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Take any curve at all, and select an arbitrary point A. Now draw the midpoint between A and every point of the curve. I conjecture that you will end up with a curve that is a translated and scaled version of the original curve. Why? What's the scaling factor and where is the translation exactly? Is my conjecture even true?

It seems like some classical problem the Greeks have solved, but I couldn't find anything online and am stuck on it myself. Any help is appreciated.

EDIT: This can also be generalized such that you don't take the the midpoint, but some point that divides the line in a given ratio. The conjecture still seems to hold.

jan
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    Using the language of Euclidean geometry, this is called "similarity". For example, Euclid proves that two triangles are similar if and only if they have the same angles. – Lee Mosher Apr 16 '20 at 21:27
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    The scaling factor in your construction is $\frac{1}{2}$: your transformation shrinks the distance between $A$ and each point on the curve by a factor of $\frac{1}{2}$, and in fact the transformation shrinks all distances by $\frac{1}{2}$. – Lee Mosher Apr 16 '20 at 21:28

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This is the definition of a homothetic transformation, with the point $A$ being its center and the given ratio being the scale factor.

See, for example, https://en.wikipedia.org/wiki/Homothetic_transformation for more information.

MPW
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Lets $x(t)$ with real $t \in [0,1]$ be the curve and let $c$ be the point we have chosen. Then the midpoint function is $y(t)=\frac{1}{2}(c+x(t))$ which is easily seen to be a scaled and translated version of the original.

CyclotomicField
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  • Thank you for your answer. Is the $x(t)$ a function? The question is about any curve so even ones that aren't functions. Also why is $t$ restricted to that specific range? – jan Apr 16 '20 at 21:39
  • @Ciprum $x(t)$ is a function from $\mathbb{R} \rightarrow \mathbb{R}^n$. – CyclotomicField Apr 16 '20 at 21:43
  • But since $c$ is the output of the function $x(t)$ doesn't that mean that it lies on the curve (because the curve is defined by $x(t)$ )? By the problem statement, it doesn't have to. – jan Apr 16 '20 at 22:03
  • @Ciprum I misunderstood and thought you wanted a point on the curve, but as you can see from the substitution $x(t_0)=c$ that it can be arbitrary. I'll edit to reflect this. – CyclotomicField Apr 16 '20 at 22:26
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Using Homothety you can prove this.

Draw line from point $A$ to every point of the curve. Now the midpoints of the lines creates new curve which is homothetic similar to the original curve. The homothetic center would be point $A$ and new curve would be at $\dfrac{1}{2}$ of the original curve.

Soyeb Jim
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    I see. However it seems like an ad hoc argument. I would like to know why exactly do those Homeothetic transformations create similar shapes? – jan Apr 16 '20 at 21:37
  • Think of this as you are just shrinking the whole graph from every side and you are shrinking towards point $A$.

    A more geometric approach would be- Think of two points $B$ and $C$ other than $A$, where $BC \approx 0$. Now for triangle $ABC$, taking the midpoints we can create new triangle $AB'C'$ which is similar to triangle $ABC$. Now the curve can be made by infinity many triangle as $ABC$. all of which would be similar as well. In turns preserving the curve shape.(I am not a native English speaker, so my use of words are not great, sorry for that)

     – Soyeb Jim Apr 16 '20 at 21:51