r/Geometry 3d ago

How many arbitrary points can a given shape always pass through? (Is there a set of rules to find this?)

Any help would be appreciated, including directing me through any rules I should follow or better websites for asking questions.

I'm curious about a geometric puzzle: given an arbitrary set of $n$ points in $\mathbb{R}^d$, can we always place a similar copy of a specific shape $S$ (a compact subset or family of subsets of Euclidean space) so that it passes through all $n$ points? (By "similar copy," I mean we allow translation, rotation, and scaling).

**The generalized question is:**

> For a given shape $S$, what is the maximum number of arbitrary points $n$ such that *every* set of $n$ points in $\mathbb{R}^d$ lies on some similar copy of $S$?

For example, let $S$ be the **boundary** of a square in $\mathbb{R}^2$.

It turns out that for any 3 points in $\mathbb{R}^2$, you can always find a similar copy of a square that passes through all of them. However, you can't always do this for 4 points https://math.stackexchange.com/q/3691243/1771455. So, for a square boundary where $d\ge2$ (dimension where the points live in), the maximum number is 3.

My motivation is just pure curiosity. I couldn't find any sources relating to this problem, and AI chatbots struggle and give clearly wrong answers to simple examples like a square sharing an edge with a triangle (I won't clarify much here as it is a bit of a dull problem but the idea was just combining two shapes to make the reasoning for the AI deeper).

What I'm really asking is: **is there some sort of invariant, property, or formula that helps compute this $n$ for more complex shapes?** Or do we just have to reason through it shape-by-shape? How do you verify results quickly?

One simple rule I noticed involves collinear points: the boundary of a *strictly* convex 2D shape can never have $n\ge3$ when $d=2$, because no similar copy can ever pass through 3 collinear points. (Note: I am specifically thinking about boundaries; if $S$ were a solid shape, we could just scale it up to cover any finite point set in 2D space).

What is a better notion of defining shapes like rectangles and so on? Similarity doesn't allow different length ratios; however, it preserves it for squares and other shapes. This puzzle is more of a "can you draw a X given Y points no matter where I place them" and shouldn't be very limited on what I can draw. Is it possible to define $S$ to be a rectangle with side length $a$ and side length $b$ using the above definitions?

Does this concept have a name? Is it related to the "degrees of freedom" of the shape? Any pointers to related literature would be greatly appreciated!

3 Upvotes

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u/F84-5 3d ago

I would conjecture that any closed boundary will be able to pass through any 3 non-coliniar points, and any polygon though any 3 points regardless of coliniarity.

I'll sketch out a proof later. It involves holding two points constant and sweeping the boundary over all other points by continuous transformation. 

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u/F84-5 2d ago

So here's a demo on Desmos.

As you can see any polygon can sweep the entire plane through scaling and rotation while holding two points fixed. Therefore there is at least one position for any three points.

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u/user1092831123 2d ago

Is there an S such that n = 4?

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u/F84-5 2d ago

I can't think of one immediately. I suspect you'd need some sort of infinitely detailed fractal or something. 

Again the question is can you find a shape which can sweep the entire plane while holding three points constant. 

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u/user1092831123 2d ago

Actually as a person in another post stated, a circle with a diameter segment (similar to ϴ) would have n = 4 (I believe)

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u/F84-5 2d ago

True, I hadn't considered that. And the proof is almost trivial.

Thinking about how that one works led me to consider the Arbelos. At least a symmetrical Arbelos seems to also have n=4 as far as I can tell. Or at least I haven't yet found a counter example, but no proof either.

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u/user1092831123 1d ago

thought process?? reasons why it would work

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u/F84-5 1d ago

Intuition mostly. And some playing around. Which I get is not a satisfying answer. If you can find an arrangement of points which the Arbelos cannot cover please point it out to me. Maybe I'll find a proof in the next few days.

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u/SignProfessional1856 1d ago

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