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I'm curious about shapes which just one of them is determined by a number of points. From an amazing theorem in plain curves geometry we know that vertices of triangles similar to arbitrary triangles $T$ is dense on every closed jordan curve in a plain, so If $J$ be such curve and $A,B,C$ be 3 noncolinear points on plain then at least one curve similar to $J$ contains $A,B,C$ ,if $J$ be a circle then exactly one circle passes $A,B,C$.

Question: is circle the only shape on a Euclidean plain (not only from closed curves mentioned above),which just one similar to it defined by 3 non-colinear points?(Means not two or more similar of the shape fits $A,B,C$ ,just one of it).

Note: here the mentioned "shape" can be any subset of $\mathbb{R}^2$. The "trivial triangle" trivially could not be an answer because we can find many similar of it passing through the 3 points creates it as vertices.

What about generalization to $n$ points in $\mathbb{R}^m$ which exactly define $k$ similar shapes?

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    $\begingroup$ Your statement needs clarification. For example an ellipse can be defined by three points, two foci and one end of the major axis. $\endgroup$ Jul 19 '18 at 21:11
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    $\begingroup$ @lulu. Isn't the parabola actually determined by 3 points plus its axis of symmetry? But the circle only needs 3 points. $\endgroup$
    – md2perpe
    Jul 19 '18 at 21:11
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    $\begingroup$ If $v_1-v_0,v_2-v_0,\ldots,v_{n+1}-v_0\in\mathbb R^{n+1}$ linearly independent, then the sphere $S^n=\{x\in\mathbb R^{n+1}~:~|x|_2=1\}$ is uniquely determined by $v_0,v_1,v_2,\ldots,v_{n+1}$. $\endgroup$ Jul 19 '18 at 21:12
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    $\begingroup$ A triangle comes to mind. $\endgroup$ Jul 19 '18 at 21:13
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    $\begingroup$ Maybe this is a more rigorous interpretation of the question. Let $\mathcal{C}$ be a collection of shapes (connected 1 dimensional manifolds with corners equipped with an embedding into $\mathbb{R}^2$). We say that $\mathcal{C}$ has property $\mathcal{P}$ if for every three distinct points $x, y, z \in \mathbb{R}^2$ there exists a unique element of $\mathcal{C}$ that contains $x$, $y$, and $z$. What are some interesting $\mathcal{C}$ that have property $\mathcal{P}$? $\endgroup$
    – RghtHndSd
    Jul 19 '18 at 22:25
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A circle in the plane is determined by three numbers. For example, you can use the two coordinates of the center and the radius. A point on a curve gives you one number because it takes two numbers to specify the point but you have a degree of freedom in where the point is on the curve. An axis-aligned square is also determined by three numbers and three points are enough to specify it as long as they are on different sides. I believe a parabola with a given axis direction will work as well because you can specify it with the coordinates of the vertex and one number for the width. Three points should suffice here as well.

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  • $\begingroup$ For a rigth angled isosceles triangle there are many $y=x^2$ parabola passes through its vertices, pay attention the question states just one of the shape passes through each 3 points you may choose. So parabola is not an answer. $\endgroup$
    – MasM
    Jul 20 '18 at 1:53
  • $\begingroup$ I think the spirit of the question is three points in non-special position. $\endgroup$ Jul 20 '18 at 2:16
  • $\begingroup$ Appreciate your approuch of required numbers to form a shape but here the problem seems more profound than using just this approach. $\endgroup$
    – MasM
    Jul 20 '18 at 3:09
  • $\begingroup$ @MasM If I give you three collinear points, then there are infinitely many equilateral triangles that pass through all three. So if the points are allowed to be arbitrarily bad, even a triangle is not an answer. $\endgroup$ Jul 20 '18 at 4:37
  • $\begingroup$ @Misha Lavrov I mentioned the 3 points $A,B,C$ which we consider are not colinear. $\endgroup$
    – MasM
    Jul 20 '18 at 10:17
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I will attempt to make the question rigorous and then answer it. Note that "the trivial triangle" will not be an answer to the question under the following formulation.


Let a Euclidean plane be given, and let $\mathcal S$ be the collection of sets defined as follows: a set $\mathcal J$ is a member of $\mathcal S$ if and only if there exists closed Jordan curve $J$ in the given Euclidean plane such that every member of $\mathcal J$ is a closed Jordan curve in the same plane similar to $J,$ and every closed Jordan curve in that plane similar to $J$ is a member of $\mathcal J.$

Let $\mathcal T$ be the collection of all sets of three non-collinear points in the given plane. That is, a set $T$ is in $\mathcal T$ if and only if $T$ contains exactly three members, every member of $T$ is a point in the given plane, and the three points in $T$ are not collinear.

Define a function $k$ on the domain $\mathcal S \times \mathcal T$ as follows: $k(\mathcal J, T)$ is a set of closed Jordan curves in the given plane, and a closed Jordan curve $J$ in that plane is in $k(\mathcal J, T)$ if and only if $J \in \mathcal J$ and $T \subset J.$ In other words, $k$ takes any set of similar closed Jordan curves in the plane and any set of three non-collinear points in the plane and returns all the curves within the given set that pass through all the given points.

I interpret the question as asking whether the following claim is true.

Claim: Let the set $\mathcal C$ contain all the circles in the given plane and no other members. Then

  • the set $k(\mathcal C, T)$ has exactly one member for every $T \in \mathcal T,$ and
  • no other set $\mathcal J \in \mathcal S$ has the property that $k(\mathcal J, T)$ has exactly one member for every $T \in \mathcal T.$

That is, there is one and only one member of the set of all circles that passes through any given set of three distinct non-collinear points, and there is no other set of mutually similar closed Jordan curves (other than circles) that has this property.

Proof: Let $\mathcal J$ be a set with the given property. Let $\{A,B,C\}$ be vertices of an equilateral triangle in the plane, and let $J$ be a curve in $\mathcal J$ such that $\{A,B,C\} \subset J.$

Now let $J'$ be the rotation of $J$ that takes $A$ to $B,$ $B$ to $C,$ and $C$ to $A.$ Then $J$ and $J'$ are in $k(\mathcal J, \{A,B,C\}).$ But by assumption, $k(\mathcal J, \{A,B,C\})$ has exactly one element. Therefore $J' = J,$ and therefore $J$ has three-way rotational symmetry.

Now from $D,$ the center of triangle $\triangle ABC,$ construct a ray $R$ at an arbitrary angle to the ray $DA.$ Let $A'$ be a point where the ray $R$ intersects $J.$ Then there is a transformation consisting of a rotation about $D$ (and possibly a dilation about $D$) that takes $A$ to $A'$ and takes $B$ and $C$ to points on $J.$ Let $J''$ be the image of $J$ under the inverse of that transformation. Then $\{A,B,C\} \subset J'',$ and therefore the assumptions imply that $J'' = J.$

We now know that $J$ is fixed under a set of transformations that include rotation-dilations about $D$ through all possible angles. The only such closed Jordan curve is a circle with center at $D.$

Therefore $\mathcal J$ must contain a circle, and therefore $\mathcal J = \mathcal C.$

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  • $\begingroup$ It seems that this proof only works for equilateral triangle vertices but not for an arbitrarily triple points in euclidean plane. $\endgroup$
    – MasM
    Apr 7 '19 at 13:51
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    $\begingroup$ The idea here is basically a proof by contradiction. If $\mathcal J$ is a set with the properties you describe, it has to work for all non-collinear sets of three points, including equilateral triangles. I did not even say anything about equilateral triangles in general, I just picked one of them. One should be enough for a counterexample. $\endgroup$
    – David K
    Apr 7 '19 at 16:33
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    $\begingroup$ Of course, in order to invalidate this proof, you just have to add a few words to the question; instead of "three non-collinear points," make it "three non-collinear points which are not vertices of an equilateral triangle." So this is not a very satisfying proof, and one that didn't rely on this particular counter-example would be preferable. $\endgroup$
    – David K
    Apr 7 '19 at 16:35
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    $\begingroup$ Perhaps if you would rephrase the question in a way that makes it mathematically well-defined, you might get a better answer. $\endgroup$
    – David K
    Apr 7 '19 at 16:38

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