A city has $4$ horizontal and $n\geq3$ vertical boulevards which intersect at $4n$ crossroads. The crossroads divide every horizontal boulevard into $n-1$ streets and every vertical boulevard into $3$ streets. The mayor of the city decides to close the minimum possible number of crossroads so that the city doesn't have a closed path(this means that starting from any street and going only through open crossroads without turning back you can't return to the same street). $a)$Prove that exactly $n$ crossroads are closed. $b)$Prove that if from any street you can go to any other street and none of the $4$ corner crossroads are closed then exactly $3$ crossroads on the border are closed(A crossroad is on the border if it lies either on the first or fourth horizontal boulevard, or on the first or the n-th vertical boulevard).

A point $T$ is given on the altitude through point $C$ in the acute triangle $ABC$ with circumcenter $O$, such that $\measuredangle TBA=\measuredangle ACB$. If the line $CO$ intersects side $AB$ at point $K$, prove that the perpendicular bisector of $AB$, the altitude through $A$ and the segment $KT$ are concurrent.

Find all $f:R^+ \rightarrow R^+$ such that $f(f(x) + y)f(x) = f(xy + 1)\ \ \forall x, y \in R^+$ @below: https://artofproblemsolving.com/community/c6h2254883_2020_imoc_problems Quote: Feel free to start individual threads for the problems as usual

Two infinite arithmetic sequences with positive integers are given:$$a_1<a_2<a_3<\cdots ; b_1<b_2<b_3<\cdots$$It is known that there are infinitely many pairs of positive integers $(i,j)$ for which $i\leq j\leq i+2021$ and $a_i$ divides $b_j$. Prove that for every positive integer $i$ there exists a positive integer $j$ such that $a_i$ divides $b_j$.

Does there exist a set $S$ of $100$ points in a plane such that the center of mass of any $10$ points in $S$ is also a point in $S$?

Point $S$ is the midpoint of arc $ACB$ of the circumscribed circle $k$ around triangle $ABC$ with $AC>BC$. Let $I$ be the incenter of triangle $ABC$. Line $SI$ intersects $k$ again at point $T$. Let $D$ be the reflection of $I$ across $T$ and $M$ be the midpoint of side $AB$. Line $IM$ intersects the line through $D$, parallel to $AB$, at point $E$. Prove that $AE=BD$.