# Monthly Archives: July 2012

## IMO 1983 – Problem 3

Let and be positive integers and let be a set of points in the plane such that

no three points of are collinear, and

for any point of , there are at least points of equidistant from

Prove that

Try the question …

Solution will be updated soon

## IMO 2012 problems

This year IMO problems !!!!

**Problem 1 :
**

Given triangle the point is the centre of the excircle opposite the vertex This excircle is tangent to the side at , and to the lines and at and , respectively. The lines and meet at , and the lines and meet at Let be the point of intersection of the lines and , and let be the point of intersection of the lines and Prove that is the midpoint of .

**Problem 2 :**

Let be an integer, and let be positive real numbers such that Prove that

**Problem 3 :**

The *liar’s guessing game* is a game played between two players and . The rules of the game depend on two positive integers and which are known to both players.

At the start of the game chooses integers and with Player keeps secret, and truthfully tells to player . Player now tries to obtain information about by asking player questions as follows: each question consists of specifying an arbitrary set of positive integers (possibly one specified in some previous question), and asking whether belongs to . Player may ask as many questions as he wishes. After each question, player 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 consecutive answers, at least one answer must be truthful.

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

1. If then can guarantee a win. 2. For all sufficiently large , there exists an integer such that cannot guarantee a win.

**Problem 4 :**

Find all functions such that, for all integers that satisfy , the following equality holds:

** Problem 5 :**

Let be a triangle with , and let be the foot of the altitude from . Let be a point in the interior of the segment . Let be the point on the segment such that . Similarly, let be the point on the segment such that . Let be the point of intersection of and .

Show that .

**Problem 6 :**

Find all positive integers for which there exist non-negative integers such that

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 such that

for all x,y

**Solution :**

For any positive real numbers , we have that

and by Cauchy in positive reals, then for all

Now it’s easy to see that , then for all positive real numbers

## 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 p_{n} that is the largest prime.

(3) Let p_{1}, p_{2}, …, p_{n} be the list of all primes that exist.

(4) Let x = p_{1}*p_{2}*…*p_{n} + 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 p_{1} … p_{n} divide x since x ≡ 1 (mod p_{i}) for any of the primes p_{1} … p_{n}

(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:

**ax ^{2} + bx + c = 0**

The equation above is known as the quadratic equation.

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

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

4a^{2}x^{2} + 4abx + 4ac = 0

(2) Next, we add b^{2} – b^{2} to the equation:

4a^{2}x^{2} + 4abx + b^{2} + 4ac – b^{2} = 0

(3) Now, we add b^{2} – 4ac to both sides which gives us:

4a^{2}x^{2} + 4abx + b^{2} = b^{2} – 4ac

(4) Further, we know that:

(2ax + b)^{2} = 4a^{2}x^{2} + 4axb + b^{2}

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

(2ax + b)^{2} = b^{2} – 4ac

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

2ax + b = ±√b^{2} – 4ac

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

**x = (-b ±√(b ^{2} – 4ac))/2a.**