## Balkan Mathematical Olympiad 2017 – Problems

**Problem 1.** Find all ordered pairs of positive integers such that:

**Problem 2.** Consider an acute-angled triangle with and let be its circumscribed circle. Let and be the tangents to the circle at points and , respectively, and let be their intersection. The straight line passing through the point and parallel to intersects in point . The straight line passing through the point and parallel to intersects in point . The circumcircle of the triangle intersects in , where is located between and . The circumcircle of the triangle intersects the line (or its extension) in , where is located between and .

Prove that , , and are concurrent.

**Problem 3.** Let denote the set of positive integers. Find all functions such that

for all

**Problem 4.** On a circular table sit students. First, each student has just one candy. At each step, each student chooses one of the following actions:

- (A) Gives a candy to the student sitting on his left or to the student sitting on his right.
- (B) Separates all its candies in two, possibly empty, sets and gives one set to the student sitting on his left and the other to the student sitting on his right.

At each step, students perform the actions they have chosen at the same time. A distribution of candy is called legitimate if it can occur after a finite number of steps. Find the number of legitimate distributions.

(Two distributions are different if there is a student who has a different number of candy in each of these distributions.)

Source: AoPS

## Missing digit – short puzzle

The number has digits, all different; which digit is missing?

*Mathematical Mind-Benders, Peter Winkler*

## Monotonic bijection from naturals to pairs of natural numbers

This is a cute problem I found this evening.

Suppose is a bijection such that if and , then .

Prove that if then .

*Proof:* The trick is to divide the pairs of positive integers into families with the same product.

Note that the -th column contains as many elements as the number of divisors of . Now we just just use a simple observation. Let be on the -th column (i.e. ). If then cannot be on one of the first columns. Indeed, the monotonicity property implies . The fact that is a bijection assures us that cover the first columns. Moreover, one element from the -th column is surely covered, namely . This means that

where we have denoted by the number of positive divisors of .

## IMC 2014 Day 1 Problem 4

Let be a perfect number and be its prime factorization with . Prove that is an even number.

A number is *perfect* if , where is the sum of the divisors of .

**IMC 2014 Day 1 Problem 4**

## Group actions and applications to number theory

**Theorem.** Let be a finite group and be a prime number. Then . (we denote by the cardinal of ).

*Proof:* Denote . Then since if we cnoose the first elements arbitrarily in then the last element can be chosen in a unique way such that the equation is satisfied. Note that is stable under cyclic permutations of the elements of a vector. Therefore the action of on by

is well defined.

We will need the following lemma:

*Lemma.* If is a group (with prime) and acts on then

## IMO 1981 Day 1

**Problem 1.** Let be a point inside a given triangle and denote the feet of the perpendiculars from to the lines respectively. Find such that the quantity

is minimal.

**Problem 2.** Let and consider all subsets of elements of the set . Each of these subsets has a smallest member. Let denote the arithmetic mean of these smallest numbers. Prove that

**Problem 3.** Determine the maximum value of where with .

## Best approximation of a certain square root

Let be a real number such that the inequality

holds for an infinity of pairs of natural numbers. Prove that .