anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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UnkleRhaukus
  • UnkleRhaukus
|dw:1350302056370:dw|
anonymous
  • anonymous
Yes, like that, thank u for drawing it:-)
UnkleRhaukus
  • UnkleRhaukus
each row must have either more red, or more blue dots

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

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