Category Archives: Problems

NMTC 2012 ( CLASS 9th and 10th) Question paper

The AMTI is a pioneer organisation in promoting and conducting Maths Talent Tests in India. Last year (43rd TC Data) (in the 43rd National level tests) 54058 students from 332 institutions spread all over India, participated at the screening level; 10% of them insitutionwise were selected for the final test. For the benefit of final level contestants and the chosen few for INMO, special orientation camps were conducted. Merit certificates and prizes were awarded to the deserving students.

NMTC Question Paper

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

Find the number of integers k in the set \{0, 1, 2,\cdots, 2012\}   such that  \binom{2012}{k} is a multiple of 2012

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

Let  \{a_n\} be a sequence of nondecreasing positive integers such that  \frac{r}{a_{r}}= k+1 for some positive integers k and r. Prove that there exists a positive integer s such that  \frac{s}{a_s} = k

CGMO 2012 ( China Girls Math Olympiad 2012) Problem 6

There are n cities, 2 airline companies in a country. Between any two cities, there is exactly one 2-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  n

CGMO 2012 ( China Girls Math Olympiad 2012) Problem 5

As shown in the figure below, the in-circle of ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of BCI. Prove that \angle ODB = \angle OEC.
import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xma...

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

There is a stone at each vertex of a given regular 13-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 13-gon

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

Find all pairs (a,b) of integers satisfying: there exists an integer  d \ge 2 such that a^n + b^n+1  is divisible by d for all positive integers n

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

Circles Q_1 and Q_2 are tangent to each other externally at T. Points A and E are on Q_1, lines AB and DE are tangent to Q_2 at B and D, respectively, lines AE and BD meet at point P
Prove that

(1) \frac{AB}{AT}=\frac{ED}{ET};
(2) \angle ATP + \angle ETP = 180^{\circ}.
import graph; size(5.97cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real ...

CGMO 2012 (China Girls Math Olympiad 2012) Problem 1

Let  a_{1}, a_{2},\ldots, a_{n} be non-negative real numbers. Prove that                                                                          \LARGE \frac{1}{1+a_{1}}+\frac{ a_{1}}{(1+a_{1})(1+a_{2})}+\frac{ a_{1}a_{2}}{(1+a_{1})(1+a_{2})(1+a_{3})}\cdots+\frac{ a_{1}a_{2}\cdots a_{n-1}}{(1+a_{1})(1+a_{2})\cdots (1+a_{n})}\le 1.

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)
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)

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

3) general solution of g(g(x))=-g(-x) with g(x) continuous

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
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