Baltic Way 1993 · Problem 14
Combinatorics
A square is divided into 16 equal squares, obtaining the set of 25 different vertices. What is the least number of vertices one must remove from this set, so that no 4 points of the remaining set are the vertices of any square with sides parallel to the sides of the initial square?
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Review
Topics
Colorings and configurations · Pigeonhole and extremal arguments
Solutions
Solution
Solution:
The example in Figure 3a demonstrates that it suffices to remove 8 vertices to "destroy" all squares. Assume now that we have managed to do that by removing only 6 vertices. Denote the horizontal and vertical lines by and respectively. Obviously, one of the removed vertices must be a vertex of the big square - let this be vertex . Then, in order to "destroy" all the squares shown in Figure we have to remove vertices and . Thus we have removed 6 vertices without having any choice but a square shown in Figure is still left intact.
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Figure 3
Contest context
Results from Baltic Way 1993
8 teams
- Mean score
- 0.9 / 5
- Scores of 4 or 5
- 0 / 8
- Estonia
- 1 / 5
Score distribution
All team scores
| Team | Score |
|---|---|
| Poland | 0 / 5 |
| Latvia | 0 / 5 |
| Estonia | 1 / 5 |
| Sweden | 2 / 5 |
| Lithuania | 1 / 5 |
| Finland | 1 / 5 |
| Iceland | 1 / 5 |
| Denmark | 1 / 5 |