# Indian Statistical Institute, ISI BStat & BMath 2020 Solutions & Discussions

## Problem 1.

Let be a root of the equation and let be a root of the equation . Construct a polynomial

where are all integers such that .

# Solution:

We compute powers of

Then

Equate all coefficients to zero:

Find an integer solution:

## Problem 2.

Let be a fixed real number. Consider the equation

where is the set of real numbers. For what values of , will the equation have exactly one double root?

# Solution:

Let

if and only if Values and are roots of the equation This equation has two double roots. Value is the root of the equation

and this is the only double root of this equation.

## Problem 3.

Let and be variable points on -axis and -axis respectively such that the line segment AB is in the first quadrant and of a fixed length . Let be the mid-point of and be a point such that

(a) and the origin are on the opposite sides of and,
(b) is a line segment of length which is perpendicular to .
Find the locus of .

# Solution:

Let has coordinates and has coordinates Then Coordinates of are The vector

The unit orthogonal vector is So, we find

and coordinates of are

The point is situated on the line The maximal value of subject to is achieved when and is equal to The value of subject to is .

Answer: The locus of is a segment

## Problem 4.

Let a real-valued sequence be such that

Find all possible real values of such that .

# Solution:

We check that for all and all

Indeed,

So, the function is increasing on and decreasing on

Let and choose Then for all

It follows that

## Problem 5.

Prove that the largest pentagon (in terms of area) that can be inscribed in a circle of radius is regular (i.e., has equal sides).

# Solution:

Separate the inscribed pentagon into 5 isosceles triangles.

Let be angles of triangles between equal sides. The length of these sides are equal to the radius of the circle The area of each triangle is so, the area of the pentagon is

We have and the function is strictly concave of In particular

since So, the area of the pentagon is

the equality is achieved only when i.e. when the pentagon is regular.

## Problem 6.

Prove that the family of curves

satisfies

# Solution:

We differentiate the equality of the family of curves:

Then

and

It follows that

## Problem 7.

Consider a right-angled triangle with integer-valued sides where are pairwise co-prime. Let . Suppose divides . \\
Then

(a) Prove that
(b) Find all such triangles (i.e., all possible triplets a,b,c) with perimeter less than 100.

# Solution 1:

By the Pythagoras’ Theorem,

divides hence

Let be a prime factor of Then divides divides consequently divides But and are co-prime, and we get a contradiction. So, the only possible prime factor of is The same reasoning show that the power of in the prime factorization of is at most We have proved that either or

Let Then

It follows that is odd,

We have

The perimeter is

Corresponding triangles are

Let Thne

It follows that is even,

The perimeter is

For we have If is odd, then is even and and are not co-prime. So, in fact
Corresponding triangles are

Answer: there are six triangles that satisfy the condition:

## Problem 8.

A finite sequence of numbers is said to be if

How many alternating sequences of length , with distinct numbers can be formed such that for ?

# Solution :

We will say that a permutation of the set is an “up-down” permutation, if

We will say that a permutation of the set is a “down-up” permutation, if

Correspondence is a one-to-one correspondence between the set of “up-down” permutations and the set of “down-up” permutations. Let denote the number of “up-down” permutation. Then is the number of all alternating permutations of Consider an alternating permutation Let so that If is even, then is an “up-down” permutation of numbers (there are such permutations), and is an “up-down” permutation of numbers (there are such permutations). If is odd, then is a “down-up” permutation of numbers (there are such permutations), and is a “up-down” permutation of numbers (there are such permutations). There are ways to split numbers into groups by and elements. Hence,

Since we compute

There are permutations of the set of five numbers.

Additionally we have to choose numbers from the set

## Video Solution:

The sequence given is alternating so let us take ak=p such that k is even ,then for all ai will be less than p for i is even and for all aj will be greater than p for j is odd .
Then comes the sequence of length L and there will be [L/2] no of ais(i=even) in the sequence and [L/2] or [L/2]+1]( even and odd respectively )
Then the partition of the set {1,2…,n} in suchall ai way such ai will be chosen from {1,2..p} and aj from {p+1…n}
And that p will range from [L/2] upto n – ([L/2]+1)
The bound is there for ai and ajs
By combinatorial arguemnt, since L is odd ,we will have [L/2]+ 1 no of ajs

Summation i = [L/2] upto n-([L/2]+1) (i choose [L/2] ) (20-i choose [L/2]+1) ([L/2])!. ([L/2]1)!

Regret for making the solution look messy ,but it would have looked better in copy format ,this is something i tried to derive ,might be wrong, thank in advanced for checking for my solution.

• The first statement is wrong, “ak=p such that k is even ,then for all ai will be less than p for i is even” as we can’t compare them. Check my video solutions, I have even explained it there that we cannot compare them.