olymmath

NMC 2022

Nordic Mathematical Contest 2022.

Other problems

Problem 1

Find all functions f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} such that

f(f(x)f(1−x))=f(x) and f(f(x))=1−f(x)f(f(x) f(1-x))=f(x) \text { and } f(f(x))=1-f(x)

for all real xx.

Problem 2

In Wonderland, the towns are connected by roads, and whenever there is a direct road between two towns there is also a route between these two towns that does not use that road. (There is at most one direct road between any two towns.) The Queen of Hearts ordered the Spades to provide a list of all "even" subsystems of the system of roads, that is, systems formed by subsets of the set of roads, where each town is connected to an even number of roads (possibly none). For each such subsystem they should list its roads. If there are totally nn roads in Wonderland and xx subsystems on the Spades' list, what is the number of roads on their list when each road is counted as many times as it is listed?

Problem 3

Anton and Britta play a game with the set M={1,2,3,…,n−1}M=\{1,2,3, \ldots, n-1\} where n≥5n \geq 5 is an odd integer. In each step Anton removes a number from MM and puts it in his set AA, and Britta removes a number from MM and puts it in her set BB (both AA and BB are empty to begin with). When MM is empty, Anton picks two distinct numbers x1,x2x_{1}, x_{2} from AA and shows them to Britta. Britta then picks two distinct numbers y1,y2y_{1}, y_{2} from BB. Britta wins if

(x1x2(x1−y1)(x2−y2))n−12≡1 mod n\left(x_{1} x_{2}\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\right)^{\frac{n-1}{2}} \equiv 1 \bmod n

otherwise Anton wins. Find all nn for which Britta has a winning strategy.

Problem 4

Let ABCA B C be an acute-angled triangle with circumscribed circle kk and centre of the circumscribed circle OO. A line through OO intersects the sides ABA B and ACA C at DD and EE. Denote by B′B^{\prime} and C′C^{\prime} the reflections of BB and CC over OO, respectively. Prove that the circumscribed circles of ODC′O D C^{\prime} and OEB′O E B^{\prime} concur on kk.