HongKong 2022-23 IMO Selection Tests

a. Answer: $65$ Note that $$\left(\genfrac{}{}{0ex}{}{64}{2}\right)=2016<2022<2080=\left(\genfrac{}{}{0ex}{}{65}{2}\right).$$ As each pair of straight lines produces at most one intersection point, $64$ lines produce at most $2016$ intersection points, which is insufficient. Therefore, at least $65$ lines are needed. It is possible to have $65$ lines. For example, suppose there are $65$ pairwise non-parallel lines $l_1,l_2,\dots ,l_{65}$ such that for each $k=1,2,\dots ,29$, the lines $l_{2k-1}$ and $l_{2k}$ intersect at a distinct point on $l_{65}$, and there are no concurrent lines other than $l_{2k-1},l_{2k},l_{65}$ for $k=1,2,\dots ,29$. Then the number of intersection points is $$\left(\genfrac{}{}{0ex}{}{65}{2}\right)-2\cdot 29=2022.$$

b. Answer: $65$ As in part (a), at least $65$ lines are needed. An example of $65$ lines is given below. Suppose there are $65$ lines $l_1,l_2,\dots ,l_{65}$ where $l_{64}\parallel l_{65}$ and there is no other pair of parallel lines. Also, for each $k=1,2,\dots ,28$, the lines $l_{2k-1}$ and $l_{2k}$ intersect at a distinct point on $l_{65}$, and there are no concurrent lines other than $l_{2k-1},l_{2k},l_{65}$ for $k=1,2,\dots ,28$. Then the number of intersection points is $$\left(\genfrac{}{}{0ex}{}{65}{2}\right)-1-2\cdot 28=2023.$$