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INMO 2007

Try to solve these interesting INMO 2007 Questions.

  1. In a triangle ”ABC” right-angled at ”C”, the median through ”B” bisects the angle between ”BA” and the bisector of ”\(\angle B\)”. Prove that \(\frac{5}{2} < \frac{AB}{BC} < 3 \).
  2. Let ”n” be a natural number such that \( n = a^2 +b^2 +c^2 \) , for some natural numbers \( a, b, c \). Prove that \( 9n = (p_{1a} + q_{1b} + r_{1c})^2 + (p_{2a} + q_{2b} + r_{2c})^2 + (p_{3a}+ q_{3b} + r_{3c})^2 \),where \(p_{j} ‘s\), \(q_{j} \)’s, \( r_{j}\)’s are all nonzero integers. Further, of 3 does not divide at least one of ”a, b, c,” prove that 9n can be expressed in the form \( x^2 + y^2 + z^2 \), where ”x, y, z” are natural numbers none of which is divisible by 3.
  3. Let ”m” and ”n” be positive integers such that the equation \( x^2−mx+n = 0 \) has real roots \(\alpha\) and  \(\beta \). Prove that \(\alpha \) and \(\beta\) are integers if and only if  \([m\alpha]+[m\beta] \) is the square of an integer. (Here [x] denotes the largest integer not exceeding x.)
  4. Let \(\sigma = (a_{1}, a_{2}, a_{3}, . . . , a_{n}\)  be a permutation of  (1, 2, 3, . . . , n) . A pair     \(a_{i}, a_{j}\) is said to correspond to an inversion of \(\sigma \), if  < j  but \(a_{i} > a_{j}\) . (Example: In the permutation  (2, 4, 5, 3, 1) , there are 6 inversions corresponding to the pairs  (2, 1), (4, 3), (4, 1), (5, 3), (5, 1), (3, 1) How many permutations of (1, 2, 3, . . . , n), (n>3) , have exactly two inversions.
  5. Let ABC be a triangle in which AB = AC. Let D be the midpoint of BC and P be a point on AD. Suppose E is the foot of the perpendicular from P on AC. If \(\frac{AP}{PD}=\frac{BP}{PE}= \lambda\),\(\frac{BD}{AD}= m\) and \(z = m^2(1 + \lambda) \), prove that \(z^2− (\sigma^3− \sigma^2− 2)z + 1 = 0\). Hence show that \(\sigma ≥ 2 \) and \(\lambda = 2\) if and only if ABC is equilateral.
  6. If x, y, z are positive real numbers, prove that \( (x+y+z)^2(yz+zx+xy)^2≤ 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2) \).
  7. Let  f :Z mapsto Z be a function satisfying \(f(0) \neq 0\) , \(f(1)=0\) and 1.f(xy) + f(x)f(y) = f(x) + f(y)\), 2. \((f(x-y) – f(0))f(x)f(y) = 0\) for all x , y in Z  simultaneously. 1. Find the set of all possible values of the function f. 2. If \(f(10) \neq 0\) and  f(2) = 0 , find the set of all integers n such that \( f(n)\neq 0\) .

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A trigonometric polynomial ( INMO 2020 Problem 2)

Suppose $P(x)$ is a polynomial with real coefficients, satisfying the condition $P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta)$, for every real $\theta$. Prove that $P(x)$ can be expressed in the form$$P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}$$for some real numbers $a_0, a_1, \dots, a_n$ and non-negative integer $n$.

Using a very standard trigometric identity , we can easily convert the following ,
\begin{align*} P(\cos\theta + \sin\theta) &= P(\cos\theta - \sin\theta) \\ \implies P\left(\sqrt{2}\sin\left(\frac{\pi}{4} + \theta\right)\right) &= P\left(\sqrt{2}\cos\left(\frac{\pi}{4} + \theta\right)\right) \\ \implies P(\sqrt{2}\sin x) &= P(\sqrt{2}\cos x) \\ \end{align*} Assuming ,  $ (\frac{\pi}{4}+\theta) = x$ for all reals \(x\). So,
\[P(-\sqrt{2}\sin(x)) = P(\sqrt{2}\sin(-x)) = P(\sqrt{2}\cos(-x)) = P(\sqrt{2}\cos(x)) = P(\sqrt{2}\sin(x))\]for all \(x\in\mathbb{R}\). Since \(P(x) = P(-x)\) holds for infinitely many \(x\), it must hold for all \(x\) (since \(P(x)\) is a polynomial). so we get that ,  $P(x)$ is a even polynomial .

Also
\[P(\sqrt{2}\cos(x)) = P(\sqrt{2}\sin(x))\] implies that
\[P(t) = P(\sqrt{2}\sin(\cos^{-1}(t/\sqrt{2})))\]putting , $x=\cos^{-1}(t/\sqrt{2})$
for infinitely many \(t\) $\in [-\sqrt2 ,\sqrt2]$.
\[\sqrt{2}\sin(\cos^{-1}(t/\sqrt{2})) = \sqrt{2 - t^2}\]so we get , \(P(x) = P(\sqrt{2-t^2})\)
Again as it is a polynomial function we can extend it all $\mathbb{R} $. And we get , \(P(x) = P(\sqrt{2-x^2})\) for all reals \(x\)
Since \(P(x)\) is even , we can choose a even polynomial $Q(x)$ such that ,\(Q(x) = P(\sqrt{x+1})\). \[P(\sqrt{1+x}) = Q(x) = a_0 + a_1x^2 + a_2x^4 + \cdots + a_nx^{2n}\]now take , $\sqrt{1+x} = y$ and you get the polynomial of required form .

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Kites in Geometry | INMO 2020 Problem 1

Understand the problem

Let \( \Gamma_1 \) and \( \Gamma_2 \) be two circles with unequal radii, with centers \(  O_1 \) and \( O_2 \) respectively, in the plane intersecting in two distinct points A and B. Assume that the center of each of the circles \( \Gamma_1 \) and \( \Gamma_2 \) are outside each other. The tangent to \( \Gamma_ 1 \) at B intersects \( \Gamma_2 \) again at C, different from B; the tangent to \(   \Gamma_2 \) at B intersects \(  \Gamma_1 \) again in D different from B. The bisectors of \( \angle DAB \) and \( \angle CAB \) meet \( \Gamma_1 \) and \( \Gamma_2 \) again in X and Y, respectively. different from A. Let P and Q be the circumcenters of the triangles ACD and XAY, respectively. Prove that PQ is perpendicular bisector of the line segment \( O_1 O_2 \). 

Tutorial Problems… try these before watching the video.

1. Suppose \( P O_1 Q O_2 \) be a kite (that is \( PO_1 = PO_2 \)  and \(  QO_1 1 = QO_2 \). Show that PQ is perpendicular bisector of the other diagonal $ O_1 O_2 $.$.

2. Show that for any two circles intersecting each other at two distinct points, the common chord is bisected perpendicularly by the line joining the center.

You may send solutions to support@cheenta.com. Though we usually look into internal students work, we will try to give you some feedback.

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INMO 2020 Problem 4

Let n\ge 3 be an integer and a_1,a_2,\cdots a_n be real numbers satisfying 1<a_2\le a_2\le a_3\cdots \le a_n. If \Sigma_ia_i=2n then prove that 2+a_1+a_1a_2+a_1a_2a_3+\cdots +a_1a_2\cdots a_{n-1}\le a_1a_2\cdots a_n.

The conditions hint at inequalities involving an order, such as the rearrangement and Chebychev inequalities. Also note that a_i=2 for all i is an equality case, hence we should try to use inequalities in such a way that matches the equality case.

The RHS can be rewritten as a_1a_2\cdots a_n= a_1a_2\cdots a_n-a_1a_2\cdots a_{n-1}+a_1a_2\cdots a_{n-1}-a_1a_2\cdots a_{n-2}+a_1a_2\cdots a_{n-2}\cdots -a_1+a_1=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+a_1. That is, a_1a_2\cdots a_n-1= a_1a_2\cdots a_n-a_1a_2\cdots a_{n-1}+a_1a_2\cdots a_{n-1}-a_1a_2\cdots a_{n-2}+a_1a_2\cdots a_{n-2}\cdots -a_1+a_1=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+(a_1-1).

Now Chebychev inequality gives \frac{a_1a_2\cdots a_n-1}{n}= \frac{a_1a_2\cdot a_{n-1}(a_n-1)+a_1a_2\cdot a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+(a_1-1)}{n}\ge \frac{1+a_1+a_1a_2+\cdots +a_1a_2\cdots a_{n-1}}{n}\cdot\frac{(a_1-1+a_2-1+\cdots +a_n-1)}{n}=\frac{1+a_1+a_1a_2+\cdots +a_1a_2\cdots a_{n-1}}{n}. Cancelling the denominators, we get the desired result.

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Number Theory, Ireland MO 2018, Problem 9

Understand the problem

The sequence of positive integers $a_1, a_2, a_3, ...$ satisfies $a_{n+1} = a^2_{n} + 2018$ for $n \ge 1$.
Prove that there exists at most one $n$ for which $a_n$ is the cube of an integer.

Source of the problem

Ireland MO 2018, Problem 9

Topic
Number Theory
Difficulty Level
8/10
Suggested Book
Excursion in Mathematics by Bhaskaryacharya Prathisthan

Start with hints

Do you really need a hint? Try it first!

, wIt is so important to know and use the modulo technqiue at the right time.  We will use the modulo technique, i.e. we will see the problem through the lens of modulo some number. What is that number? If you visit this website, you will understand that to handle cubes modulo something is 9. So, we will deal the whole equation modulo 9.  

Definition: kth power residue of a number n is the complete residue system modulo n. For eg: Quadratic Residue (2nd power) of 4 is {0,1}.

  • Cubic(3rd) Power Residue of 9 is {0,1,-1}.
  • 6th Power Residue of 9 is {0,1}
  • Quadratic(2nd Power) Residue of 9 is {0,1,4,7}
We will use these ideas here.  
Let $a_k$ be the smallest integer which is a cube; let $a_k=a^3$. Note that, $a_{k+1}=a^6+2018$.  Now, the modulo picture comes in. Starting from this cube. We will observe the sequence modulo 9. Case 1: \( a_k = 0\) mod 9 Then, the sequence modulo 9 will be  $0 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9  are {0,1,-1}.
Case 2: \( a_k = 1,-1\) mod 9 Then, the sequence modulo 9 will be  $\pm 1 \mapsto 3 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9  are {0,1,-1}. QED

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Number Theory, France IMO TST 2012, Problem 3

Understand the problem

Let $p$ be a prime number. Find all positive integers $a,b,c\ge 1$ such that:
\[a^p+b^p=p^c.\]
Source of the problem
France IMO TST 2012, Problem 3
Topic
Number Theory
Difficulty Level
7/10

 

Suggested Book
Challenges and Thrills of Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

Observe that  we will try to fundamental solutions to \( a^p + b^p = p^c\). 

A fundamental solution (a,b,c,p) gives infinitely many solutions \( (a.p^k, b.p^k, c+k, p)\).

A fundamental solution is, therefore (a,b,c,p) if gcd(a,b) = 1.

We will focus on the fundamental solutions.

We will deal with two cases:

Case 1: p = 2

The equation reduces to \( a^2 + b^2 = p^2 \). 

As gcd(a,b) = 1, it implies a and b are odd.

Now any odd square = 1 mod 4.

So, \( a^2 + b^2 = 2 mod 4 \).

Hence, the only fundamental solution is (1,1,1) = (a,b,c) 

We have that the following solutions are: $(1, 1, 1)$ and $\left(2^k, 2^k, 2^{2k+1}\right)$.

 

Case 2: p is an odd prime

This now requires the idea of Lifting the Exponents. Please read here if you don’t know it.

It is an advanced technique to deal with Diophantine Equations.

Let’s check that the conditions of the LTE are satisfying here.

p is an odd prime. gcd(a,b) = 1.

p doesn’t divide a or b as we are looking for fundamental solutions.

\( a^p = a mod p; b^p = b mod p \).

Hence, \( a^p + b^p = a + b mod p \).

So, p | a+b, and p don’t divide a or b.

 Hence, we can apply LTE.

Using the LTE idea, we get 

$1+v_p(a+b)=v_p\left(a^p+b^p\right) = c$

Then since $a+b | a^p+b^p$, we have that $a+b = p^{c-1}$, so $a^p+b^p = pa+pb$

Now, you see this can’t happen for large p, as the LHS is exploding too fast like exponential as p increases and RHS is linear in p.

So, we will apply inequality to prove this and find a bound for p for which it works and search in that bound.

Note that $a^p \geq pa, b^p \geq pb$ or $a^{p-1} \geq p, b^{p-1} \geq p$ if $a, b \geq 2$.

Therefore, we must have that $a=1, b=p^{c-1}-1$ (or vice versa). But clearly $\left(p^{c-1}-1\right)^p > p^c$, so no solution.

QED

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Algebra, Austria MO 2016, Problem 4

Understand the problem

Let $a,b,c\ge-1$ be real numbers with $a^3+b^3+c^3=1$.
Prove that $a+b+c+a^2+b^2+c^2\le4$, and determine the cases of equality.
Source of the problem
Austria MO 2016. Final Round, Problem 4
Topic
Inequality
Difficulty Level
Suggested Book
Challenges and Thrills of Pre-College Mathematics

Start with hints

Do you really need a hint? Try it first!

The idea is that you have to capture the symmetry in the equations and correspondingly find it.

Observe that the inequality $a+b+c+a^2+b^2+c^2\le4$ with the constraint $a^3+b^3+c^3=1$ can be written as 

\( (a^3 – a ^2 – a +1)  + (b^3 – b ^2 – b +1) + (c^3 – c ^2 – c +1) \ge 0  \) using the constraint. 

 

Now, \( (a^3 – a ^2 – a +1)  + (b^3 – b ^2 – b +1) + (c^3 – c ^2 – c +1) \ge 0  \) demands you to look into the polynomial  $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$.

Thus, the problem reduces to show that if $a,b,c\ge-1$ are real numbers, then  $P(a)+P(b)+P(c)\ge0$.

What if we can show that individually if  $x\ge-1$ we always have $P(x)\ge0$?

Then our problem will be solved right?

We have $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$, observe that it automatically implies that if \( x+1 \geq 0 \) then we will have \( P(x) \geq 0\).

 

Equality Cases:

For equality we must have $P(a)=P(b)=P(c)=0$, and hence $a,b,c\in\{-1,+1\}$.
Hence equality holds if and only if one of the three variables is $-1$ and the other two are $+1$.

QED

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Number Theory, Cyprus IMO TST 2018, Problem 1

Understand the problem

Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square.
Source of the problem
Cyprus IMO TST 2018, Problem 1
Topic
Number Theory
Difficulty Level
7/10
Suggested Book
Challenges and Thrills of Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

Let us write the problem in Mathematical Language i.e. in the form of equations. 

\( (11111)_n\) in base n \( = 1 + n + n^2 + n^3 + n^4 \).

So, the problem reduces to finding positive integer solutions to \( m^2 = 1 + n + n^2 + n^3 + n^4 \).

 

The idea is that we will try to bound the \( 1 + n + n^2 + n^3 + n^4 \) in between some squares and from that we will try to estimate the values of m in terms of n.

Observe that$(2m)^2=4n^4+4n^3+4n^2+4n+4.$

Now, can you form squares from the right side?  If not can you bound it by two squares?

 

 

First of all to form, you take the max terms \(4n^4 = (2n)^2 \). So, that term must be included in the square.

Also, try to find a, b, c such that \( (2n^2 + an + b)^2\) can be made greater or lesser the given expression. 

Observe that you will get the following.

$(2n^2+n)^2<4n^4+4n^3+4n^2+4n+4<(2n^2+n+2)^2$

Now, guess that  $(2n^2+n)^2<(2m)^2<(2n^2+n+2)^2$

So, what we get the relationship of m and n?

We get that \((2m) = (2n^2 + n + 1) \). Hence, 

$(2m)^2=(2n^2+n+1)^2 \Leftrightarrow 4n^4+4n^3+4n^2+4n+4=(2n^2+n+1)^2.$

Observe that, this results in a lot of cancellation of terms and we are left with: 

$n^2-2n-3=0.$This gives the solution (m,n) = (11, 3)

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Categories
Math Olympiad

Number Theory, South Africa 2019, Problem 6

Understand the problem

Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$$
 20^m - 10m^2 + 1 = 19^n.
$$
Source of the problem
South Africa MO 2019, Problem 6
Topic
Number Theory
Difficulty Level
7/10
Suggested Book
Excursion in Mathematics by Bhaskaracharya Prathisthan

Start with hints

Do you really need a hint? Try it first!

If you read this, you will get to know some techniques to explore Diophantine Equations.

Let’s get an idea of m and n  by using the modulo technique.

To get an idea of n, we must remove or eliminate m, to do that we take modulo 10.

Observe that the equation demands to be taken modulo 10, given the numbers and it turns out that \( 19^n = (-1)^n = 1 mod 10 \). It implies that n must be even.

Try to get an idea of m now.

Also, (0,0) is a solution. So, we take both m and n as non-zero.

To remove n, using the information that n is even, we can remove the variable n, taking modulo 4.

So, \( 2m^2 + 1 = 1 mod 4  \) implies m must be even. Let m = 2p.

Observe that RHS is a square and LHS \( < 20^{2p} \).

The largest square \( < 20^{2p}\) is \( (20^{p} – 1)^2\).

Thus,  \( LHS \leq (20^{p} – 1)^2 \).

Hence $20^{2p} - 10(2p)^2 + 1 ~\le~ (20^p-1)^2 ~=~ 20^{2p}-2\cdot20^p+1$,
which simplifies to   $p^2\ge20^{p-1}$. (*)

Now, this turns out to be inequality and this results in the solution p = 1.

This gives the only solution (2,2) as (m,n).

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