olymmath

IT-CESENATICO 2021

Italian Mathematical Olympiad 2021.

Other problems

Problem 1

A positive integer mm is said to be zero taker\emph{zero taker} if there exists a positive integer kk such that: kk is a perfect square; mm divides kk ; the decimal expression of kk contains at least 20212021 '0' digits, but the last digit of kk is not equal to 00 .

Find all positive integers that are zero takers.

Problem 2

Let ABCABC a triangle and let II be the center of its inscribed circle. Let DD be the symmetric point of II with respect to ABAB and EE be the symmetric point of II with respect to ACAC . Show that the circumcircles of the triangles BIDBID and CIECIE are eachother tangent.

Problem 3

A grid consists of n×nn\times n points, with n∈Z+n\in\mathbb{Z}^+ . In some of these points is a sentry. Every sentry chooses two directions, one perpendicular to the other (one vertical and the other horizontal) and watches over all the points that are found in the two chosen directions. Each sentry watches over her own point as well and the sentries on the edge of the grid can also watch the vacuum, depending on the directions they have chosen.

For instance, in the figure below, representing a disposition of 55 sentries in a 4×44\times 4 grid, the sentries in A,B,C,D,EA, B, C, D, E watch over 1,3,4,5,71, 3, 4, 5, 7 points, respectively; points DD and EE are watched by one sentry, point CC is watched by two sentries, points A,BA, B and FF are watched by three sentries.

(a) Prove that we can place 1212 sentries in a 4×44\times 4 grid in such a way that every point of the grid is watched by at most two sentries. (b) Let S(n)S(n) be the maximum number of sentries we can place on an n×nn\times n grid in such a way that every point of the grid is watched by at most two sentries. Prove that 3n≤S(n)≤4n3n\leq S(n)\leq 4n for all n≥3n\geq 3 .

Attachments:

Screenshot_20220321-220700_Drive.jpg

Problem 4

Given two fractions a/ba/b and c/dc/d we define their pirate sum as: ab⋆cd=a+cb+d\frac{a}{b} \star \frac{c}{d} = \frac{a+c}{b+d} where the two initial fractions are simplified the most possible, like the result. For example, the pirate sum of 2/72/7 and 4/54/5 is 1/21/2 . Given an integer n≥3n \ge 3 , initially on a blackboard there are the fractions: 11,12,13,...,1n\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{n} . At each step we choose two fractions written on the blackboard, we delete them and write at their place their pirate sum. Continue doing the same thing until on the blackboard there is only one fraction. Determine, in function of nn , the maximum and the minimum possible value for the last fraction.

Problem 5

Let ABCABC be an acute-angled triangle, let MM be the midpoint of BCBC and let HH be the foot of the BB -altitude. Let QQ be the circumcenter of ABMABM and let XX be the intersection point between BHBH and the axis of BCBC .

Show that the circumcircles of the two triangles ACMACM , AXHAXH and the line CQCQ pass through a same point if and only if BQBQ is perpendicular to CQCQ .

Problem 6

A sequence x1,x2,...,xn,...x_1, x_2, ..., x_n, ... consists of an initial block of pp positive distinct integers that then repeat periodically. This means that {x1,x2,…,xp}\{x_1, x_2, \dots, x_p\} are pp distinct positive integers and xn+p=xnx_{n+p}=x_n for every positive integer nn . The terms of the sequence are not known and the goal is to find the period pp . To do this, at each move it possible to reveal the value of a term of the sequence at your choice.

(a) Knowing that 1≤p≤101 \le p \le 10 , find the least nn such that there is a strategy which allows to find pp revealing at most nn terms of the sequence. (b) Knowing that pp is one of the first kk prime numbers, find for which values of kk there exist a strategy that allows to find pp revealing at most 55 terms of the sequence.