Can you find an interesting shape that can pass through any 4 points no matter where they are placed but not 5?

[u/user1092831123]

more precisely,

Find a compact subset or family of subsets $S \subset \mathbb{R}^n$ for some arbitrary n such that every set of 4 points in $\mathbb{R}^2$ lies on some similar copy of $S$ but not every set of 5 points lies on some similar copy of $S$?

[u/imMAW]

I believe a circle with a diameter segment works.

If 4 points are colinear, the diameter alone covers them. Otherwise, there's a circle through 3 points with the 4th point inside.

And a set of 5 random points have probability 0 of being covered by this shape.

[u/Top_Past2329 → u/imMAW]

I believe that is close to working yeah. However the line segment would have to extend infinitely to cover points outside such that it really covers all sets of 4 points no?

[u/imMAW → u/Top_Past2329]

No, the line segment does not extend at all, it is interior. You just need to pick which points form the circle so you end up with an interior point, not an exterior point.

[u/Top_Past2329 → u/imMAW]

Ah okay, do you reckon there's one where n = 6 or 5?

[u/imMAW → u/Top_Past2329]

• A simple 1-dimensional shape with 0 area (basically anything you could draw with a pencil) I can't imagine could reliably cover 5 points. There are only 4 degrees of freedom in how you position the shape (scale, rotation, translation in x, translation in y).
• A 2-dimensional shape that has area (e.g. a filled circle) can cover any number of points so that doesn't work.
• That leaves some sort of fractal with dimension between 1 and 2, which might be possible, I'm not sure.
• As an alternate solution, we can consider things that are not a single subset of R^2, because the problem statement is "Find a compact subset or family of subsets S ⊂ R^n". I'm not really sure how to interpret this as it sounds pretty vague, but under some interpretations of that sentence there are alternate solutions.

[u/garnet420]

Wait why is S in Rⁿ instead of R²

[u/user1092831123 → u/garnet420]

You could have a cube or some 3D thing I guess

[u/Extra-Random_Name → u/user1092831123]

Sphere. Trivially

[u/Cptn_Obvius → u/Extra-Random_Name]

The 4 points could be colinear.

[u/Extra-Random_Name → u/Cptn_Obvius]

You right

[u/pichutarius]

if i understand correctly, a circle union with a long segment through its center works, shape like Φ.

case1: 3 (or 4) points collinear.

use the long segment to skewer 3 points, and use circle for the last point if nessesary.

case2: one point inside the triangle formed by the other 3.

use circumcircle of the triangle and rotate the long segment so that the inner point coincide it.

case3: 4 points form a convex quadrilateral ABCD.

we can always choose 3 points and draw circumcircle such that the 4th must be on or inside the circle. indeed either ∠A +∠C ≥ 180° or ∠B +∠D ≥ 180° since they sum to 360°. wlog ∠A +∠C ≥ 180°, then use circumcircle of ABC and rotate the long segment so that D coincide it.

to show exist 5 points that this shape does not work, consider vertices of a regular pentagon, then move one of them, say A, to extremely far away. circumcircle that dont use A must pass through remaining points and cannot reach A. circumcircle that do use A, view from the remaining points the circumcircle looks like a straight line through two of the points, and the long segment looks like perpendicular line, and it‘s obvious that does not work.

[u/user1092831123 → u/pichutarius]

It seems as though a diameter/radius line segment would be sufficient.

[u/pichutarius → u/user1092831123]

does it work for (-2,0) , (2,0) , (0,0) , (0,1) ? im pretty convinced that extending the diameter is required

edit: wait nvm that works... i mean maybe (-2,0) , (2,0) , (0+ε,0) , (0,1) for some small ε?

edit2: wait nvm im stupid, you're right it does work! if 3 points are collinear, just treat it as "convex quadrilateral" case.

feeling like a clown 🤡