# Monthly Archives: August 2012

## CGMO-2012 (China Girls Math Olympiad 2012) Problem 8

Find the number of integers in the set such that is a multiple of

## CGMO – 2012 ( China Girls Math Olympiad 2012 ) Problem 7

Let be a sequence of nondecreasing positive integers such that for some positive integers and . Prove that there exists a positive integer such that

## CGMO 2012 ( China Girls Math Olympiad 2012) Problem 6

There are cities, airline companies in a country. Between any two cities, there is exactly one -way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of

## CGMO 2012 ( China Girls Math Olympiad 2012) Problem 5

As shown in the figure below, the in-circle of is tangent to sides and at and respectively, and is the circumcenter of . Prove that .

## CGMO-2012 (China Girls Math Olympiad 2012) Problem 4

There is a stone at each vertex of a given regular -gon, and the color of each stone is black or white. Prove that we may exchange the position of two stones such that the coloring of these stones are symmetric with respect to some symmetric axis of the -gon

## CGMO-2012 (China Girls Math Olympiad 2012) Problem-3

Find all pairs of integers satisfying: there exists an integer such that is divisible by for all positive integers

## CGMO 2012 (China Girls Math Olympiad 2012)- Problem 2

Circles and are tangent to each other externally at . Points and are on , lines and are tangent to at and , respectively, lines and meet at point

Prove that

(1) ;

(2) .

## Functional equations challenge 1

This is a challenging problem !!

Find all continuous functions such that for all :

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

1) replacing the equation by a simpler one :

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Let and so and the equation becomes and so :

is a continuous function such that

Q.E.D.

Let

From 1) above, we get that and so, since is continuous :

Either , either for some .

Let then : the restriction of to :

and so is a bijection, so is monotonous

implies then is increasing and so is increasing.

So, and so which is impossible and so

Same, and so which is impossible and so

3) general solution of with continuous

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Then the original equation becomes (since ) and so :

The general continuous solutions of are any odd continuous function such that

A simple way to build them is :

3.1) If :

3.2) If for some :

Let any continuous fonction from such that and define as :

:

:

:

Notice that gives the solution

4) general solution of with continuous

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The general solution of original equation is where is any function described in 3) above

5) examples of solutions

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gives the solution