IMO 2025 · 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 \(n\geq 3\) 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 \(a+b\leq n+1\), the point \((a,b)\) is on at least one of the lines; and
  • exactly \(k\) of the \(n\) lines are sunny.

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{k\in\{0,1,3\}.} \]

Solution

Write \[ S_m=\{(a,b):a,b\text{ are positive integers and }a+b\leq m+1\}. \] For \(m\geq2\), these points lie in the triangle with vertices \[ (1,1),\qquad (m,1),\qquad (1,m). \] Its three side lines are \[ x=1,\qquad y=1,\qquad x+y=m+1, \] all of which are not sunny.

We first explain a reduction. Suppose \(m\) distinct lines cover \(S_m\), and one of them is a side line. Remove that line and the points of \(S_m\) lying on it. The remaining points are a translated copy of \(S_{m-1}\):

  • If the removed line is \(x=1\), translate by \((-1,0)\).
  • If it is \(y=1\), translate by \((0,-1)\).
  • If it is \(x+y=m+1\), no translation is needed.

Every remaining point is covered by one of the remaining \(m-1\) lines. Translating those lines preserves their distinctness and directions. Thus the reduction gives a covering of \(S_{m-1}\) by \(m-1\) distinct lines, with exactly the same number of sunny lines.

For \(m\geq4\), such a reduction is always possible. Indeed, the boundary of the triangle contains \(3m-3\) points of \(S_m\): each side contains \(m\) points, and each vertex is counted twice. Any line other than a side line meets this boundary in at most two points, since its intersection with the triangle is either empty, a single point, or a segment whose only boundary points are its endpoints. Consequently, if none of the \(m\) covering lines were a side line, they could cover at most \(2m\) boundary points. This contradicts \[ 3m-3>2m\qquad(m\geq4). \]

Repeated reduction therefore leaves three distinct lines covering \(S_3\), without changing the number of sunny lines. We now classify this case.

If one of these three lines is a side line, reduce once more to two lines covering \[ S_2=\{(1,1),(1,2),(2,1)\}. \] At least one of the two lines contains two of these three points. Every line through two of them is horizontal, vertical, or parallel to \(x+y=0\), so it is not sunny. Hence there is at most one sunny line, giving \(k=0\) or \(k=1\).

Otherwise, none of the three lines covering \(S_3\) is a side line. All six points of \(S_3\) lie on the triangle’s boundary. Each covering line contains at most two of them, so each must contain exactly two, and each point lies on exactly one covering line.

Consider a vertex. Four of the other five points lie on its two incident sides, so pairing it with any of those points would give a side line. Its covering line must therefore join it to the midpoint of the opposite side. Applying this to all three vertices forces precisely the lines \[ x=y,\qquad 2x+y=5,\qquad x+2y=5. \] Their slopes are \(1,-2,-\tfrac12\), respectively, so all three are sunny. Thus \(k=3\).

This proves that only \(0,1,3\) are possible. We finish by constructing each value for every \(n\geq3\).

  • \(k=0\): Take the \(n\) vertical lines \[ x=1,2,\ldots,n. \]

  • \(k=1\): Take the \(n-1\) vertical lines \[ x=1,2,\ldots,n-1 \] together with the sunny line \[ y=x-n+1. \] The vertical lines cover every point of \(S_n\) except \((n,1)\), which lies on the last line.

  • \(k=3\): Take \[ x=y,\qquad 2x+y=5,\qquad x+2y=5, \] together with \[ x+y=s\qquad(s=5,6,\ldots,n+1). \] The first three lines cover \(S_3\), respectively in the pairs \[ \{(1,1),(2,2)\},\quad \{(1,3),(2,1)\},\quad \{(3,1),(1,2)\}. \] Every other point of \(S_n\) lies on one of the listed diagonals. There are \(n-3\) diagonals, with none when \(n=3\); all \(n\) lines are distinct, and exactly the first three are sunny.

The solution was checked against the official IMO shortlist solutions.