**Problem 1.**

Given any set of points within a unit square, show that there always exists three distinct points in such that the area of the triangle formed by the three points is less than or equal to . (Combinatorics)

Given any set of points within a unit square, show that there always exists three distinct points in such that the area of the triangle formed by the three points is less than or equal to . (Combinatorics)

For every positive integer , prove that

where is the sum of all the divisors of . (Number Theory)

Does there exist non-linear polynomials and such that (Polynomials)

If is a twice differentiable function and is continuous on an open interval in , then prove that

(Calculus)

Thanks sir

Sir..can i submit solution??

Yes, you can type them here.

For the first problem , pigeonhole principle solves the problem quite easily .

We divide the given square into smaller squares of sides of length each . By pigeonhole principle , we can say that when 9 points are divided amongst 4 smaller squares , one sqaure must contain points .

Let’s call the square M . In the sqaure M, we join any of the diagonals , and we have a right angle triangle whose maximum base is and maximum height is

So we can conlude that the area has an upper bound of

Yes, you are correct but the result that you are using for the maximum area of a triangle inside a square has to be explicitly explained as to why some other triangles will not have area more than where is the side length of the square.

For the second problem

We start by plugging in values .Like for

We notice that all of the fractions are in the form of .

Then we have the fraction occurring times.

Note that , we are basically counting the number of times 1/k occurs.

So the sum

And thus proved .

Although the bound is probably even less than

The latex equation got messed up in this one .

I think it’s fine now. Thanks a lot for the discussion. I will comment on the posts tomorrow early morning. Till then solve the rest.

You are absolutely correct. Well done.