olymmath

RMM 2021

Romanian Master of Mathematics 2021.

Day 1

Problem 1

  • Day 1

Let T1,T2,T3,T4T_{1}, T_{2}, T_{3}, T_{4} be pairwise distinct collinear points such that T2T_{2} lies between T1T_{1} and T3T_{3}, and T3T_{3} lies between T2T_{2} and T4T_{4}. Let ω1\omega_{1} be a circle through T1T_{1} and T4T_{4}; let ω2\omega_{2} be the circle through T2T_{2} and internally tangent to ω1\omega_{1} at T1T_{1}; let ω3\omega_{3} be the circle through T3T_{3} and externally tangent to ω2\omega_{2} at T2T_{2}; and let ω4\omega_{4} be the circle through T4T_{4} and externally tangent to ω3\omega_{3} at T3T_{3}. A line crosses ω1\omega_{1} at PP and W,ω2W, \omega_{2} at QQ and R,ω3R, \omega_{3} at SS and TT, and ω4\omega_{4} at UU and VV, the order of these points along the line being P,Q,R,S,T,U,V,WP, Q, R, S, T, U, V, W. Prove that PQ+TU=RS+VWP Q+T U=R S+V W.

Hungary, Geza Kos

Answer, solutions and source

Problem 2

  • Day 1
  • Answer

Xenia and Sergey play the following game. Xenia thinks of a positive integer NN not exceeding 5000. Then she fixes 20 distinct positive integers a1,a2,…,a20a_{1}, a_{2}, \ldots, a_{20} such that, for each k=1,2,…,20k=1,2, \ldots, 20, the numbers NN and aka_{k} are congruent modulo kk. By a move, Sergey tells Xenia a set SS of positive integers not exceeding 20, and she tells him back the set {ak:k∈S}\left\{a_{k}: k \in S\right\} without spelling out which number corresponds to which index. How many moves does Sergey need to determine for sure the number Xenia thought of?

Russia, Sergey Kudrya

Answer, solutions and source

Problem 3

  • Day 1

A number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4. Prove that the number of such ways to assign the leaders is divisible by 17 .

Russia, Mikhail Antipov

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Consider an integer n≥2n \geq 2 and write the numbers 1,2,…,n1,2, \ldots, n down on a board. A move consists in erasing any two numbers aa and bb, and, for each cc in {a+b,∣a−b∣}\{a+b,|a-b|\}, writing cc down on the board, unless cc is already there; if cc is already on the board, do nothing. For all integers n≥2n \geq 2, determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves.

Answer, solutions and source

Problem 5

  • Day 2

Let nn be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter 6n6 n, and 60∘60^{\circ} rotational symmetry (that is, there is a point OO such that a 60∘60^{\circ} rotation about OO maps the polygon to itself). In light of the pandemic, the government of Zoomtopia would like to relocate its 3n2+3n+13 n^{2}+3 n+1 citizens at 3n2+3n+13 n^{2}+3 n+1 points in the kingdom so that every two citizens have a distance of at least 1 for proper social distancing. Prove that this is possible. (The kingdom is assumed to contain its boundary.)

USA, Ankan Bhattacharya

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Problem 6

  • Day 2

Initially, a non-constant polynomial S(x)S(x) with real coefficients is written down on a board. Whenever the board contains a polynomial P(x)P(x), not necessarily alone, one can write down on the board any polynomial of the form P(C+x)P(C+x) or C+P(x)C+P(x), where CC is a real constant. Moreover, if the board contains two (not necessarily distinct) polynomials P(x)P(x) and Q(x)Q(x), one can write P(Q(x))P(Q(x)) and P(x)+Q(x)P(x)+Q(x) down on the board. No polynomial is ever erased from the board.

Given two sets of real numbers, A={a1,a2,…,an}A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\} and B={b1,b2,…,bn}B=\left\{b_{1}, b_{2}, \ldots, b_{n}\right\}, a polynomial f(x)f(x) with real coefficients is (A,B)(A, B)-nice if f(A)=Bf(A)=B, where f(A)={f(ai):i=1,2,…,n}f(A)=\left\{f\left(a_{i}\right): i=1,2, \ldots, n\right\}.

Determine all polynomials S(x)S(x) that can initially be written down on the board such that, for any two finite sets AA and BB of real numbers, with ∣A∣=∣B∣|A|=|B|, one can produce an (A,B)(A, B)-nice polynomial in a finite number of steps.

Iran, Navid SafaEi

Answer, solutions and source