The (smaller) question is the area of the smallest triangle defined by the set of points.
Then, above that, there's the question of which set of points defines the largest smallest triangle.
Adding points to the set can only ever shrink the size of the smallest triangle that the set defines. So if you're free to add or remove points, you will always maximize the size of the smallest triangle by going all the way down to three points. At that point, you can't remove any more, but you're still free to rearrange them, so you're asking the question "what is the largest triangle that can be inscribed in a square?".
Why did you respond to me? What did you think you were saying?
> Adding points to the set can only ever shrink the size of the smallest triangle that the set defines.
I don’t think this is true. Any 3 points almost in a line will have a very small area. Adding points inside the triangle would help, and bisecting a small angle, a point on that line would work would work even outside of the triangle, for a ways anyway. I can think of a few other examples.
>> Adding points to the set can only ever shrink the size of the smallest triangle that the set defines.
> I don’t think this is true. Any 3 points almost in a line will have a very small area. Adding points inside the triangle would help
Come on. Adding points can never help. This is not a difficult proof:
Let A be a set of points, and let T be the set of all triangles whose vertices are drawn from A.
Let A' be any set of points that is a superset of A.
Let T' be the set of all triangles whose vertices are drawn from A'.
We can immediately observe that, since A is a subset of A', T is likewise a subset of T'.
Therefore, the smallest triangle in T' can never be larger than the smallest triangle in T, because the smallest triangle in T is a member of T'.
> I can think of a few other examples.
You can claim to, but you won't be telling the truth. This is a simple case of the obvious principle that if you're playing against an omniscient opponent, giving more options to them can't help you.
And yet still some point arrangements make that triangle size zero, others make the area 1/2 (assuming unit square). Most something in between. It's a problem similar to sphere packing, just not quite as close to practical application. My brain refuses to auto-resolve already at n=5. Is the obvious distribution (four corners, one in the center) the best according to size of the smallest triangle? I think it is, but no quick answers from me on that one, sorry.
I love this story, it reads like a description of academic arcadia: scientists who when the day's work of teaching is over don't Netflix'n'chill (or go maximise their share of administrative leakage) but gang up on odd mindbenders.
The (smaller) question is the area of the smallest triangle defined by the set of points.
Then, above that, there's the question of which set of points defines the largest smallest triangle.
Adding points to the set can only ever shrink the size of the smallest triangle that the set defines. So if you're free to add or remove points, you will always maximize the size of the smallest triangle by going all the way down to three points. At that point, you can't remove any more, but you're still free to rearrange them, so you're asking the question "what is the largest triangle that can be inscribed in a square?".
Why did you respond to me? What did you think you were saying?
> Adding points to the set can only ever shrink the size of the smallest triangle that the set defines.
I don’t think this is true. Any 3 points almost in a line will have a very small area. Adding points inside the triangle would help, and bisecting a small angle, a point on that line would work would work even outside of the triangle, for a ways anyway. I can think of a few other examples.
Most points won’t help though.
>> Adding points to the set can only ever shrink the size of the smallest triangle that the set defines.
> I don’t think this is true. Any 3 points almost in a line will have a very small area. Adding points inside the triangle would help
Come on. Adding points can never help. This is not a difficult proof:
Let A be a set of points, and let T be the set of all triangles whose vertices are drawn from A.
Let A' be any set of points that is a superset of A.
Let T' be the set of all triangles whose vertices are drawn from A'.
We can immediately observe that, since A is a subset of A', T is likewise a subset of T'.
Therefore, the smallest triangle in T' can never be larger than the smallest triangle in T, because the smallest triangle in T is a member of T'.
> I can think of a few other examples.
You can claim to, but you won't be telling the truth. This is a simple case of the obvious principle that if you're playing against an omniscient opponent, giving more options to them can't help you.
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if there're only three points, how many triangles can you get from that? one. so the biggest triangle is the smallest. thought that was obvious
And yet still some point arrangements make that triangle size zero, others make the area 1/2 (assuming unit square). Most something in between. It's a problem similar to sphere packing, just not quite as close to practical application. My brain refuses to auto-resolve already at n=5. Is the obvious distribution (four corners, one in the center) the best according to size of the smallest triangle? I think it is, but no quick answers from me on that one, sorry.
I love this story, it reads like a description of academic arcadia: scientists who when the day's work of teaching is over don't Netflix'n'chill (or go maximise their share of administrative leakage) but gang up on odd mindbenders.
That "obvious" distribution is actually the worst, three points are in a line so have area of zero.
1 reply →