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estudier
 3 years ago
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.
estudier
 3 years ago
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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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350302056370:dw

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, like that, thank u for drawing it:)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0each row must have either more red, or more blue dots

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0therefore two of the rows has four or more of a certain colour

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0In total, there are 21 dots, so at least 11 are red or at least 11 are blue..

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i get a minimum of 8

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350302644089:dw

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0OK, maybe we can assume that 11 are blue and go from there....

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350302783459:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350302833654:dw

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Clearly, there are some different cases to consider.....

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i dont know how to be general

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0There is some symmetry so we could assume that the top row has 7,6,5 or 4 blue dots.

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.01 blue 20 red is also possible is it, it wont work.. just trying to understand the question

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Then you are just working with reds instead of blues but the problem is still the same.

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0is this some sort of probability question?

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0looks it is related to counting

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0is it necessary that the dots along the edge of the rectangle be the same colour as the dots at the vertices?

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Only that 4 dots of 1 colour are the vertices..

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350303695070:dw

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0atleast 2 are of red or blue dots in each column

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350303798138:dw

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0That's right, you may just as well stick to one colour, it is simpler......

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0in the second column only one way is possible for not to have a rectangle

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0there's an odd number of rows and an odd number of columns, is that a clue to anything?

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350303894487:dw

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0but third column we cant escape a rectangle !

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0I think we should first start with the simplest case, which is when the complete row (say the top one) is all one colour......

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0oh yeah we need to exhaust all cases

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0begin from right side

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350304093421:dw

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0For that case, the question becomes where are the 4 remaining dots?

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0next, to escape rectangle, we can have all 3b (or) 1r and 2 b

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0so next column we cant have a rectangle, so lets see 3rd column for each of those

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350304238693:dw

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0anthing in third column would make a rectangle when 3bs are there in 2nd column

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0Do we count square as a rectangle?

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0For the first case, we can say that the second row has at least 2 dots, right?

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0For if not, they are in the third row instead....

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0or else if we just prove two exact same columns exist that would be enough

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Take these 2 and the 2 above them in the first row and you get a rectangle.

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0what'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

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0The other 3 cases are not quite as easy but the reasoning is similar kind to that in the first case

sara12345
 3 years ago
Best ResponseYou've already chosen the best response.0i see we can have all 7 unique columns

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0For the second case, you may as well assume that the first 6 are blue and then try to visualize where the remaining 5 are....

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm...not much interest in this, I will close it.....

MrMoose
 3 years ago
Best ResponseYou've already chosen the best response.1If 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

MrMoose
 3 years ago
Best ResponseYou've already chosen the best response.1so, throughout the whole figure, you may only have 4 columns with consecutive colors: dw:1350619522881:dw (points arbitrarily chosen for demonstration)

MrMoose
 3 years ago
Best ResponseYou've already chosen the best response.1Therefore, you must have at least 3 columns with alternating colors. These come in 2 states: R and B B R R B

MrMoose
 3 years ago
Best ResponseYou've already chosen the best response.1therefore, by the pigeonhole principle, at least 2 of the columns share the same state

MrMoose
 3 years ago
Best ResponseYou've already chosen the best response.1If two columns share the same state, they form a rectangle
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