diff --git "a/IMO/md/en-IMO-2025-notes.md" "b/IMO/md/en-IMO-2025-notes.md"
--- "a/IMO/md/en-IMO-2025-notes.md"
+++ "b/IMO/md/en-IMO-2025-notes.md"
@@ -105,11 +105,11 @@ Hence, by induction we may repeatedly delete a long line without changing the nu
We now classify all the ways to cover the \(1 + 2 + 3 = 6\) points in an \(n = 3\) grid with 3 lines.
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Long line present
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No long line
@@ -150,7 +150,7 @@ Proof. \(\angle AEC = \angle ABC = \angle CAB = 90^{\circ} - \alpha\) .
Hence, if we let \(A' := \overline{CE} \cap \overline{DF}\) , we have a parallelogram \(ACA'D\) . Note in particular that \(\overline{BA'} \parallel \overline{CD}\) .
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Next, let \(T\) denote the circumcenter of \(\triangle A^{\prime}EF\) . (This will be the tangency point later in the problem.)
@@ -435,7 +435,7 @@ Construction. We show a general construction when \(n = k^{2}\) illustrated belo
tiles as promised.
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@@ -457,7 +457,7 @@ To apply this to the present problem, take the \(n\) uncovered squares which we
The figure below shows two examples of the process, each for a board with \(n = 9\) , for two choices of LIS and LDS. The cells in the LIS and LDS have been marked with green circles, and the boundaries of the quadrants are drawn in green lines. In the left example, the LIS and LDS have a black square in common (that cell has all four directions labeled). In the right example, the LIS and do not have common squares.
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We observe that: