Problem 1. A line in the plane is called $sunny$ if it is not parallel to any of the $x$–axis, the $y$–axis, or the line $x+y=0$.

Let $n\ge3$ be a given integer. Determine all nonnegative integers $k$ such that there exist $n$ distinct lines in the plane satisfying both of the following:
  • for all positive integers $a$ and $b$ with $a+b\le n+1$, the point $(a,b)$ lies on at least one of the lines; and
  • exactly $k$ of the $n$ lines are sunny.

Solution 1Solution 2Solution 3

Solution 1 Call the set of points $(a,b), a,b \in \mathbb{Z}_{>0}$ with $a+b \leq n +1$ the Turbo triangle $T_n$ of depth $n$ (because the answer to the problem is independent of $n$, but this time the problem is well positioned). Clearly, a line is sunny if it is not parallel (or coincides) with any of the sides of $T_n$.

Assume $n\geq 4$. If one of the lines contains a side of $T_n$, then it is not sunny and so we reduce the problem to $n-1\geq 3$ without a change of the required answer. Now for a while, suppose this is not the case. Then each of the $n-1$ boundary points of the vertical side of $T_n$ (excluding the vertex $(1,1)$) must lie on a different line and each of the $n-1$ boundary points on the horizontal side of $T_n$ (excluding the vertex $(1,1)$) must lie on a different line. These are $2n-2$ points and a line can contain at most $2$ of them (otherwise go back to the first sentence), so there are two cases:

- These points need at least $n$ lines to be covered - but in the latter scenario there are no more lines available while we have still not covered $(1,1)$, contradiction.

- These points are such that there are exactly $n-1$ distinct lines covering them, a line covering two. Here the points are divided into pairs, each pair lying on one of the $n-1$ distinct lines. Then none of $(1,1)$ and the $n-2$ boundary points of the inclined side of $T_n$ are currently covered, since a line cannot intersect all three sides of a triangle internally. But these points are not all collinear for $n\geq 4$, contradiction.

Therefore, we have shown that we must have a line containing a side of $T_n$ and hence that the answer to the problem for all $n\geq 3$ is the same. So finally let $n=3$ and work manually through all $k\leq 3$. For $k=0$ take the three lines (at depth $1,2,3$) parallel to the inclined side of $T_3$. For $k=1$ take the inclined side, the parallel line below it and the altitude through the lower vertex (in coordinates: $x+y = 4, x+y = 3, x = y$).
For $k=3$ take the three medians of $T_3$. Finally, $k=2$ does not work: any line must contain at least two of the $6$ points of $T_3$ and if a line is a side of $T_3$ or a midsegment of $T_3$, then the remaining two cannot both be not a side or a midsegment (I need a few mini pictures to fully explain this so have to omit it, sorry).
Solution 2 The answer is $0, 1, 3$. We will prove this through a few claims.

For a positive integer $n \geq 3$, call a nonnegative integer $k \leq n$ $n$-sunny if there exists $n$ distinct lines in the plane, such that exactly $k$ lines are sunny and all points $(a, b)$ with $1 \leq a, b, a+b \leq n+1$ lie on at least one line.

For the first three claims, let $n$ be a positive integer at least $3$. For the entirety of this solution, by points we will refer to the points $(a, b)$, where $a, b$ are positive integers not exceeding $n+1$, unless stated otherwise.

Claim 1. If $k$ is $n$-sunny then $k$ is $n+1$-sunny.
Proof. Construct a configuration that proves $k$ is $n$-sunny, then add the line $x + y = n+2$. Since this line is not sunny, we obtain a configuration that proves $k$ is $(n+1)$-sunny.

Claim 2. If $k$ is $(n+1)$-sunny, and $k \leq n$, then $k$ is $n$-sunny.
Proof. Construct a configuration that proves $k$ is $(n+1)$-sunny.

Assume that in this configuration, none of the three lines $x = 1$, $y = 1$, $x + y = n+2$ are used. Then each of the $n+1$ lines go through at most $2$ points that lie on the lines $x = 1$, $y = 1$, $x + y = n+2$. Since there are $3n$ such points, we obtain that $3n \leq 2(n+1)$, which is impossible since $n \leq 2$.

Therefore, at least one of the three lines $x = 1$, $y = 1$, $x + y = n+2$ show up in the configuration. Remove that line, and apply a translation of the remaining lines if necessary, we obtain a configuration that proves $k$ is $n$-sunny.

Claim 3. If $n$ is $n$-sunny, then $n = 3$.
Proof. Construct a configuration that proves $n$ is $n$-sunny. Notice that because all lines are sunny, none of them is $x = 1$, $y = 1$ or $x + y = n+1$. Proceed similarly to the proof for Claim 2, we see that $3(n-1) \leq 2n$, so $n \leq 3$, giving us $n = 3$.

Claim 4. The only $3$-sunny numbers are $0, 1, 3$.
Proof. Assume that a configuration exists that proves $2$ is $3$-sunny.
If both of the sunny lines intersect at least $2$ points, then one of them must go through the points $(1, 3), (2, 1)$ and the other one must go through the points $(1, 2), (3, 1)$. Then the remaining line must go through the points $(1, 1), (2, 2)$, but then this last line is also sunny, giving us $3$ sunny lines, a contradiction.
So at least one of the two sunny lines pass through exactly one point. This means that the non-sunny line pass through at least $3$ points, so it must be $x = 1$, $y = 1$ or $x + y = 4$. However, a brute-force check will show that none of these cases can actually happen.

It remains to show that $0, 1, 3$ are $3$-sunny. The diagram (in the attached files) will give the configurations to do so.

Combining all claims, we see that if $k$ is $n$-sunny then $k$ is $3$-sunny, and thus $k \in \{0, 1, 3\}$.
Solution 3 Call a line moony if it is not sunny. We claim $k=0,1,3$ are the only possible values of $k$
Proof that $k \leq 3, k\neq 2$ :
We induct on $n$ for $n=3$, obviously $k \leq 3$ and $k\neq 2$.
consider the region $T_n$ of points defined as a function of $n$ as follows:
$$T_n=\{(a,b) : a+b\le n+1 , \{a,b\} \in {Z^+}^2\}$$
Call the 3 lines passing through
  • $(1,n)$ and $(n,1)$
  • $(1,n), (1,1)$
  • $(n,1), (1,1)$
as sides of $T_n$
Now, we rephrase the definition of moony and sunny as follows:

A line in the plane is called $moony$ if it is parallel to any of the sides of $T_n$ and sunny otherwise.
Now, we induct on $n$
Base Case: $n=3$ is trivial
Inductive step Suppose our claim is true for $n \in [1,m]$. For $n=m+1$, consider $T_{m+1}$, let the $m+1$, distinct lines satisfying the properties be $l_1, l_2 \dots l_{m+1}$
Claim: One of $l_1, l_2 \dots l_{m+1}$ is a side of $T_{m+1}$
Proof: Assume not then since each side of $T_{m+1}$ points has $m+1$ points and there are exactly $m+1$ lines and no line can pass through 2 points of a side, also since all lines must pass through all points, we can say for each point on a side, there is a unique line in $$\{l_1,l_2 \dots l_{m+1}\}$$passing through it, but then some line must pass through 3 points all on different sides which is a contradiction!
Hence, one of the lines is a side of $T_{m+1}$, if we remove that side, the remaining region is isomorphic to $T_m$ and
we have that $m$ lines which must cover the entire region, left and of which exactly $k$ are sunny, hence $k \neq 2$ and $k \leq 3$ by inductive hypothesis.

Proof that $k=0,1,3$ work:
For $n=3$, it is trivial.
For $n>3$, draw $n-3$ parallel moony lines, starting from a side of $T_n$ which passes through $(n,1)$ and $(1,n)$
and each subsequent moony line being 1 unit apart from the previous, then the remaining region is $T_3$ which we cover with $k$ sunny lines and $3-k$ moony lines. This construction satisfies both conditions.
Solution 4 Solution4