At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

|dw:1350302056370:dw|

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

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

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...

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 ?

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.

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

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

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