## Romanian Masters in Mathematics contest – 2018

**Problem 1.** Let be a cyclic quadrilateral an let be a point on the side The diagonals meets the segments at The line through parallel to mmets the extension of the side beyond at The line through parallel to meets the extension of the side beyond at Prove that the circumcircles of the triangles and are tangent .

**Problem 2.** Determine whether there exist non-constant polynomials and with real coefficients satisfying

**Problem 3.** Ann and Bob play a game on the edges of an infinite square grid, playing in turns. Ann plays the first move. A move consists of orienting any edge that has not yet been given an orientation. Bob wins if at any point a cycle has been created. Does Bob have a winning strategy?

**Problem 4.** Let be positive integers such that and . Let be the set of values attained by as runs through the positive integers. Show that is the set of all positive divisors of some positive integer.

**Problem 5.** Let be positive integer and fix distinct points on a circle. Determine the number of ways to connect the points with arrows (oriented line segments) such that all of the following conditions hold:

- each of the points is a startpoint or endpoint of an arrow;
- no two arrows intersect;
- there are no two arrows and such that , , and appear in clockwise order around the circle (not necessarily consecutively).

**Problem 6.** Fix a circle , a line to tangent , and another circle disjoint from such that and lie on opposite sides of . The tangents to from a variable point on meet at and . Prove that, as varies over , the circumcircle of is tangent to two fixed circles.

Source: Art of Problem Solving forums

**Some quick ideas:** For Problem 1 just consider the intersection of the circle with the circle . You’ll notice immediately that this point belongs to the circle . Furthermore, there is a common tangent to the two circles at this point.

For Problem 2 we have . Eliminate the highest order term from both sides and look at the next one to get a contradiction.

Problem 4 becomes easy after noticing that if divides and then divides .

In Problem 5 try to prove that the choice of start points determines that of the endpoints. Then you have a simple combinatorial proof.

Problem 6 is interesting and official solutions use inversions. Those are quite nice, but it may be worthwhile to understand what happens in the non-inverted configuration.

I will come back to some of these problems in some future posts.

## 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*

## IMO 2015 Problem 1

**Problem 1.** We say that a finite set of points in the plane is *balanced* if, for any two different points and in , there is a point in such that . We say that is *center-free* if for any three different points , and in , there is no points in such that .

(a) Show that for all integers , there exists a balanced set having points.

(b) Determine all integers for which there exists a balanced center-free set having points.

**Problem 2.** Find all triples of positive integers such that are all powers of 2.

**Problem 3.** Let be an acute triangle with . Let be its cirumcircle., its orthocenter, and the foot of the altitude from . Let be the midpoint of . Let be the point on such that and let be the point on such that . Assume that the points and are all different and lie on in this order.

Prove that the circumcircles of triangles and are tangent to each other.

*Source: AoPS*

## 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