Question bank IMO 2019

Let $\mathbb Z$ be the set of integers. Determine all functions $f:\mathbb Z\to\mathbb Z$ such that, for all integers $a,b$,

$$f(2a)+2f(b)=f(f(a+b)).$$

In triangle $ABC$ point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $\overline{PQ} \parallel \overline{AB}$. Point $P_1$ is chosen on ray $PB_1$ beyond $B_1$ such that $\angle PP_1C = \angle BAC$. Point $Q_1$ is chosen on ray $QA_1$ beyond $A_1$ such that $\angle CQ_1Q = \angle CBA$. Prove that points $P_1$, $Q_1$, $P$, $Q$ are cyclic.

A social network has $2019$ users, some pairs of which are friends (friendship is symmetric). If $A$, $B$, $C$ are three users such that $AB$ are friends and $AC$ are friends but $BC$ is not, then the administrator may perform the following operation: change the friendships such that $BC$ are friends, but $AB$ and $AC$ are no longer friends.

Initially, $1009$ users have $1010$ friends and $1010$ users have $1009$ friends. Prove that the administrator can make a sequence of operations such that all users have at most $1$ friend.

Solve over positive integers the equation $$
k! = \prod_{i=0}^{n-1} (2^n-2^i) = (2^n-1)(2^n-2)(2^n-4) \dots (2^n-2^{n-1}).
$$

Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, which can show either heads or tails. He does the following operation: if there are $k$ coins which show heads and $k \gt 0$, then he flips the $k$th coin over; otherwise he stops the process. (For example, the process starting with $THT$ would be $THT \to HHT \to HTT \to TTT$, which takes three steps.)

Prove the process will always terminate, and determine the average number of steps this takes over all $2^n$ configurations.

Let $ABC$ be an acute triangle with $AB\ne AC$, incenter $I$ and incircle $\omega$. Let $D$, $E$, $F$ denote the tangency points of $\omega$ with $\overline{BC}$, $\overline{CA}$, $\overline{AB}$. The line through $D$ perpendicular to $\overline{EF}$ meets $\omega$ again at $R$ (other than $D$), and line $AR$ meets $\omega$ again at $P$ (other than $R$). Suppose the circumcircles of $\triangle PCE$ and $\triangle PBF$ meet again at $Q$ (other than $P$). Prove that lines $DI$ and $PQ$ meet on the external $\angle A$-bisector. The external $\angle A$-bisector is the line through $A$ perpendicular to $AI$.