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Difference between revisions of "2025 IMO Problems/Problem 1"

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* For all positive integers <math>a</math> and <math>b</math> with <math>a+b\le n+1</math>, the point <math>(a,b)</math> lies on at least one of the lines; and
 
* For all positive integers <math>a</math> and <math>b</math> with <math>a+b\le n+1</math>, the point <math>(a,b)</math> lies on at least one of the lines; and
 
* Exactly <math>k</math> of the <math>n</math> lines are sunny.
 
* Exactly <math>k</math> of the <math>n</math> lines are sunny.
 +
 +
 +
==Solution 1==
 +
Consider a valid construction for <math>k=4</math>.
 +
<cmath>
 +
\text{Claim: One of the } n \text{ lines must be } x=1, y=1, \text{ or } x+y=n.
 +
</cmath>
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\textbf{Proof}: Assume for the sake of contradiction not. Then, the following holds:
 +
<cmath>
 +
\quad \text{1. } x=1 \text{ is not in our lines.}
 +
</cmath>
 +
Then, two points with <math>x=1</math> are on the same line. This implies that each point with <math>x</math>-coordinate <math>1</math> must lie on distinct lines, hence there exists a bijection between the lines and points with <math>x</math>-coordinate <math>1</math>. It follows with similar reasoning that:
 +
<cmath>
 +
\quad \text{2. Lines are bijective with points with } y \text{-coordinate.}
 +
</cmath>
 +
<cmath>
 +
\quad \text{3. Lines are bijective with points } x+y=n+1.
 +
</cmath>
 +
Consider the points on <math>x+y=n+1</math> that are not <math>(1,n)</math> or <math>(n,1)</math>. Then, because there exists a bijection, any such point must have a line through a point with <math>x</math>-coordinate <math>1</math> and <math>y</math>-coordinate <math>1</math> that are not <math>(1,n)</math> or <math>(n,1)</math> (otherwise <math>x+y=n+1</math> exists). But this cannot be possible if the point is not <math>(1,1)</math>. Since <math>n \ge 3</math>, by the Pigeonhole Principle there must be at least <math>1</math> point that has to pass through this condition, hence we have proved the claim. <math>\square</math>
 +
 +
---
 +
 +
Hence, remove one of the <math>x=1, y=1,</math> or <math>x+y=n+1</math> lines. We then get a valid covering for <math>n-1</math> with the same number of sunny lines! Thus, any possible number of sunny lines for <math>n</math> must be possible for <math>n-1</math>.
 +
For <math>n=3</math>, we have possibilities <math>k=0, k=1, \text{ or } k=3</math>. By our induction above, we conclude that for any <math>n</math>, the possible <math>k</math> is a subset of <math>\{0,1,3\}</math>. <math>\blacksquare</math>
  
  
 
== Video Solution ==
 
== Video Solution ==
 
https://www.youtube.com/watch?v=kJVQqugw_JI [includes motivational discussion]
 
https://www.youtube.com/watch?v=kJVQqugw_JI [includes motivational discussion]
 +
 +
https://youtu.be/4K6wbEuNooI [includes exploration to show motivation behind arguement]

Revision as of 20:42, 16 July 2025

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 1

Consider a valid construction for $k=4$. \[\text{Claim: One of the } n \text{ lines must be } x=1, y=1, \text{ or } x+y=n.\] \textbf{Proof}: Assume for the sake of contradiction not. Then, the following holds: \[\quad \text{1. } x=1 \text{ is not in our lines.}\] Then, 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. $\square$

---

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 $\{0,1,3\}$. $\blacksquare$


Video Solution

https://www.youtube.com/watch?v=kJVQqugw_JI [includes motivational discussion]

https://youtu.be/4K6wbEuNooI [includes exploration to show motivation behind arguement]

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