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The part of the problem left out is that you cannot choose the number of points, only their arrangement. Given n, choose an arrangement of n points in the squar

The part of the problem left out is that you cannot choose the number of points, only their arrangement. Given n, choose an arrangement of n points in the square that maximises the area of the largest smallest triangle. This is not trivial. It sounds like one can just arrange 3 points to form a very large triangle and then put the rest of the points away in a corner, but then the smallest triangle will be between some of the points in the corner, not your conveniently large one.

thaumasiotes 3y ago
> This is not trivial. I never thought it was! I'm having enough trouble trying to prove that the largest triangle that will fit in a square is half the size of the square.
- tomsmeding 3y ago
  Claim one: all 3 points lie on the edges. Proof: if one doesn't, one can enlarge the area of the triangle by moving it along one of the edges of the current triangle, away from the other point on that edge, until it hits a border of the square. Then take one vertex that is not in a square corner yet. The area of the triangle is 1/2 * (length of opposite edge) * (distance from opposite edge to this vertex). This can be written as a linear function of the position of this vertex along the square edge it is on. Thus over all positions of the vertex on this square-edge, the area either is always the same (if the opposite triangle edge is parallel to this square-edge), or the area is maximised by putting this vertex in a square corner. Repeat this process three times. Conclusion: the obvious half-square triangle has maximal area (even if it is not unique with that property), so the maximal area is half the square's area.
  - thaumasiotes 3y ago
    > Thus over all positions of the vertex on this square-edge, the area either is always the same (if the opposite triangle edge is parallel to this square-edge), or the area is maximised by putting this vertex in a square corner. I'm sure this is correct, but the proof doesn't seem ironclad. As far as I see, you're relying on the idea that moving two vertices jointly, as one operation, can't produce any benefits relative to moving the two vertices in sequence one after the other. And while I have no reason to doubt that that's correct, it isn't obvious to me. (Wording it another way - how do you prove that [the position of vertex B on the edge of the square] that maximizes the area of the triangle holding the position of vertex C fixed is the same as [the position of vertex B on the edge of the square] that maximizes the area of the triangle holding only the edge to which vertex C is attached fixed?)
    - rbanffy 3y ago
      I know this is not how proofs work, but is is quite intuitive. What is the point we stop proving things that look obvious and axiomatic?

thaumasiotes 3y ago

> This is not trivial. I never thought it was! I'm having enough trouble trying to prove that the largest triangle that will fit in a square is half the size of the square.

tomsmeding 3y ago
Claim one: all 3 points lie on the edges. Proof: if one doesn't, one can enlarge the area of the triangle by moving it along one of the edges of the current triangle, away from the other point on that edge, until it hits a border of the square. Then take one vertex that is not in a square corner yet. The area of the triangle is 1/2 * (length of opposite edge) * (distance from opposite edge to this vertex). This can be written as a linear function of the position of this vertex along the square edge it is on. Thus over all positions of the vertex on this square-edge, the area either is always the same (if the opposite triangle edge is parallel to this square-edge), or the area is maximised by putting this vertex in a square corner. Repeat this process three times. Conclusion: the obvious half-square triangle has maximal area (even if it is not unique with that property), so the maximal area is half the square's area.
- thaumasiotes 3y ago
  > Thus over all positions of the vertex on this square-edge, the area either is always the same (if the opposite triangle edge is parallel to this square-edge), or the area is maximised by putting this vertex in a square corner. I'm sure this is correct, but the proof doesn't seem ironclad. As far as I see, you're relying on the idea that moving two vertices jointly, as one operation, can't produce any benefits relative to moving the two vertices in sequence one after the other. And while I have no reason to doubt that that's correct, it isn't obvious to me. (Wording it another way - how do you prove that [the position of vertex B on the edge of the square] that maximizes the area of the triangle holding the position of vertex C fixed is the same as [the position of vertex B on the edge of the square] that maximizes the area of the triangle holding only the edge to which vertex C is attached fixed?)
  - rbanffy 3y ago
    I know this is not how proofs work, but is is quite intuitive. What is the point we stop proving things that look obvious and axiomatic?