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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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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 A,B,…,EA, B, \ldots, E and 1,2,…,51,2, \ldots, 5 respectively. Obviously, one of the removed vertices must be a vertex of the big square - let this be vertex A1A 1. Then, in order to "destroy" all the squares shown in Figure 3 b−e3 \mathrm{~b}-\mathrm{e} we have to remove vertices B2,C3,D4,D2B 2, C 3, D 4, D 2 and B4B 4. Thus we have removed 6 vertices without having any choice but a square shown in Figure 3f3 \mathrm{f} is still left intact.

Diagram for the mathnet 00xp 1 of bw-1993-14. a\mathrm{a} Diagram for the mathnet 00xp 1 of bw-1993-14. b\mathrm{b} Diagram for the mathnet 00xp 1 of bw-1993-14. c Diagram for the mathnet 00xp 1 of bw-1993-14. d Diagram for the mathnet 00xp 1 of bw-1993-14. e Diagram for the mathnet 00xp 1 of bw-1993-14. f 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

02
15
21
30
40
50
All team scores
TeamScore
Poland0 / 5
Latvia0 / 5
Estonia1 / 5
Sweden2 / 5
Lithuania1 / 5
Finland1 / 5
Iceland1 / 5
Denmark1 / 5