BALTIC-WAY 1993 Problem 14
- Answer
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?
Remark. The proposed solution to this problem claimed that it is enough to remove 7 vertices but the example to demonstrate this appeared to be incorrect. Below we show that removing 6 vertices is not sufficient but removing 8 vertices is. It seems that removing 7 vertices is not sufficient but we currently know no potential way to prove this, apart from a tedious case study.
Dataset text NuminaMath olympiads. Quoted with source.
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Solutions
No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.