of the rectangle be the same colour as the dots at the vertices?
Only that 4 dots of 1 colour are the vertices..
|dw:1350303695070:dw|
atleast 2 are of red or blue dots in each column
|dw:1350303798138:dw|
That's right, you may just as well stick to one colour, it is simpler......
in the second column only one way is possible for not to have a rectangle
there's an odd number of rows and an odd number of columns, is that a clue to anything?
|dw:1350303894487:dw|
but third column we cant escape a rectangle !
is this ok proof ?
I think we should first start with the simplest case, which is when the complete row (say the top one) is all one colour......
oh yeah we need to exhaust all cases
begin from right side
|dw:1350304093421:dw|
For that case, the question becomes where are the 4 remaining dots?
next, to escape rectangle, we can have all 3b (or) 1r and 2 b
so next column we cant have a rectangle, so lets see 3rd column for each of those
|dw:1350304238693:dw|
anthing in third column would make a rectangle when 3bs are there in 2nd column
Do we count square as a rectangle?
Yes.
For the first case, we can say that the second row has at least 2 dots, right?
For if not, they are in the third row instead....
or else if we just prove two exact same columns exist that would be enough
Take these 2 and the 2 above them in the first row and you get a rectangle.
what's the worst case scenario? in terms of number of dots for both colours... 11 of colour A and 10 of colour B... one colour must have even dots, and another odd dots the square would result from the colour with even dots what's the smallest rectangle we can have, what's the largest..... i'm only thinking, idk if it helps or makes any sense but it's what i can contribute to the solution xD hope it helps
The other 3 cases are not quite as easy but the reasoning is similar kind to that in the first case
i see we can have all 7 unique columns
For the second case, you may as well assume that the first 6 are blue and then try to visualize where the remaining 5 are....
Hmm...not much interest in this, I will close it.....
If we take an arbitrary row of reds and blues, only 2 of the colors of the row on top of it may be the same (1 red and 1 blue), otherwise you get a rectangles: |dw:1350619459441:dw| leads to a rectangle
so, throughout the whole figure, you may only have 4 columns with consecutive colors: |dw:1350619522881:dw| (points arbitrarily chosen for demonstration)
Therefore, you must have at least 3 columns with alternating colors. These come in 2 states: R and B B R R B
therefore, by the pigeonhole principle, at least 2 of the columns share the same state
If two columns share the same state, they form a rectangle
QED

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