4 ms·
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 tri
by 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 agoI 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?