olymmath

TUYMAADA 2024

Tuymaada Olympiad 2024.

Official paper

The official paper in 4 languages

English (senior) · English (junior) · English (senior) · English (junior)

Junior

J1

  • Junior

Triangular numbers are numbers of the form 1+2+...+n1 + 2 + . . . + n with positive integer nn , that is 1,3,6,101, 3, 6, 10 , . . . . Find the largest non-triangular positive integer number that cannot be represented as the sum of distinct triangular numbers. Proposed by A. Golovanov

Answer, solutions and source

J2

  • Junior

We will call a hedgehog a graph in which one vertex is connected to all the others and there are no other edges; the number of vertices of this graph will be called the size of the hedgehog. A graph GG is given on nn vertices (where n>1n > 1 ). For each edge ee , we denote by s(e)s(e) the size of the maximum hedgehog in graph GG , which contains this edge. Prove the inequality (summation is carried out over all edges of the graph GG ): ∑e1s(e)⩽n2.\sum_e \frac{1}{s(e)} \leqslant \frac{n}{2}. Proposed by D. Malec, C. Tompkins

Answer, solutions and source

J3

  • Junior

Three athletes ran at different constant speeds along a track of length 11 . They started moving at the same time at one end of the track. Having reached one of the ends of the track, the athlete immediately turned around and continued running in the opposite direction. After a while, all three athletes met at the start and finished training. At what maximum SS can we knowingly say that at some point the sum of the pairwise distances between athletes was at least SS ? Proposed by A. Golovanov, I. Rubanov

Answer, solutions and source

J4

  • Junior

A triangle ABCABC is given. NN and MM are the midpoints of ABAB and BCBC , respectively. The bisector of angle BB meets the segment MNMN at EE . HH is the base of the altitude drawn from BB in the triangle ABCABC . The point TT on the circumcircle of ABCABC is such that the circumcircles of TMNTMN and ABCABC are tangent. Prove that points T,H,E,BT, H, E, B are concyclic. Proposed by M. Yumatov

Answer, solutions and source

J6

  • Junior

Extension of angle bisector BLBL of the triangle ABCABC (where AB<BCAB < BC ) meets its circumcircle at NN . Let MM be the midpoint of BLBL . Isosceles triangle BDCBDC with base BCBC and angle equal to ABCABC at DD is constructed outside the triangle ABCABC . Prove that CM⊥DNCM \perp DN . Proposed by А. Mardanov

Answer, solutions and source

J7

  • Junior

Given are quadratic trinomials ff and gg with integral coefficients. For each positive integer nn there is an integer kk such that f(k)g(k)=n+1n.\frac{f(k)}{g(k)}=\frac{n + 1}{n}. Prove that ff and gg have a common root. Proposed by A. Golovanov

Answer, solutions and source

J8

  • Junior

A toy factory produces several kinds of clay toys. The toys are painted in kk colours. Diversity of a colour is the number of different toys of that colour. (Thus, if there are 55 blue cats, 77 blue mice and nothing else is blue, the diversity of colour blue is 22 .) The painting protocol requires that each colour is used and the diversities of each two colours are different. The toys in the store could be painted according to the protocol. However, a batch of clay Cheburashkas arrived at the store before painting (there were no Cheburashkas before). The number of Cheburashkas is not less that the number of the toys of any other kind. The total number of all toys, including Cheburashkas, is at least (k+1)(k+2)2\frac{(k+1)(k+2)}{2} . Prove that now the toys can be painted in k+1k + 1 colours according to the protocol. Proposed by F. Petrov

Answer, solutions and source

Senior

S1

  • Senior

Prove that a positive integer of the form n4+1n^4 +1 can have more than 10001000 divisors of the form a4+1a^4 +1 with integral aa .

Answer, solutions and source

S2

  • Senior

Chip and Dale play on a 100×100100 \times 100 table. In the beginning, a chess king stands in the upper left corner of the table. At each move the king is moved one square right, down or right-down diagonally. A player cannot move in the direction used by his opponent in the previous move. The players move in turn, Chip begins. The player that cannot move loses. Which player has a winning strategy?

Answer, solutions and source

S3

  • Senior

All perfect squares, and all perfect squares multiplied by two, are written in a row in increasing order. let f(n)f(n) be the nn -th number in this sequence. (For instance, f(1)=1,f(2)=2,f(3)=4,f(4)=8f(1)=1,f(2)=2,f(3)=4,f(4)=8 .) Is there an integer nn such that all the numbers f(n),f(2n),f(3n),…,f(10n2)f(n),f(2n),f(3n),\dots,f(10n^2) are perfect squares?

Answer, solutions and source

S4

  • Senior

A triangle ABCABC is given. The segment connecting the points where the excircles touch ABAB and ACAC meets the bisector of angle CC at XX . The segment connecting the points where the excircles touch BCBC and ACAC meets the bisector of angle AA at YY . Prove that the midpoint of XYXY is equidistant from AA and CC .

Answer, solutions and source

S6

  • Senior

The triangle ABCABC is given. On the arc BCBC of its circumscribed circle, which does not contain point AA , the variable point XX is selected, and on the rays XBXB and XCXC , the variable points YY and ZZ , respectively, so that XA=XY=XZXA = XY = XZ . Prove that the line YZYZ passes through a fixed point. Proposed by A. Kuznetsov

Answer, solutions and source

S7

  • Senior

Given are two polynomial ff and gg of degree 100100 with real coefficients. For each positive integer nn there is an integer kk such that f(k)g(k)=n+1n.\frac{f(k)}{g(k)}=\frac{n+1}{n}. Prove that ff and gg have a common non-constant factor.

Answer, solutions and source

S8

  • Senior

A graph GG has nn vertices ( n>1n>1 ). For each edge ee let c(e)c(e) be the number of vertices of the largest complete subgraph containing ee . Prove that the inequality (the summation is over all edges of GG ): ∑ec(e)c(e)−1≤n22.\sum_{e} \frac{c(e)}{c(e)-1}\le \frac{n^2}{2}.

Answer, solutions and source