Does there exist a continuous function $f:[0,1]\rightarrow\mathbb{R}$ such that for any two points P,Q on the curve, there exists a point R on the curve such that PQR is an equilateral triangle? If so, can we find a smooth one?

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  • $\begingroup$ is $f : [0,1] \to {\mathbb R}^2$? $\endgroup$ – sdcvvc May 4 '12 at 14:14
  • $\begingroup$ Well you can construct it either way. It's the same thing for the purposes of this question. I guess for a proof it's easier to use f:[0,1]-->R^2 because the geometry is in R^2. Edit: it might actually be easier to work in R. Whatever. It doesn't matter. $\endgroup$ – Adam Rubinson May 4 '12 at 14:23
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    $\begingroup$ My only guess for the answer would be to use the Intermediate Value Theorem for a relatively flat (e.g. |gradient| <= 2/3) function to show that it is never true. To this end I'm not making progress, and even so, it says nothing about a function that has |gradient| > 2/3 in places. $\endgroup$ – Adam Rubinson May 4 '12 at 14:35
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    $\begingroup$ What curve? The graph of $f$? $\endgroup$ – Chris Eagle May 4 '12 at 14:48
  • $\begingroup$ Do you mean to fill an equilaterial triangle? $\endgroup$ – Jim Hefferon May 4 '12 at 14:50

Here is the answer to the generalized form of your question in the case of bounded sets. As it turns out, the convexity assumption is unnecessary.

Proposition. Let $A\subseteq\Bbb R^2$ be a bounded set, such that for each pair of distinct points $x,y\in A$ there is a point $z\in A$, such that $x,y,z$ are the vertices of an equilateral triangle. Then $A$ is one of the following:

  • empty set,
  • a set containing only one point,
  • the set of vertices of some equilateral triangle.

Proof. Let $A$ be a set satisfying the hypotheses of the proposition. Suppose $A$ contains more than one point. We will show that then $A$ must be the set of vertices of some equilateral triangle.

First, we shall prove the proposition in the case that $A$ is closed. (In the end we shall show that the general case follows easily from this one.) So, let's assume $A$ is closed, i.e. $A$ contains all its limit points. Then $A$ is compact, so there exist points $a,b\in A$ such that $$d(a,b)=\operatorname{diam}A=\sup\lbrace d(x,y)|\text{ }x,y\in A\rbrace,$$ where $d$ is the metric in $\Bbb R^2$ and $\operatorname{diam}$ stands for diameter.

Let $c$ be a point such that $a,b,c$ are the vertices of an equilateral triangle. This exists by the hypotheses. Let $r=d(a,b)$ and let $S=\overline{K}(a,r)\cap\overline{K}(b,r)\cap\overline{K}(c,r)$, where $\overline{K}(x,r)$ denotes the closed ball centered at $x$ with radius $r$. Then $A\subseteq S$, because otherwise we would have $\operatorname{diam} A>r$ which is not the case. By the way, $S$ is a Reuleaux triangle:

enter image description here

We already know that $A$ contains the vertices of this Reuleaux triangle. We shall now show that this is all that $A$ contains. We shall first define some more points. Let $a'$ be the reflection of $a$ across the line through $b$ and $c$. Define $b'$ and $c'$ analogously. (As reflections of $b$ and $c$ across the lines opposite to them.) Let $$S_{a}:=S\setminus(\overline{K}(b',r)\cup\overline{K}(c',r))\\S_{b}:=S\setminus(\overline{K}(c',r)\cup\overline{K}(a',r))\\S_{c}:=S\setminus(\overline{K}(a',r)\cup\overline{K}(b',r))$$ Now, note that every point of $S\setminus\lbrace a,b,c\rbrace$ is contained in at least one of the sets $S_a,S_b,S_c$. So, to complete the proof in the case where $A$ is closed, it suffices to show that $A\cap S_a=\emptyset,A\cap S_b=\emptyset$ and $A\cap S_c=\emptyset$, which we will now do.

Actually, we will only prove the case of $S_a$. The other two can be proved by a completely symmetric argument, so we leave them to the reader. Let $x\in S_a$. Let $u(x)$ be the point that forms an equilateral triangle together with $a$ and $x$, for which the triangle $a,x,u(x)$ is oriented clockwise (i.e. negatively). Let $v(x)$ be the other point that forms an equilateral triangle with $a$ and $x$, i.e. $a,x,v(x)$ is oriented counterclockwise (positively). This defines two functions $u:S_a\to\Bbb R^2$, $v:S_a\to\Bbb R^2$. By definition, $u$ is a rotation around the point $a$ by $-\frac{\pi}{3}$, and $v$ is the rotation around the point $a$ by $\frac{\pi}3$.

By exploiting this geometry, one can now easily see that $u$ rotates the set $S_a$ outside of $S$, i.e. $u(S_a)\subseteq \Bbb R^2\setminus S$, and for $v$ the same holds: $v(S_a)\subseteq \Bbb R^2\setminus S$. But this means that for each $x\in S_a$ the only vertices that could form an equilateral triangle with $x$ and $a$ are outside of $S$, hence not in $A$. But then $x$ cannot be an element of $A$. Here's a picture showing $S_a$ (middle), and its two rotations (the Reuleaux triangle around $S_a$ is $S$, and the two rotations lie outside of $S$):

enter image description here

This completes the proof of the closed case.

In the general case of bounded (not necessarily closed) $A$ with at least two points, we proceed as follows. Let $\overline{A}$ be the closure of $A$. Since $\Bbb R^2$ is such a nice (i.e. first-countable) space, this means that $$\overline{A}=\lbrace x\in\Bbb R^2|\text{ there is a sequence } (x_n)_n\text{ with terms in } A\text{, such that } \lim_{n\to\infty}x_n=x\rbrace.$$ Now, clearly $A\subseteq\overline{A}$. Furthermore, $\overline{A}$ also satisfies the hypotheses of the proposition. To see this, let $x,y\in\overline{A}$. Then there are sequences $(x_n)_n$ and $(y_n)_n$ in $A$, such that $\lim_{n\to\infty}x_n=x$, $\lim_{n\to\infty}x_n=x$. But since $x_n,y_n\in A$ for each $n$, we can associate a point $z_n\in A$ to each pair $x_n,y_n\in A$ such that $x_n,y_n,z_n$ are the vertices of an equilateral triangle. Because $\overline{A}$ is compact, we can choose a subsequence of $(z_n)_n$ that converges to a point $z\in\overline{A}$. By continuity of the metric, $x,y,z$ again form an equilateral triangle. So, $\overline{A}$ indeed satisfies the hypotheses of the proposition and is therefore the set of vertices of some equilateral triangle. From this it easily follows that $\overline{A}=A$, which concludes the proof. $\square$

This proves among other things, that there is no continuous function $f:[0,1]\to\Bbb R$, whose graph contains for every pair of points a third point that forms an equilateral triangle with them. It proves even more: there is also no such discontinuous function $f:[0,1]\to\Bbb R$. Why? Suppose there is. Then it must be bounded, since, otherwise its graph would contain two points $x,y$ such that $d(x,y)>100$. There would then have to be a third point on the graph whose first coordinate would lie outside of $[0,1]$, which is not the case. But for bounded sets, the proposition applies. The same applies for any function $f:B\to\Bbb R$, where $B\subseteq\Bbb R$ is a bounded set.

The situation might be more interesting with functions $f:\Bbb R\to\Bbb R$, however. But in this case again, there is no such bounded function. (Same argument as in the previous paragraph.) It might be possible to construct something unbounded and probably horribly discontinuous, though, but I haven't thought much about that.

  • $\begingroup$ I did suspect that convexity was unnecessary for proofs (especially general ones). And for geometric purposes, it seems as though you "need" convexity, even though you have shown you don't if you use compactness of R^n. In fact, the compactness method you use adds to the beauty of the problem in my opinion. HOWEVER, in my question I require for you to also show the equilateral triangle property holds/does not hold for cts, unbounded f:R-->R functions. If you can do this you will achieve the bounty... I don't think this is unreasonable as it clearly states it in my question at the bottom. $\endgroup$ – Adam Rubinson May 12 '12 at 2:35
  • $\begingroup$ I can think of ideas that might work for "horrible discontinuous unbounded functions f:R-->R with the property" I am thinking: take the "equilateral triangle grid": exchangedownloads.smarttech.com/public/content/29/… considering only the vertices, not the lines joining the vertices. Rotate the whole graph by pi/4 radians anti-clockwise (for example). Then I think every point is no on top any other point. Maybe it doesn't fill up R in the domain but maybe we can add another point somewhere and "expand out" so that it does. $\endgroup$ – Adam Rubinson May 12 '12 at 2:46
  • $\begingroup$ @AdamRubinson: If I think of something, I'll let you know. I won't be around for two days, though, since I have some stuff to do ... $\endgroup$ – Dejan Govc May 12 '12 at 3:01
  • $\begingroup$ thanks v. much. To try to be fair, if someone else finds the answer before you, then if possible I'll see if I can share the bounty between you two. I'll ask a mod if I can do that if that situation arises... $\endgroup$ – Adam Rubinson May 12 '12 at 3:03

Not in the smooth case. Here we suppose that $f$ is $C^1$.

Let $s=\frac{\sqrt{3}}3=\tan{\fracπ6}.$

Suppose that for some $x_0$, $|f'(x_0)|<s$ then around $x_0$ the graph of $f$ looks like a straight line so horizontal that the third point of a equilateral triangle containing $(a,f(a))$ and $(b,f(b))$ would be of abscissa $x∈[a,b]$.

In the following picture, the case on the left is the one that is impossible.


Then $|f'|≥s$ and $f'$ is continuous so $f$ is monotonous so $f$ can't meet your criteria.

I don't know how to generalize.

  • $\begingroup$ For your line of reasoning, we only need |f'(x_0)| < 2/3, not necc. 1/2. Thanks though, I think I can construct an epsilon-delta proof now for functions that have |f'(x_0)| < 2/3 for some x_0. If f is such that there is no x_0 with |f'(x_0)| < 2/3, then the result is kind of obvious. So that is the smooth case done. The "cts but non-smooth" case still remains to be solved. $\endgroup$ – Adam Rubinson May 4 '12 at 15:32
  • $\begingroup$ I made an error, it was not $\frac12=\sin{\fracπ6}$ but $\tan{\fracπ6}$. However $\frac23$ is not small enough. $\endgroup$ – jmad May 5 '12 at 15:55
  • $\begingroup$ Thanks for the diagram. Just for anyone reading: the shaded-in part of the diagram (equilateral triangles) are not supposed to be part of the function. Only the vertices of the equilateral triangle are meant to be part of the function. The only other option (for the smooth case) is if we have a function which has |f(x)| >= 2/3 everywhere. f must be monotone otherwise there must be a turning point which would mean |f|<=2/3 --><--. But f monotone and gradient >= 2/3 clearly won't work. $\endgroup$ – Adam Rubinson May 5 '12 at 18:52
  • $\begingroup$ However if we don't require smoothness then we can try some crazy fractal functions like the Weierstrass function: en.wikipedia.org/wiki/Weierstrass_function $\endgroup$ – Adam Rubinson May 5 '12 at 18:53
  • $\begingroup$ However if we don't require smoothness then we can try some crazy fractal functions like the Weierstrass function: en.wikipedia.org/wiki/Weierstrass_function I guess this is what I am after for the non-smooth case. Also, I don't think we can find a subset of R^2 (e.g. shaded in circle, shaded in square etc.) which works. Suppose there were such a shape, call it a set X in R^2. Just find (z1,z2) in X that maximizes {|zA-zB|: zA,zB in X}}. Then you would need the 3rd vertex (say z3) to be in X and when you colour in the equilateral triangle joining z1, z2 and z3, this shaded triangle $\endgroup$ – Adam Rubinson May 5 '12 at 19:04

Suppose that the set $S$ is bounded, and suppose that it has more than one point. Let $A$ and $B$ be two points in $S$. Then it must have a third point $C$ which is (another) vertex of the equilateral triangle with one side as $AB$, and by our condition of convexity the entire triangle $ABC$ must be in $S$.

Now let $D$ be the point of intersection of the perpendicular bisector of $\angle BAC$ and ${BC}$. Then there must be a point $E$ which is (another) vertex of the equilateral triangle with one side as $AD$. Let the distance of $AB$ be equal to $d$. Then, by several applications of the pythagorean theorem, we have that either the length of $CE=\frac {\sqrt 7} 2 d$ or $BE=\frac{\sqrt 7} 2 d$. Either way, this shows that the distance between points is unbounded, which is a contradiction with our hypothesis. Hence $S$ may not have more than one point.

If the sets are unbounded, it need not be all of $\mathbb R ^2$. For instance, any half-plane should do.

Here's a nice simple proof for why a continuous function is not possible. I've tried to be more laconic, and just give the idea of why it will not work, as I believe is in line with your preference (from the comments). Please let me know if you would like more details.

enter image description here

For any point (in the picture I place the point at the origin, but it's no matter) we cannot have any point of our graph on the lines emanating from this point at a $\pi/6$ angle with the horizontal, for if we did it would quickly follow that the only way to create an equilateral triangle would be to place a point above/below one of these points, and thus $f$ would not be a function. By the intermediate value theorem, this means that we cannot place any points in the shaded area, for in order to create an equilateral triangle with the point at the origin and a point in the shaded area we would need to place a point outside the shaded area, and at some point $f$ would have to cross the line. Now place a point anywhere in the non-shaded area. Immediately, from our discussion earlier, we know that $f$ cannot cross the dotted lines emanating from this point either, and furthermore the point needed to create an equilateral triangle will lie in between these lines, which we know (by the intermediate value theorem) is not a point which $f$ can reach. Hence, such a continuous function is impossible.

Also, my proof at the beginning, that a bounded set $S$ which is convex is not possible, does not really deal with showing that $S$ cannot be between a circle and a triangle, it is much simpler than that. Read it again, and note that all it shows is that, given two points $A, B$ with distance $d$, we are guaranteed that there are points $C, D$ which have distance $\frac {\sqrt 7} 2 d$. The reason this shows that the set $S$ cannot be bounded is that if it were bounded by some ball with diameter $h$, we are guaranteed that there exist points in $S$ with a distance greater than $h$. We simply pick any two points $A, B$ with distance $d$, and then repeat the process above $k$ times, where $\left(\frac {\sqrt 7} 2\right) ^k d>h$. Therefore $S$ cannot be bounded. Note that nowhere do we assume anything about bounding $S$ by the smallest ball possible, or choosing points yielding the maximum distance because (in particular, if $S$ were open) such points may not exist.

  • $\begingroup$ Assuming what you wrote is correct... I still want to know if there is a $cts$ function f:R-->R with the equilateral triangle properties (I think this is equivalent to asking the same question for f:[0,infinity)-->R ) $\endgroup$ – Adam Rubinson May 11 '12 at 21:41
  • $\begingroup$ @AdamRubinson Do you mean that the image of $f$ is a set $S$ which has these properties? $\endgroup$ – process91 May 12 '12 at 1:45
  • $\begingroup$ @AdamRubinson It doesn't make sense to ask for a continuous function $f:\mathbb R \to \mathbb R$ such that the image set satisfies these properties because the image of $f$ will be one dimensional. What I think you mean is to ask for a function $f:\mathbb R \to \mathbb R ^2$. A simple answer to your question is that a Peano curve can be made with domain $\mathbb R$, whose range is all of $\mathbb R ^2$. Such a function would be continuous (read the linked page on Peano curves to see why)... $\endgroup$ – process91 May 12 '12 at 1:55
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    $\begingroup$ Yes. That is actually an extremely nice proof in my opinion. I'll read the second half tomorrow. Also, are there any moderators on this site? I had a quick glance at all users yesterday and it was not obvious that there were... $\endgroup$ – Adam Rubinson May 13 '12 at 2:32
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    $\begingroup$ @AdamRubinson: Michael and Dejan are the respective owners of their answers, even though this is your question. I think it's inappropriate to combine their answers without asking both users first, as well as unnecessarily redundant. There are four mods (Qiaochu, Mariano, Willy, Zev) and a couple more on the way, but IIRC moderators do not have the power to split bounties, and the only way to award a second bounty is to increase the amount on it. (Related.) $\endgroup$ – anon May 13 '12 at 15:02

Edit: The bounty relates to this post. Basically, fill in all the details. I'm doing this because I am interested in the problem, however I don't have time atm to investigate.

There is no convex subset of R^2 other than:

  1. (The empty set)
  2. One point

which is bounded and has the property that, given any two points in the set, the 3rd point is in the set. This can be proven as such:

  1. Suppose there is such a convex bounded set in R^2 (i.e. we can stick a closed circle (i.e. closed ball) round it)
  2. Stick the smallest possible closed circle/closed ball round it.
  3. Take the two points in the set which have max distance.
  4. If there is a 3rd point in the set then the set must be a convex shaded region between an equilateral triangle and the circle.
  5. An equilateral triangle fails (think about the perpendicular bisector).
  6. Circle fails (diameter)
  7. Anything in between fails (the perpendicular from the equilateral triangle is actually outside the circle by construction (I think)).

Contradiction. Shaded concave sets... part of the bounty.

Now to unbounded sets in R^2...

We can have an unbounded set that works. E.g. take a small straight line and "extend it" by taking two points on the line and filling up the equilateral triangles. I don't know what types of sets we will get but they will be crazy and whether or not they fill up R^2 I don't know...

Still I can't work out if a cts function R-->R exists. Or if such a function R-->R exists at all. if we fill up the equilateral grid and then shift it slightly is that a function? Bounty is in response to this question

  • $\begingroup$ Step 3 is wrong: If the set is itself a (filled) eqilateral triangle, then two maximally distant points in it do make an equilateral triangle with a third point. (A vertex and the mid-point of the opposite side don't, so your conclusion may be true -- but you haven't proved it here.). $\endgroup$ – TonyK May 10 '12 at 16:50
  • $\begingroup$ Yes. Well I can show it for convex sets of R^2. Concave sets I haven't actually shown yet. Let's change the bounty question... $\endgroup$ – Adam Rubinson May 10 '12 at 17:15
  • $\begingroup$ basically, if anyone can fill in all the details and make some concrete proofs or even better some examples of functions, then they get 100 points. I'm going back to revision. Good luck $\endgroup$ – Adam Rubinson May 10 '12 at 17:32
  • $\begingroup$ Are you using a definition of convexity that admits isolated points, such as those which are the vertices of an equilateral triangle, as given by your 3rd example? If so, then there are unbounded sets which do not take up all of $\mathbb R ^2$. Consider that, given any two points, we can find exactly two other points which are vertices of an equilateral triangle with the first two. Start with two points, say $(-1,0)$ and $(1,0)$, and repeat this process recursively. The infinite set of points will be unbounded, will satisfy the other requirements, but will not contain the origin. $\endgroup$ – process91 May 10 '12 at 19:04
  • $\begingroup$ It does not, however, satisfy the typical definition of convexity, and neither does your third example, I believe. $\endgroup$ – process91 May 10 '12 at 19:08

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