Week 1 Challenging Problems

Fractions > Week 1 Challenging Problems

Problems for the week

Deadline 19.04.2020

Problem 1.

Given any set $S$ of $9$ points within a unit square, show that there always exists three distinct points in $S$ such that the area of the triangle formed by the three points is less than or equal to $\frac{1}{8}$ . (Combinatorics)

Problem 2.

For every positive integer $n$ , prove that

$\frac{\sigma(1)}{1} + \frac{\sigma(2)}{2}+\dots+\frac{\sigma(n)}{n} \leq 2n$

where $\sigma(n)$ is the sum of all the divisors of $n$ . (Number Theory)

Problem 3.

Does there exist non-linear polynomials $P$ and $Q$ such that $P(Q(x))=(x-1)(x-2)\dots (x-15)?$ (Polynomials)

Problem 4.

If $f$ is a twice differentiable function and $f''$ is continuous on an open interval in $\mathbb{R}$ , then prove that

$\lim_{h\to{0}} \frac{f(x+h)-2f(x)+f(x-h)}{h^2} = f''(x).$

(Calculus)

Go to Week 2 to challenging problems: Click Here

9 Comments

Shakshi raj

April 14, 2020 at 10:33 pm Reply

Thanks sir
Sounak Das

April 17, 2020 at 5:42 pm Reply

Sir..can i submit solution??
- FractionsHub
  
  April 17, 2020 at 5:46 pm Reply
  
  Yes, you can type them here.
Chowman

April 17, 2020 at 7:04 pm Reply

For the first problem , pigeonhole principle solves the problem quite easily .
We divide the given square into $4$ smaller squares of sides of length $\frac{1}{2}$ each . By pigeonhole principle , we can say that when 9 points are divided amongst 4 smaller squares , one sqaure must contain $3$ 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 $\frac{1}{2}$ and maximum height is $\frac{1}{2} .$
So we can conlude that the area has an upper bound of $\frac{1}{2}* \frac{1}{2} * \frac{1}{2} \leq \frac{1}{8}$
- FractionsHub
  
  April 18, 2020 at 5:39 am Reply
  
  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 $\frac{1}{2}\times s^2$ where $s$ is the side length of the square.
Chowman

April 17, 2020 at 7:09 pm Reply

For the second problem
We start by plugging in values .Like for $\frac{\sigma (6)}{6}=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{6}{6}$
We notice that all of the fractions are in the form of $\frac{1}{k}$ .
Then we have the fraction $\frac{1}{k}$ occurring $[\Floor{\frac{n}{k}}]$ times.

Note that $k\times[{\frac{n}{k}}]$ , we are basically counting the number of times 1/k occurs.

So the sum

$\frac{n}{1} + .... + \frac{n}{n^2} \leq \frac{n}{1} + \frac{n}{4} + .... + \frac{n}{n^2} + .... = n(\frac{\pi^2} { 6}) \leq 2n$

And thus proved .

Although the bound is probably even less than $n(\frac{\pi^2} { 6})$
- Chowman
  
  April 17, 2020 at 7:13 pm Reply
  
  The latex equation got messed up in this one .
  - FractionsHub
    
    April 17, 2020 at 7:28 pm Reply
    
    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.
- FractionsHub
  
  April 18, 2020 at 5:36 am Reply
  
  You are absolutely correct. Well done.