MathDB

Problem 4

Part of 2024 Kyiv City MO Round 2

Problems(5)

Genius strikes again: nobody solved this in contest!

Source: Kyiv City MO 2024 Round 2, Problem 7.4

2/4/2024
Points XX and YY are chosen inside an acute-angled triangle ABCABC with altitude ADAD so that ∠BXA+∠ACB=180∘,∠CYA+∠ABC=180∘\angle BXA + \angle ACB = 180^\circ , \angle CYA + \angle ABC = 180^\circ, and CD+AY=BD+AXCD + AY = BD + AX. Point MM is chosen on the ray BXBX so that XX lies on segment BMBM and XM=ACXM = AC, and point NN is chosen on the ray CYCY so that YY lies on segment CNCN and YN=ABYN = AB. Prove that AM=ANAM = AN.
Proposed by Mykhailo Shtandenko
geometry
Economic crisis :(

Source: Kyiv City MO 2024 Round 2, Problem 8.4/9.3

2/4/2024
In a certain magical country, there are banknotes in denominations of 20,21,22,…2^0, 2^1, 2^2, \ldots UAH. Businessman Victor has to make cash payments to 4444 different companies totaling 4400044000 UAH, but he does not remember how much he has to pay to each company. What is the smallest number of banknotes Victor should withdraw from an ATM (totaling exactly 4400044000 UAH) to guarantee that he would be able to pay all the companies without leaving any change?
Proposed by Oleksii Masalitin
combinatoricsnumber theory
Beautiful geometry from Kyiv MO

Source: Kyiv City MO 2024 Round 2, Problem 9.4

2/4/2024
Let BDBD be an altitude of △ABC\triangle ABC with AB<BCAB < BC and ∠B>90∘\angle B > 90^\circ. Let MM be the midpoint of ACAC, and point KK be symmetric to point DD with respect to point MM. A perpendicular drawn from point MM to the line BCBC intersects line ABAB at point LL. Prove that ∠MBL=∠MKL\angle MBL = \angle MKL.
Proposed by Oleksandra Yakovenko
geometry
Combinatorics from IMO 3rd absolute place

Source: Kyiv City MO 2024 Round 2, Problem 10.4

2/4/2024
There are n≥1n \geq 1 notebooks, numbered from 11 to nn, stacked in a pile. Zahar repeats the following operation: he randomly chooses a notebook whose number kk does not correspond to its location in this stack, counting from top to bottom, and returns it to the kkth position, counting from the top, without changing the location of the other notebooks. If there is no such notebook, he stops.
Is it guaranteed that Zahar will arrange all the notebooks in ascending order of numbers in a finite number of operations?
Proposed by Zahar Naumets
combinatoricspermutations
Problem rejected from IMO, EGMO, USAMO. But I still like it!

Source: Kyiv City MO 2024 Round 2, Problem 11.4

2/4/2024
Let ABCABC be an acute triangle with circumcenter OO and orthocenter HH. Rays AOAO, COCO intersect sides BC,BABC, BA in points A1,C1A_1, C_1 respectively, KK is the projection of OO onto the segment A1C1A_1C_1, MM is the midpoint of ACAC. Prove that ∠HMA=∠BKC1\angle HMA = \angle BKC_1.
Proposed by Anton Trygub
geometry