## 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

## IMC 2016 Problems – Day 2

**Problem 6.** Let be a sequence of positive real numbers satisfying . Prove that

**Problem 7.** Today, Ivan the Confessor prefers continuous functions satisfying for all . Fin the minimum of over all preferred functions.

**Problem 8.** Let be a positive integer and denote by the ring of integers modulo . Suppose that there exists a function satisfying the following three properties:

- (i) ,
- (ii) ,
- (iii) for all .

Prove that modulo .

**Problem 9.** Let be a positive integer. For each nonnegative integer let be the number of solutions of the inequality . Prove that for every we have .

**Problem 10.** Let be a complex matrix whose eigenvalues have absolute value at most . Prove that

(Here for every matrix and for every complex vector .)

Official source and more infos here.

## Balkan Mathematical Olympiad – 2016 Problems

**Problem 1.** Find all injective functions such that for every real number and every positive integer ,

**Problem 2.** Let be a cyclic quadrilateral with . The diagonals intersect at the point and lines and intersect at the point . Let and be the orthogonal projections of onto lines and respectively, and let , and be the midpoints of , and respectively. Prove that the second intersection point of the circumcircles of triangles and lies on the segment .

**Problem 3.** Find all monic polynomials with integer coefficients satisfying the following condition: there exists a positive integer such that divides for every prime for which is a positive integer.

**Problem 4.** The plane is divided into squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of colours so that no rectangle with perimeter contains two squares of the same colour. Show that no rectangle of size or contains two squares of the same colour.

## SEEMOUS 2016 Problem 4 – Solution

## ICPC 2015 World Final Problem B

This the the solution to Problem B from the International Collegiate Programming Contest. The list of problems can be found here. The idea is to find the maximal area of the intersection of two moving polygons. The inputs give the initial positions of two convex polygons and the vectors giving their speed in the plane.

One idea of a solution is as follows: take a maximal time and a discretization of the interval by dividing it into parts. Iterate over the translations at times given by this discretization, compute at each step the area of the intersection of the two polygons (if there is any intersection at all), and in the end find the time for which this area is maximized. Now, even if the discretization step is large (greater than the demanded precision of ), we can conclude that is an approximation of the final time with an error smaller than . This is due to the fact that the function representing the area of the intersection has two monotonicity intervals, as a consequence of the fact that the polygons are convex. On the first interval, the intersection area is increasing, on the second one it is decreasing. Thus, once we have a discrete maximum, we are close to the real maximum.

Now, all we are left to do in order to achieve any desired precision is to refine the search near this discrete maximum, find a new, better approximation, and refine again, until we have enough precision. Of course, one first problem with this algorithm is the initial search using a maximal time . It is possible that if or are not large enough, then we do not detect any intersection of the two polygons. Thus, an initial guess, based on a mathematical argument is needed in order to reduce the search of the optimal time to an interval which is small enough to have enough initial precision to detect the discrete maximum.

The algorithm presented below, uses Matlab’s predefined function polybool which can compute the intersection of two polygons. Of course, this is an overkill, since dealing with intersections of convex polygons is not that complicated (but still, I didn’t have enough time to play with the problem, in order to provide a more optimized version). I do not treat the search for the initial time interval. As I think about it, I guess a n argument based on finding some line intersections should give us a narrow enough time interval (with some care for the case when the two speed directions are collinear). The algorithm presented below solves the sample cases, but could fail in more general situations.

function Prob1_2015(p1,p2) % choose initial polygons; see the text of the problem p1 = [6 3 2 2 4 3 6 6 6 7 4 6 2 2 2]; p2 = [4 18 5 22 9 26 5 22 1 -2 1]; %p1 = [4 0 0 0 2 2 2 2 0 1 1]; %p2 = [4 10 0 10 2 12 2 12 0 -1 1]; np1 = p1(1); np2 = p2(1); cp1 = p1(2:1+2*np1); cp2 = p2(2:1+2*np2); vp1 = p1(end-1:end); vp2 = p2(end-1:end); cp1 = cp1(:); cp1 = reshape(cp1,2,np1); xp1 = cp1(1,:); yp1 = cp1(2,:); cp2 = cp2(:); cp2 = reshape(cp2,2,np2); xp2 = cp2(1,:); yp2 = cp2(2,:); %set precision prec = 1; %here there should be a clever initial choice of %the starting time and stopping time %this choice works well in this case start = 0; stop = 5; while prec>1e-6 n = 100; timex = linspace(start,stop,n); areas = zeros(size(timex)); for i = 1:n t = timex(i); xt1 = xp1+t*vp1(1); yt1 = yp1+t*vp1(2); xt2 = xp2+t*vp2(1); yt2 = yp2+t*vp2(2); [x,y] = polybool('intersection',xt1,yt1,xt2,yt2); areas(i) = polyarea(x,y); end [m,I] = max(areas); if m<1e-6 % if no polygonal intersection is detected % there is nothing further to do display('never'); break end tapp = timex(I); fprintf('Precision %f | Time %f\n',prec,tapp); start = tapp-prec; stop = tapp+prec; prec = prec/10; end clf %this draws the final position of the polygon %as well as their intersection fill(xt1,yt1,'blue') axis equal hold on fill(xt2,yt2,'red') fill(x,y,'green') axis equal hold off

Here’s what you get for the first sample test provided in the questions:

>> Prob1_2015 res = 4.193548

## Balkan Mathematical Olympiad – 2015 Problems

**Problem 1.** If and are positive real numbers, prove that

(Montenegro)

**Problem 2.** Let be a scalene triangle with incentre and circumcircle . Lines intersect for the second time at points , respectively. The parallel lines from to the sides intersect at points , respectively. Prove that the points are collinear. (Cyprus)

**Problem 3.** A committee of film critics are voting for the Oscars. Every critic voted just an actor and just one actress. After the voting, it was found that for every positive integer , there is some actor or some actress who was voted exactly times. Prove that there are two critics who voted the same actor and the same actress. (Cyprus)

**Problem 4.** Prove that among consecutive positive integers there is an integer such that for every positive integer the following inequality holds

where by denotes the fractional part of the real number . The fractional part of the real number is defined as the difference between the largest integer that is less than or equal to to the actual number . (Serbia)

Source: AoPS

## Seemous 2015 Problems

**Problem 1.** Prove that for every the following inequality holds:

**Problem 2.** For any positive integer , let the functions be defined by , where . Solve the equation .

**Problem 3.** For an integer , let be matrices satisfying:

where is the identity matrix and is the zero matrix in .

Prove that:

- (a) and .
- (b) and .

**Problem 4.** Let be an open interval which contains and be a function of class such that .

- (i) Prove that there exists such that for .
- (ii) With determined at (i) define the sequence by
Study the convergence of the series for .

**Hints: **