2025 IMO Problems/Problem 1

A line in the plane is called sunny if it is not parallel to any of the $x$ –axis, the $y$ –axis, and 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 $(a,b)$ is on at least one of the lines; and
exactly $k$ of the $n$ lines are sunny.

Video Solution

https://www.youtube.com/watch?v=kJVQqugw_JI (includes motivational discussion)

https://youtu.be/4K6wbEuNooI (includes exploration to show motivation behind arguement)

Solution from Edutube: https://www.youtube.com/watch?v=n2Ct4z0eUhg

Solution 1

Consider a valid construction for $k \ge 4$ . $\text{Claim: One of the } n \text{ lines must be } x=1, y=1, \text{ or } x+y=n.$ Proof: Assume for the sake of contradiction not. Then, the following holds: $\quad \text{1. } x=1 \text{ is not in our lines.}$ Otherwise, two points with $x=1$ are on the same line. This implies that each point with $x$ -coordinate $1$ must lie on distinct lines, hence there exists a bijection between the lines and points with $x$ -coordinate $1$ . It follows with similar reasoning that: $\quad \text{2. Lines are bijective with points with } y \text{-coordinate.}$ $\quad \text{3. Lines are bijective with points } x+y=n+1.$ Consider the points on $x+y=n+1$ that are not $(1,n)$ or $(n,1)$ . Then, because there exists a bijection, any such point must have a line through a point with $x$ -coordinate $1$ and $y$ -coordinate $1$ that are not $(1,n)$ or $(n,1)$ (otherwise $x+y=n+1$ exists). But this cannot be possible if the point is not $(1,1)$ . Since $n \ge 3$ , by the Pigeonhole Principle there must be at least $1$ point that has to pass through this condition, hence we have proved the claim. $\square$

We can find the constructions for $0,1,3$ easily.

Hence, remove one of the $x=1, y=1,$ or $x+y=n+1$ lines. We then get a valid covering for $n-1$ with the same number of sunny lines! Thus, any possible number of sunny lines for $n$ must be possible for $n-1$ . For $n=3$ , we have possibilities $k=0, k=1, \text{ or } k=3$ . By our induction above, we conclude that for any $n$ , the possible $k$ is a subset of $\{0,1,3\}$ . $\blacksquare$

~MC

Solution 2

A Short Story

Please don't read this solution if you have already perfectly understood other solutions suggested :)) After solving this problem on my own, I LaTeXed the solution as I thought it is an original solution... It turned out to be the exact same solution as listed above :( Anyhow, hoping this would benefit others, I am just posting this :)

Finding all $k$ for each $n$ value will lead to the answer. Starting with $n=3$ , it is easy to see that $k \in \{0, 1, 3\}$ are the only solutions.

Notice that for the rest of the cases, if a line is a "long line" ( $x=1, y=1, x+y=n+1$ ), the case would be identical to its previous case $n-1$ . Therefore, investigating a case where no "long lines" are included will lead to a new potential value of $k$ .

Notice that each line must contain exactly one points from each long lines. However, no such case exists as most triples of dots will form a triangle. Therefore, by induction, all cases lead to $n=3$ , when $k \in \{0, 1, 3\}$ . $\blacksquare$

~MaPhyCom (Renamed: Thinking Tree)

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2025 IMO Problems/Problem 1

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Solution 1

Solution 2

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