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.

- anonymous

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- UnkleRhaukus

|dw:1350302056370:dw|

- anonymous

Yes, like that, thank u for drawing it:-)

- UnkleRhaukus

each row must have either more red, or more blue dots

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- UnkleRhaukus

therefore two of the rows has four or more of a certain colour

- anonymous

In total, there are 21 dots, so at least 11 are red or at least 11 are blue..

- UnkleRhaukus

i get a minimum of 8

- UnkleRhaukus

|dw:1350302644089:dw|

- anonymous

OK, maybe we can assume that 11 are blue and go from there....

- UnkleRhaukus

|dw:1350302783459:dw|

- UnkleRhaukus

|dw:1350302833654:dw|

- anonymous

Clearly, there are some different cases to consider.....

- UnkleRhaukus

i dont know how to be general

- anonymous

There is some symmetry so we could assume that the top row has 7,6,5 or 4 blue dots.

- anonymous

1 blue 20 red is also possible is it, it wont work.. just trying to understand the question

- anonymous

Then you are just working with reds instead of blues but the problem is still the same.

- sasogeek

is this some sort of probability question?

- anonymous

Um...no.

- anonymous

looks it is related to counting

- anonymous

Reasoning, I guess...

- sasogeek

is it necessary that the dots along the edge of the rectangle be the same colour as the dots at the vertices?

- anonymous

Only that 4 dots of 1 colour are the vertices..

- anonymous

|dw:1350303695070:dw|

- anonymous

atleast 2 are of red or blue dots in each column

- anonymous

|dw:1350303798138:dw|

- anonymous

That's right, you may just as well stick to one colour, it is simpler......

- anonymous

in the second column only one way is possible for not to have a rectangle

- sasogeek

there's an odd number of rows and an odd number of columns, is that a clue to anything?

- anonymous

|dw:1350303894487:dw|

- anonymous

but third column we cant escape a rectangle !

- anonymous

is this ok proof ?

- anonymous

I think we should first start with the simplest case, which is when the complete row (say the top one) is all one colour......

- anonymous

oh yeah we need to exhaust all cases

- anonymous

begin from right side

- anonymous

|dw:1350304093421:dw|

- anonymous

For that case, the question becomes where are the 4 remaining dots?

- anonymous

next, to escape rectangle, we can have all 3b (or) 1r and 2 b

- anonymous

so next column we cant have a rectangle, so lets see 3rd column for each of those

- anonymous

|dw:1350304238693:dw|

- anonymous

anthing in third column would make a rectangle when 3bs are there in 2nd column

- anonymous

Do we count square as a rectangle?

- anonymous

Yes.

- anonymous

For the first case, we can say that the second row has at least 2 dots, right?

- anonymous

For if not, they are in the third row instead....

- anonymous

or else if we just prove two exact same columns exist that would be enough

- anonymous

Take these 2 and the 2 above them in the first row and you get a rectangle.

- sasogeek

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

- anonymous

The other 3 cases are not quite as easy but the reasoning is similar kind to that in the first case

- anonymous

i see we can have all 7 unique columns

- anonymous

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....

- anonymous

Hmm...not much interest in this, I will close it.....

- anonymous

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

- anonymous

so, throughout the whole figure, you may only have 4 columns with consecutive colors:
|dw:1350619522881:dw|
(points arbitrarily chosen for demonstration)

- anonymous

Therefore, you must have at least 3 columns with alternating colors. These come in 2 states:
R and B
B R
R B

- anonymous

therefore, by the pigeonhole principle, at least 2 of the columns share the same state

- anonymous

If two columns share the same state, they form a rectangle

- anonymous

QED

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