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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> is 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> is 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.
 +
 +
==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==
 
==Solution 1==
−
Consider a valid construction for <math>k=4</math>.
+
Consider a valid construction for <math>k \ge 4</math>.
 
<cmath>
 
<cmath>
 
\text{Claim: One of the } n \text{ lines must be } x=1, y=1, \text{ or } x+y=n.
 
\text{Claim: One of the } n \text{ lines must be } x=1, y=1, \text{ or } x+y=n.
Line 23: Line 30:
 
</cmath>
 
</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>
 
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>
−
 
+
*We can find the constructions for <math>0,1,3</math> easily.
 
----
 
----
  
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~MC
 
~MC
  
−
==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
 
  
 
==See Also==
 
==See Also==

Latest revision as of 09:52, 22 July 2025

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


See Also

2025 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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