Romanian Regional Olympiad 2024 – 10th grade

Problem 1. Let , , . Find the smallest real number such that

Problem 2. Consider a triangle inscribed in the circle of center and radius . For any denote by where are the orthocenters of the triangles , respectively.

a) Prove that if the triangle is equilateral then for every .

b) Show that if there exist three distinct points such that , then the triangle is equilateral. 

Problem 3. Let be three non-zero complex numbers with the same modulus for which and are real numbers. Show that for every positive integer the number is real. 

Problem 4. Let . Determine all functions which verify

for every and such that has a unique solution. 

Hints:

Problem 1: study the monotonicity of the function . Then observe that the inequality is equivalent to . 

Problem 2: Recall the identity whenever is the orthocenter of with circumcenter . This can be proved using complex numbers and recalling that , where is the center of gravity. Therefore

and the analogue equalities. Summing we get

a) When is equilateral and inscribed in a circle of radius we have . Moreover, the identity In particular, one can prove the following:

applied to the triangle equilateral triangle with centroid shows that

Thus

b) Assume there exist three distinct points such that . This implies that

The relation above concerning the center of gravity of shows that . Since the points are distinct, it follows that coincides with , the circumcenter of , therefore is equilateral.

Problem 3: Denote the common value of the modulus: . Then

Since it follows that . Then of course . Finally, we know that are roots of

Since the coefficients of this polynomial are real, an inductive argument shows that if are real then is real, finishing the proof. 

Problem 4. Take and get . Thus, is the identity mapping on its image!! Take and observe that . Therefore for any . Since the equation has a unique solution, it follows that and for . Take and get . Therefore

for any . Since takes all positive values in it follows that

for every , . Coupled with this implies that

for all . If it follows that therefore . From we obtain for all reals . It should be noted that this function obviously verifies the functional equation!

Putnam 2023 A1-A3

A1. For a positive integer , let . Find the smallest such that . 

Hints: Observe that

Differentiate again and observe that

It is straightforward now to evaluate and to answer the question. 

A2. Let be an even positive integer. Let be a monic, real polynomial of degree ; that is to say, for some real coefficients . Suppose that for all integers such that . Find all other real numbers for which . 

Hints: Denote . Then is a polynomial of degree . Moreover, for , showing that has the roots .

It follows that . Identifying coefficients one finds , and the value of is obtained by observing that the constant term in should be zero. Then the answer to the question is obtained by investigating the roots of . I guess a distinction should be made between the cases even or odd, since it seems that the sign of depends on that. 

A3. Determine the smallest positive real number such that there exist differentiable functions and satisfying

1. (a) ,
2. (b) ,
3. (c) for all ,
4. (d) for all , and
5. (e) .

Hints: Of course, an example of functions are and . This suggests that the answer could be . In any case, this example shows that the smallest verifies .

Assuming that is the smallest zero, on we have . Now let’s try to obtain some differential inequality using the hypothesis:

where the last inequality follows from .

Therefore verifies and

Therefore we have on with . The general solution is

The initial conditions show that . Therefore

and since we have on . Moreover which is also non-negative on . This implies

Thus the smallest such that is indeed .

Algebraic proof of the Finsler-Hadwiger inequality

Weitzenbock’s inequality states that if are the side lengths of a triangle with area then

A strengthening of this result due to Finsler and Hadwiger states

A variety of proofs rely on various trigonometric or geometric arguments. Below you can find a purely algebraic argument based on the classical characterization: are side lengths of a triangle if and only if there exist such that , , . If are strictly positive then the triangle will be non-degenerate.

Replacing with the above formulas replaces an inequality in a triangle with a general inequality where only positivity of the variables is involved. With this substitution, using classical notation for cyclic sums gives

and

On the other hand the area given by Heron’s formula is

Thus, Weitzenbock’s inequality is equivalent to

and the Finsler-Hadwiger inequality is equivalent to

This inequality follows at once, since squaring both sides gives

which is a well known consequence of

Equality holds, of course, if and only if . If the triangle is non-degenerate then it must be equilateral. Thus, Weitzenbock and Finsler-Hadwiger inequalities follow at once from classical inequalities, once the side lengths of a triangle are replaced with unconstrained variables.

A proof of the Hadwiger Finsler inequality

The Hadwiger-Finsler inequality states that if are the side lengths of a triangle with area then

This was discussed previously on the blog. This post shows a translation of the original paper by Hadwiger and Finsler and this post shows a surprising geometrical proof.

Various proofs of the inequality are known. However, since an equality always beats an inequality, let us prove the identity

It is immediate to see that Jensen’s inequality applied to the tangent function, which is convex on is enough to deduce the Hadwiger-Finsler inequality from the above identity. To prove the identity, simply compute

Replacing the usual formula gives

Summing these identities for the three angles gives precisely the desired result. The same proof can be found, for example, here.

IMO 2023 Problem 4

Problem 4. Let be pairwise different positive real numbers such that

is an integer for every Prove that  

IMO 2023 Problem 1

Problem 1. Determine all composite integers that satisfy the following property: if are all the positive divisors of with , then divides for every . 

Problems of the International Mathematical Olympiad 2023

Problem 1. Determine all composite integers that satisfy the following property: if are all the positive divisors of with , then divides for every . 

Problem 2. Let be an acute-angled triangle with . Let be the circumcircle of . Let be the midpoint of the arc of containing . The perpendicular from to meets at and meets again at . The line through parallel to meets line at . Denote the circumcircle of triangle by . Let meet again at . Prove that the line tangent to at meets line on the internal angle bisector of . 

Problem 3. For each integer , determine all infinite sequences of positive integers for which there exists a polynomial of the form , where are non-negative integers, such that

for every integer . 

Problem 4. Let be pairwise different positive real numbers such that

is an integer for every Prove that  

Problem 5. Let be a positive integer. A Japanese triangle consists of circles arranged in an equilateral triangular shape such that for each , , , , the row contains exactly circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with , along with a ninja path in that triangle containing two red circles.

In terms of , find the greatest such that in each Japanese triangle there is a ninja path containing at least red circles. 

Problem 6. Let be an equilateral triangle. Let be interior points of such that , , , and

Let and meet at let and meet at and let and meet at Prove that if triangle is scalene, then the three circumcircles of triangles and all pass through two common points.

(Note: a scalene triangle is one where no two sides have equal length.)

Source: imo-official.org, AOPS forums

Matrices with entries of the form f(i,j) in Matlab

It is often necessary to define a matrix with entries where are line and column indices. Using loops is possible, but is not the most elegant solution. Also, it may not be the fastest. Here is a way to construct such matrices fast and easy: simply use meshgrid to construct the arrays of indices . Take a look at the example below and construct your own:

N = 10;
[I,J] = meshgrid(1:N,1:N);
A = N+1-max(I,J);

IMO 2022 – Problem 2 – Non-standard functional equation

Problem 2. Let denote the set of positive real numbers. Find all functions such that for each , there is exactly one satisfying . 

IMC 2022 – Day 1 – Problem 2

Problem 2. Let be a positive integer. Find all real matrices having only real eigenvalues satisfying

for some integer . 

Weitzenböck’s inequality

Given a triangle , denote by the lengths of the sides opposite to angles . Also denote by the area of the triangle. Weitzenbrock’s inequality states that the following holds:

There are multiple ways of proving this inequality. Perhaps one of the most straightforward ways to approach this is to use the law of sines and trigonometry. We have where is the circumradius. Moreover, is related to by . Thus

Therefore the original inequality is equivalent to

One would be tempted to try the arithmetic geometric mean inequality here, but that cannot work, since the inequality is not homogeneous and are not free, they verify . However, note that the function is concave on therefore, applying Jensen’s inequality we get

As a consequence, using cyclic sums over , we get

where for the last inequality above we applied AM-GM inequality two times. Regrouping we obtain exactly

which is equivalent to the original inequality. Note that with the same ideas we have

showing that we also have

Observe that the equality can hold only if the triangle is equilateral.

This leads to the stronger inequality

Other proofs can be found in the Wikipedia article.

It can be observed that combining the above inequalities we can get the isoperimetric inequality for triangles

where the equality holds if and only if the triangle is equilateral.

Balkan Mathematical Olympiad 2022

Problem 1. Let be an acute triangle such that with circumcircle and circumcentre . Let and be the tangents to at and respectively, which meet at . Let be the foot of the perpendicular from onto the line segment . The line through parallel to line meets at . Prove that the line passes through the midpoint of the line segment . 

Problem 2. Let and be positive integers with such that all of the following hold:

i. divides ,

ii. divides ,

iii. 2022 divides .

Prove that there is a subset of the set of positive divisors of the number such that the sum of the elements of is divisible by 2022 but not divisible by . 

Problem 3. Find all functions such that for all . 

Problem 4. Consider an grid consisting of until cells, where is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Then, Xanthias views the colouring and next places frogs on the cells so that each of the cells can be reached by a frog in a finite number (possible zero) of hops. Find the least value of for which this is always possible regardless of the colouring chosen by Dionysus.

Source: AOPS

Putnam 2021 – Day 1

Here you can find the problems of the first day of the Putnam 2021 contest. Source.

A1. A grasshopper starts at the origin in the coordinate plane and makes a sequence of hops. Each hop has length , and after each hop the grasshopper is at a point whose coordinates are both integers; thus, there are possible locations for the grasshopper after the first hop. What is the smallest number of hops needed for the grasshopper to reach the point ? 

A2. For every positive real number , let

Find . 

A3. Determine all positive integers for which the sphere

has an inscribed regular tetrahedron whose vertices have integer coordinates. 

A4. Let

Find , or show that this limit does not exist. 

A5. Let be the set of all integers such that and . For every nonnegative integer , let

Determine all values of such that is a multiple of . 

A6. Let be a polynomial whose coefficients are all either or . Suppose that can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that is a composite integer?

IMC 2021 Problems – Day 1

Problem 1. Let be a real matrix such that .

(a) Prove that there is a unique real matrix that satisfies the equation

(b) Express in terms of  

Problem 2. Let and be fixed positive integers, and let be an arbitrary non-negative integer. Choose a random -element subset of uniformly (i.e., all -element subsets are chosen with the same probability) and, independently of , choose a random -element subset of uniformly. Prove that the probability

does not depend on . 

Problem 3. We say that a positive real number is good if there exists an infinite sequence such that for each the points partition the interval into segments of length at most each. Find

Problem 4. Let be a function. Suppose that for every there exists a function such that for every pair of real numbers if then .

Prove that is the pointwise limit of a sequence of continuous functions, i.e. there is a sequence of functions such that for every .

Putnam 2020 problems – Day 1

A1. How many positive integers satisfy all of the following three conditions?

• (i) is divisible by .
• (ii) has at most decimal digits.
• (iii) The decimal digits of are a string of consecutive ones followed by a string of consecutive zeros.

A2. Let be a nonnegative integer. Evaluate

A3. Let and for . Determine whether

converges.

A4. Consider a horizontal strip of squares in which the first and the last square are black and the remaining squares are all white. Choose a white square uniformly at random, choose one of its two neighbors with equal probability, and color this neighboring square black if it is not already black. Repeat this process untill all the remaining white squares have only black neighbors. Let be the expected number of white squares remaining. Find

A5. Let be the number of sets of positive integers for which

where the Fibonacci sequence satisfies and begins with . Find the largest integer such that .

A6. For a positive integer , let be the function defined by

Determine the smallest constant such that for all and all real .

Source: https://kskedlaya.org/putnam-archive/

Minimal squared sum minus sum of squares

a) Let be real numbers such that . Then

b) Prove that if then

Solution: a) Denote with and the sum of positive and negative terms among the :

It is obvious that (it is enough to expand the sums). Moreover, the equality is attained here only if and contain at most one non-zero term.

Therefore, if the original inequality would be proved. This is equivalent to . But , which directly implies .

Let us now discuss the equality case. Equality holds if and only if . Since it follows that . Moreover, equality should also hold in

This means that all but one elements in and in are zero, which shows that the equality case is attaned for .

b) It is enough to apply a) for noting that

Source: The solution here was inspired from the following MathOverflow answer: link.

IMO 2020 Problems

Problem 1. Consider the convex quadrilateral . The point is in the interior of . The following ratio equalities hold:

Prove that the following three lines meet in a point: the internal bisectors of angles and and the perpendicular bisector of segment .

Problem 2. The real numbers are such that and . Prove that

Problem 3. There are pebbles of weights . Each pebble is coloured in one of colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied:

• The total weights of both piles are the same.
• Each pile contains two pebbles of each colour.

Problem 4. There is an integer . There are stations on a slope of a mountain, all at different altitudes. Each of two cable car companies, and , operaters cable cars; each cable car provides a transfer from one of the stations to a higher one (with no intermediate stops). The cable cars of have different starting points and different finishing points, and a cable car which starts higher also finishes higher. The same conditions hold for . We say that two stations are linked by a company if one can start from the lower station and reach the higher one by using one or more cars of that company (no other movements between stations are allowed). Determine the smallest positive for which one can guarantee that there are two stations that are linked by both companies. 

Problem 5. A deck of cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.

For which does it follow that the numbers on all cards are all equal? 

Problem 6. Prove that there exists a positive constant such that the following statement is true:

Consider an integer , and a set of points in the plane such that the distance between any two different points in is at least . It follows that there is a line separating such that the distance for any point of to is at least .

(A line  separates a set of points if some segment joining two points in crosses .) 

Note. Weaker results with replaced with may be awarded points depending on the value of the constant .

Source: imo-official.org

IMC 2020 Day 1 – Some Hints for Problems 1-2

Problem 1. Let be a positive integer. Compute the number of words (finite sequences of letters) that satisfy all the following requirements: (1) consists of letters, all of them from the alphabet (2) contains an even number of letters (3) contains an even number of letters (For example, for there are such words: and .)

(proposed by Armend Sh. Shabani, University of Prishtina)

Hint:  In order to get a formula for the total number of words it is enough to note that the even number of a’s can be distributed in ways among the possible positions, and the even number of b’s can be distributed in ways among the remaining positions. Once the a’s and b’s are there, the rest can be filled with c’s and d’s in ways.  This gives

Next, use some tricks regarding expansions of to compress the sum above to .

Problem 2. Let and be real matrices such that

where is the identity matrix. Prove that

(proposed by Rustam Turdibaev, V.I. Romanovskiy Institute of Mathematics)

Hint: Every time you hear about a rank 1 matrix you should think of “column matrix times line matrix”. Indeed, writing with column vectors, gives . Moreover, taking trace in the above equality and using the fact that you can perform circular permutations in products in traces, you obtain that .

Moreover, squaring gives

Taking trace above gives the desired result!

IMC 2020 Online – Problems – Day 1

Problem 1. Let be a positive integer. Compute the number of words (finite sequences of letters) that satisfy all the following requirements: (1) consists of letters, all of them from the alphabet (2) contains an even number of letters (3) contains an even number of letters (For example, for there are such words: and .)

(proposed by Armend Sh. Shabani, University of Prishtina)

Problem 2. Let and be real matrices such that

where is the identity matrix. Prove that

(proposed by Rustam Turdibaev, V.I. Romanovskiy Institute of Mathematics)

Problem 3. Let be an integer. Prove that there exists a constant such that the following holds: For any convex polytope , which is symmetric about the origin, and any , there exists a convex polytope with at most vertices such that

(For a real , a set with nonempty interior is a convex polytope with at most vertices if is a convex hull of a set of at most points.)

(proposed by Fedor Petrov, St. Petersburg State University)

Problem 4. A polynomial with real coefficients satisfies the equation for all . Prove that for .

(proposed by Daniil Klyuev, St. Petersburg State University)

source: http://www.imc-math.org

Sum of floors of multiples of the Golden Ratio

Propose an algorithm for computing

for , where is the golden ratio. The sum is the sum of the first terms in the so called lower Wythoff sequence.

Solution: Computing a floor function and a multiplication is not complicated, therefore proposing a algorithm for computing is trivial. However, such an algorithm is not viable for .

The path to a much more efficient algorithm goes through the notion of Beatty sequence. For a positive irrational number the associated Beatty sequence is . This notion becomes interesting when noting the following fact. If is defined by then the Beatty sequences and cover all the positive integers. The latter sequence is called the associated complementary Beatty sequence. Two proofs of this fact can be found in the Wikipedia link above.

It is obvious now that is just the partial sum of the Beatty sequence associated to . Now let’s see what is the complementary sequence. A brief computation shows that the associated is given by , which shows that the terms which are not of the form are not far from it. In order to see how this can give an efficient algorithm, let’s follow the instructions below:

• denote by , the largest term in .
• looking at all the integers up to , in view of the result regarding the Beatty sequences, they are either of the form or of the form (in the associated complementary sequence).
• denote by the largest integer such that , which is given by . Then, it is not difficult to see that is the difference between the sum of all positive integers up to and .
• In the end we obtain

  

• by convention, state that or .

The last item above shows that a recursive algorithm can be implemented in order to compute . Moreover, since the computation of is reduced to the computation of where , the algorithm will converge very fast, since the sequence of upper bounds for converges exponentially to zero. Therefore, the algorithm obtained will have complexity and will work for extremely large values of , provided the language you are using can handle the expected precision.

The algorithm for computing shown above can be used, for example, in computing sums related to elements in the Wythoff table, which can be expressed using Fibonacci numbers and terms in .

Other values of , like for example (with ) lead to other types of sums for which the same type of algorithm can be applied. It is likely that even cases where is not explicitly related to through an integer may be handled through a similar procedure (to be confirmed).