Functional equations challenge 1


This is a challenging problem !!

Find all continuous  functions  f: R \longrightarrow R    such that for all x\inR:

f(1 - x) = 1 - f(f(x))

Solution : ( first try it , if not then you can read )

1) replacing the equation by a simpler one :   g(g(x))=-g(-x)
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Let   g(x)=\frac 12-f(\frac 12-x)  and so  f(x)=\frac 12-g(\frac 12-x) and the equation becomes -g(x-\frac 12)=g(g(\frac 12-x)) and so :

g(x) is a continuous function such that g(g(x))=-g(-x)
Q.E.D.

2) g(x)=x \forall x\in g(\mathbb R)
=======================

Let A=g(\mathbb R)
From 1) above, we get that a\in A\implies -a\in A and so, since g(x) is continuous :
Either A=\mathbb R, either A=[-u,u] for some u\ge 0.

g(g(x))=-g(-x) \implies g(g(g(g(x))))=-g(-g(g(x)))=-g(g(-x))=g(x) and so g(g(g(x)))=x \forall x\in A

Let then g_r(x) : A\to A the restriction of g(x) to A :
g(g(g(x)))=x \forall x\in A \implies g_r(g_r(g_r(x)))=x \forall x\in A and so g_r(x) is a bijection, so is monotonous
g(g(x))=-g(-x) \implies g_r(g_r(x))=-g_r(-x) implies then -g_r(-x) is increasing and so g_r(x) is increasing.

So, g_r(a)> a \implies g_r(g_r(a))>g_r(a)>a and so g_r(g_r(g_r(a)))>g_r(a)>a which is impossible and so g_r(a)\le a
Same, g_r(a)< a \implies g_r(g_r(a))<g_r(a)<a and so g_r(g_r(g_r(a)))<g_r(a)<a which is impossible and so g_r(a)\ge a

So g_r(x)=x
Q.E.D.

3) general solution of g(g(x))=-g(-x) with g(x) continuous
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Then the original equation g(g(x))=-g(-x) becomes g(x)=-g(-x) (since g(g(x))=g_r(g(x))=g(x)) and so :

The general continuous solutions of g(g(x))=-g(-x) are any odd continuous function g(x) such that g(x)=x \forall x\in g(\mathbb R)
A simple way to build them is :
3.1) If g(\mathbb R)=\mathbb R : g(x)=x \forall x

3.2) If g(\mathbb R)=[-u,u] for some u\ge 0 :
Let any continuous fonction h(x) from [u,+\infty)\to[-u,+u] such that h(u)=u and define g(x) as :
\forall x\in(-\infty,-u) : g(x)=-h(-x)
\forall x\in[-u,+u] : g(x)=x
\forall x\in(u,+\infty) : g(x)=h(x)

Notice that u=0 gives the solution g(x)=0

4) general solution of f(1-x)=1-f(f(x)) with f(x) continuous
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The general solution of original equation is f(x)=\frac 12-g(\frac 12-x)) where g(x) is any function described in 3) above

5) examples of solutions
======================
g(x)=0 gives the solution f(x)=\frac 12

g(x)=x gives the solution f(x)=x

u=1 and h(x)=\sin(\frac{\pi}2x) gives the solution :
If |x-\frac 12|<1 : f(x)=x
If |x-\frac 12|\ge 1 : f(x)=\frac 12+\sin(\frac{\pi}2x-\frac{\pi}4)

u=2 and h(x)=4|2\{\frac{x-2}8\}-1|-2 gives the solution : f(x)=\frac 52-4|2\{\frac{-2x-3}{16}\}-1| :)

Notice that the only differentiable solutions are f(x)=\frac 12 and f(x)=x

IMO 1983 – Problem 3


Let N and k be positive integers and  let S be a set of   n points in the plane such that

(i) no three points of S  are collinear, and
(ii)  for any point P of  S, there are at least k points of S equidistant from P

Prove that  k  <  \frac{1}{2}  +  \sqrt{2n}

Try the question …
Solution will be updated soon

IMO 2012 problems


This year IMO problems !!!!

Problem 1 :

Given triangle ABC the point J is the centre of the excircle opposite the vertex A. This excircle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively. The lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC. Prove that M is the midpoint of ST.

Problem 2 :

Let {n\ge 3} be an integer, and let {a_2,a_3,\ldots ,a_n} be positive real numbers such that  {a_{2}a_{3}\cdots a_{n}=1}  Prove that

\displaystyle \left(a_{2}+1\right)^{2}\left(a_{3}+1\right)^{3}\dots\left(a_{n}+1\right)^{n}>n^{n}.

Problem 3 :

The liar’s guessing game is a game played between two players {A} and {B}. The rules of the game depend on two positive integers {k} and {n} which are known to both players.

At the start of the game {A} chooses integers {x} and {N} with {1 \le x \le N.} Player {A} keeps {x}secret, and truthfully tells {N} to player {B}. Player {B} now tries to obtain information about {x} by asking player {A} questions as follows: each question consists of {B} specifying an arbitrary set {S} of positive integers (possibly one specified in some previous question), and asking {A}whether {x} belongs to {S}. Player {B} may ask as many questions as he wishes. After each question, player {A} must immediately answer it with [i]yes[/i] or [i]no[/i], but is allowed to lie as many times as she wants; the only restriction is that, among any {k+1} consecutive answers, at least one answer must be truthful.

After {B} has asked as many questions as he wants, he must specify a set {X} of at most {n}positive integers. If {x} belongs to {X}, then {B} wins; otherwise, he loses. Prove that:

1. If {n \ge 2^k,} then {B} can guarantee a win. 2. For all sufficiently large {k}, there exists an integer {n \ge (1.99)^k} such that {B} cannot guarantee a win.

Problem 4 :

Find all functions {f:\mathbb Z\rightarrow \mathbb Z} such that, for all integers {a,b,c} that satisfy {a+b+c=0}, the following equality holds:

\displaystyle f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).

 Problem 5 :

Let {ABC} be a triangle with {\angle BCA=90^{\circ}}, and let {D} be the foot of the altitude from {C}. Let {X} be a point in the interior of the segment {CD}. Let {K} be the point on the segment {AX}such that {BK=BC}. Similarly, let {L} be the point on the segment {BX} such that {AL=AC}. Let {M} be the point of intersection of {AL} and {BK}.

Show that {MK=ML}.

Problem 6 :

Find all positive integers {n} for which there exist non-negative integers {a_1, a_2, \ldots, a_n} such that

\displaystyle \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1.

You may download the PDF version :
IMO 2012


Thank you and good luck for these delicious problems

IMO 2007 Short list problem


Find all functions  \large f:\mathbb{R+} \rightarrow \mathbb{R+} such that

f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right)

for all x,y \large \in  \large \mathbb{R+}

Solution :

For any positive real numbers z, we have that

f(x+f(y))+z=f(x+y)+f(y)+z
f(f(x+f(y))+z)=f(f(x+y)+f(y)+z)
f(x+f(y)+z)+f(x+f(y))=f(x+y+f(y)+z)+f(x+y)
f(x+y+z)+f(y)+f(x+y)+f(y)=f(x+2y+z)+f(y)+f(x+y)
f(x+y+z)+f(y)=f(x+2y+z)
f(a)+f(b)=f(a+b)

and by Cauchy in positive reals, then f(x)=\alpha x for all x\in (0,\infty)
Now it’s easy to see that \alpha=2 , then f(x)=2x    for all positive real numbers x

The Folk Problem ! ! !


This is very intresting and can be said as Tough Problem !!!!

A cencus taker comes to the home of a lady and aks her,”How many children do you have and what are there ages?”The lady was crazy enough to answer like this,”I have three daughters and product of the greatest integer function of their ages is 36.”.Now the cencus taker asked for some other information too.The lady replied,”If I tell you the sum of greatest integer function of their ages ,then also you can not point out.”So the cencus taker asked for some useful information.The lady laughingly said,”My elder daughter fell in the bathroom yesterday”.The cencus taker understood what the ages of the 3 daughters were and went away.What are the ages of the 3 daughters of the crazy lady?

Solution Updated :
Best solution is given by Anirud Thyagharajan
Those who gave correct answers and solutions are :
Tom Sawyer ,  Anirud Thyagharajan  , Tarun Garg

The answer is: Elder girl’s age is 9 years and the other 2 have ages 2, and 2 each. This is expressed in GIF form. Solution: According to the first condition, We obtain that the product of GIF of the three daughter’s ages is 36. Now, we also know that it is not possible to ascertain the triplet using the sum of the GIF’s of the ages, so , the sum must be repetitive. Preparing a set of triplets, (1,36,1) (1,18,2) (1,12,3) (1,6,6) (1,4,9) (2,3,6) (2,9,2) (3,3,4) Now, the set of sums are { 38,21,16,14,13,13,10,11} 13 repeats itself in two forms {2,2,9} { 6,6,1} However, here , is where it becomes important, as the elder daughter falls, it is specified that there is ONLY one elder daughter, which cancels the second option. So, the answer is (2,2,9) 🙂

Infinite number of primes :)


In today’s post , I will present a very well known proof. What makes this proof especially appealing is that it is not too complex. Even so, it is very powerful. This theorem was first presented in Euclid’s Elements (Book IX, Proposition 20).

Theorem: There are an infinite number of primes.

Proof:

(1) Assume that there is only a finite number of primes.

(2) Then, there exists a prime pn that is the largest prime.

(3) Let p1, p2, …, pn be the list of all primes that exist.

(4) Let x = p1*p2*…*pn + 1.

(5) By the fundamental theorem of arithmetic (see Theorem 3, here), we know there is at least one prime that divides x. Let us call this prime p*.

(6) But none of the primes p1 … pn divide x since x ≡ 1 (mod pi) for any of the primes p1 … pn

(7) Therefore, we have a contradiction. We have a prime p* that is not in the complete list of primes.

(8) So, we reject our assumption in step#1 and conclude that there are an infinite number of primes that exist.

QED

Quadratic Formula


One of the most useful formulas of all time historically is the solution of the equation:
ax2 + bx + c = 0

The equation above is known as the quadratic equation.

Theorem: (-b ± √b2 – 4ac)/2a is the solution to the quadratic equation.

(1) First, we multiply both sides by 4a and get:

4a2x2 + 4abx + 4ac = 0

(2) Next, we add b2 – b2 to the equation:

4a2x2 + 4abx + b2 + 4ac – b2 = 0

(3) Now, we add b2 – 4ac to both sides which gives us:

4a2x2 + 4abx + b2 = b2 – 4ac

(4) Further, we know that:

(2ax + b)2 = 4a2x2 + 4axb + b2

(5) Combining #4 and #3, gives us:

(2ax + b)2 = b2 – 4ac

(6) Now, taking the square root of both sides gives us:

2ax + b = ±√b2 – 4ac

(7) Now, using basic algebra, we get to:

x = (-b ±√(b2 – 4ac))/2a.

Bumper Problems


These are the set of 5 problems  each of  5 marks


The person who get more than 20 points will get a mathematics book on any topic he want…
So try these beautiful problems to test your mathematics abilities and getting new things…

Ways by which you can answer :
I am adding a form in the end of this post  , you can answer there ….

Note : You have give solution of your answer too..

Problem 1)

The bisectors of the angles A and B of the \bigtriangleup ABC  meet the sides
BC and CA at the points D and E , respectively.
Assuming that AE+BD =AB, determine the angle C

Problem 2)

Given a  \bigtriangleup ABC and  D be point on side  AC such that  AB = DC,
\angle BAC= 60-2X ,   \angle DBC= 5X and  \angle BCA= 3X
Find the value of  X

Problem 3)

If p and q are natural numbers so that

Prove that p is divisible by 1979 .


Problem 4)

Find highest degree n of 1991 for which 1991ⁿ  divides the number :

Problem 5)

Let ƒ(n) denote the sum of the digits of n. Let N = 4444⁴⁴⁴⁴
Find ƒ(ƒ(ƒ(n))))

You can use these symbols to write solutions more conveniently

Mathematical Operators
Exponents  :   ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁼ ⁽ ⁾ ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋ ₌ ₍ ₎ 

Problem of the day 3


Nice problem !

Functional Equations !

Q)
Determine all functions ƒ satisfying  the functional relations

ƒ(x) + ƒ(1/(1-x)) =( 2(1-2x)/(x(1-x))

where x is a real number x ≠ 0 , x ≠ 1

Post your answer in comments with whole solution 🙂
person who give best answer of problem of the day more than 4 times will be awarded a book

Problem Of the Day-2


Here is Yesterday Problem

Yesterday Winner was Akash Agarwal

You must be knowing that If you win the Problem of the day for 4 times in a week – then I will send you a Really interesting book on any topic you want 🙂

Today’s Problem  :

Find all possible values of x satisfying :

[x]/[x-2] – [x-2]/[x] = (8{x} + 12)/([x-2][x])

(.) = Normal bracket
{.} = Fractional part function
[.] = GIF/Floor function