Consider the following shape which is produced by dividing the line between $0$ and $1$ on $x$ and $y$ axes into $n=16$ parts.

Question 1: What is the curve $f$ when $n\rightarrow \infty$?

enter image description here

Update: According to the answers this curve is not a part of a circle but with a very similar properties and behavior. In the other words this fact shows that how one can produce a "pseudo-cycle" with equation $x^{\frac{1}{2}}+y^{\frac{1}{2}}=1$ from some simple geometric objects (lines) by a limit construction.

Question 2: Is there a similar "limit construction by lines" like above drawing for producing a circle?

  • $\begingroup$ maybe something related to it : upload.wikimedia.org/wikipedia/commons/1/1a/Evolute2.gif $\endgroup$ – K. Rmth Nov 23 '13 at 23:30
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    $\begingroup$ @Jonathan: Amazingly it is not a circle! But something too similar! If you attach four arcs like above to each other you will find a pseudo-circle shape with a constant very near and similar to a circle $\pi'=\frac{10}{3}=3.3333...$! Find $f$ it is a very interesting entertainment! :-) $\endgroup$ – user108850 Nov 23 '13 at 23:49
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    $\begingroup$ Related: math.stackexchange.com/questions/551882/… $\endgroup$ – Blue Nov 24 '13 at 0:26
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    $\begingroup$ This is the envelop of the family of straight lines $\frac{x}{t} + \frac{y}{k-t} = 1$ commonly appear in string art. Look at example 2 in the wiki page of envelop. It has a derivation of this curve and show it is portion of a parabola. $\endgroup$ – achille hui Nov 24 '13 at 1:45
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    $\begingroup$ For reference, when you have a followup question, it should really be its own Question (and can link to this one). meta.math.stackexchange.com/questions/2304/… discusses this, and probably other Meta posts, too. $\endgroup$ – Mark S. Nov 24 '13 at 16:40

If we attach four curves like $f$ to each other in the following form a "pseudo-circle" shape appears. Note to its difference with a real circle. Its formula is $x^{\frac{1}{2}}+y^{\frac{1}{2}}=1$ a dual form of the circle equation $x^{2}+y^{2}=1$. You can find this equation simply by the geometric analysis of each line.

A very interesting point about this curve is that there is a kind of $\pi$ for it which doesn't change by radius! Here we have $\pi'=\frac{10}{3}=3.3333...$ which is very near to the $\pi$ of circle ($=3.1415...$) but $\pi'$ is a rational number not a non-algebraic real number like $\pi$!

enter image description here

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    $\begingroup$ $x^\frac{1}{2} + y^\frac{1}{2} = 1$ is not the equation for your pseudo-circle, it's the equation for your original curve. $\endgroup$ – Alex Kruckman Nov 27 '13 at 15:46

Each line connects $(a,0)$ with $(0,1-a)$, so has equation $y=(1-a)-\frac{1-a}ax$

At a given $x$, we can find the $a$ that results in the highest $y$ by differentiating: $\frac {dy}{da}=-1-x\frac{-a-(1-a)}{a^2}=-1+\frac x{a^2}$ which is zero when $a=\sqrt x$

Plugging this in, we get the curve $y=1-2\sqrt x +x$, plotted here by Alpha

This can be expressed as $\sqrt x + \sqrt y = 1,$ which is nicely symmetric.

  • $\begingroup$ Why did you need to find the $a$ that resulted in the highest $y$? $\endgroup$ – 1110101001 Nov 25 '13 at 3:17
  • $\begingroup$ @user2612743: we have lots of lines that pass above a given $(x,0)$ The point on the beauty curve is the one with the greatest $y$ at this $x$. $\endgroup$ – Ross Millikan Nov 25 '13 at 3:33
  • $\begingroup$ I got to remember this trick! $\endgroup$ – A. Chu Jan 3 '14 at 15:34

It is a parabola rotated through an eighth of a circle, satisfying $$(x+y) = \dfrac{(x-y)^2}{2} + \frac12$$ or $$x^2+y^2 -2xy-2x-2y+1=0$$ which has the solution in this part of the curve of $$y=(1-\sqrt{x})^2 \,\text{ i.e. }\, \sqrt{x}+\sqrt{y}=1 $$ though with other solutions in other parts of the same parabola extended. Its derivative is therefore $$\frac{dy}{dx}=-\frac{1-\sqrt{x}}{\sqrt{x}}$$ and so the tangent at $(x,(1-\sqrt{x})^2)$ crosses the $y$-axis at $(0, 1 - \sqrt{x}) $ and the $x$-axis at $(\sqrt{x},0)$: since $1 - \sqrt{x}+\sqrt{x}=1$, this satisfies the original construction. $\qquad\square$


Hint: find the general equation for the lines.

Then, find a formula for "the point $(x,y)$ lies above and to the right of a given line".

To find the general equation for the lines, we can use the description that we've divided the axes into equal pieces. If there are $n$ equal pieces, then there are $n-1$ points in those intervals $[0,1]$, located at distances $m/n$ from the origin, where $m$ is a positive integer less than $n$.

The line attached to $(m/n, 0)$ has $(0, (n-m)/n) = (0, 1 - m/n)$ as its other end point. (There are lots of ways to see this: I'll let you choose your own!)

To save writing, I'll let $a = m/n$, so the line segments connect $(a,0)$ to $(0, 1-a)$.

Now, we can any method for obtaining the equation of a line through $(a,0)$ and $(0, 1-a)$. We could use the two-point form. We could use slope intercept form. I'll demonstrate a method that doesn't rely on actually remembering something: the equation of a line always looks like

$$ c x + d y = e $$

for some $c$, $d$, and $e$. Since those two points have to lie on it, we have a system of equations:

$$ ca + d0 = e \qquad \qquad c0 + d(1-a) = e $$

or more simply

$$ ac = e \qquad \qquad d(1-a) = e $$

and we just need any solution that is not all zeroes. One such solution is

$$ c = \frac{1}{a} \qquad \qquad d = \frac{1}{1-a} \qquad \qquad e = 1 $$

Aside: we could have taken a different starting point: e.g. we could have said the equation of a non-vertical line always looks like $y = cx + d$, and taken the same approach to figure out what $c$ and $d$ are.

So, the equations for the line all look like

$$ \frac{x}{a} + \frac{y}{1-a} = 1 $$

One side of the line is given by the inequality

$$ \frac{x}{a} + \frac{y}{1-a} > 1 $$

and the other by

$$ \frac{x}{a} + \frac{y}{1-a} < 1 $$

We want the inequality that describes points being above/to the right of the line. While you could try to "reason it out", an easier approach is just to observe that we want the side containing $(1,1)$. An even easier approach is to observe we don't want the side containing $(0,0)$. If we plug in $(0,0)$ into the left hand side, we get zero. So we don't want the second one.

Thus, every point on and above your beauty curve satisfies the inequality

$$ \frac{x}{a} + \frac{y}{1-a} \geq 1 $$

for every rational value of $a \in (0,1)$. It really ought to satisfy the inequality for every $a \in (0,1)$ too.

If your curve behaves as nicely as it looks, this inequality should be an actual equality for exactly one value of $a$.

So which points $(x,y)$ satisfy this inequality for all values of $a$? Well, let's solve for which $a$'s it does satisfy!

First combine the fractions:

$$ \frac{x(1-a) + ya}{a(1-a)} \geq 1 $$

Now multiply through by the denominator. Don't forget to check the sign! It's positive, so that means the inequality stays in the same direction. So, we get

$$ x(1-a) + ya \geq a(1-a) $$

collecting the $a$'s together:

$$ a^2 + a(y-x-1) + x \geq 0 $$

(and we also have $0 < a < 1$). This is a quadratic function of $a$: for each particular value of $x,y$, it's graph is a parabola.

Since we want the points actually on the beauty curve, we desire that the actual equation

$$ a^2 + a(y-x-1) + x = 0 $$

have exactly one solution for $a$. Recall that the solutions are, by the quadratic formula,

$$ a = \frac{(1+x-y) \pm \sqrt{(1+x-y)^2 - 4x}}{2} $$

If there is only one solution, then that square root must actually be zero. That is, we need

$$ (1+x-y)^2 - 4x = 0 $$

or equivalently

$$ x^2 - 2xy + y^2 - 2x - 2y + 1 = 0 $$

If you study conic sections a lot, you'll notice the quadratic terms are

$$ x^2 - 2xy + y^2 = (x-y)^2$$

which signifies that your curve is actually part of a parabola. Here is a plot using Wolfram alpha

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    $\begingroup$ While there are some wasted efforts in this derivation, my intention was to write along as I worked through the problem using my hint. This isn't the only way my hint could be used; it's just the way that sprung to mind as I was working. $\endgroup$ – Hurkyl Nov 24 '13 at 14:39

It is a quadratic Bezier Curve with the control points (0,1) (0,0) (1,0).

It is not possible to create an exact circle using a quadratic Bezier curve, see this question.


It is most probably a superellipse, i.e., a geometric shape characterized by algebraic equations of the form $x^n+y^n=r^n$. Judging by the first picture to the right, I believe we might be dealing with the case $n=\frac12$ or $n=\frac23$, called astroid. Their two-dimensional counterparts are called superformulas, and their $3$D generalizations are known as superellipsoids, superquadrics and supereggs. Or perhaps it's some other form of hypocycloid. The bidimensional case $n=4$ is called squircle, by the way.

  • $\begingroup$ It's the case $n = \tfrac12$. $\endgroup$ – bubba Nov 25 '13 at 14:27

As Peter Kirkham said, it's a quadratic Bézier curve, which is always a parabola. The "string art" construction is de Casteljau's algorithm, which is the standard way to construct Bézier curves of any degree. There are some nice animations on this page

You can do the same thing with any three points, and you'll still get a parabola.

The general equation of the quadratic Bézier curve defined by three control points $\mathbf{P}_0$, $\mathbf{P}_a$, $\mathbf{P}_1$ is $$ \mathbf{P}(t) = (x(t), y(t)) = (1-t)^2\mathbf{P_0} + 2t(1-t)\mathbf{P}_a + t^2\mathbf{P}_1 $$ In our case (for the original curve), we have $\mathbf{P}_0 = (1,0)$, $\mathbf{P}_a = (0,0)$, $\mathbf{P}_1 = (0,1)$, so $$ x(t) = (1-t)^2 \quad ; \quad y(t) = t^2 $$ From these equations, it is immediately obvious that $$ \sqrt x + \sqrt y = 1 $$ Eliminating $t$ gives the equation in Hurkyl's and Henry's answers: $$ x^2 -2xy + y^2 -2x - 2y + 1 = 0 $$ The discriminant of this quadratic is zero, so it is indeed a parabola.

As Achille Hui's comment pointed out, this is just example 2 on the wikipedia page about envelopes.

If we put $t=\tfrac12$ in the parametric equation, we get $(x,y) = (\tfrac14, \tfrac14)$. This point is at a distance of around 1.06066 from the point $(1,1)$. So, the error between our parabola and the circle that it resembles is about 0.06066. Higher degree Bezier curves can produce better approximations of circular arcs, of course, using essentially the same "string art" construction of de Casteljau's.

Regarding question #2 (is there a similar construction that produces a circular arc). The answer is yes. You just have to use a rational quadratic Bezier curve, rather than a polynomial one. There is a similar form of the the de Casteljau algorithm for rational curves. Or, looking at it another way, you construct the parabolic string-art as outlined above, but in 3D, and then do a central projection onto the plane $z=1$ to get a new string-art construction that will produce a circle.

In more detail ... suppose we do the string-art construction in 3D, using the three points $$ \mathbf{Q}_0 = (1,0,1) \quad ; \quad \mathbf{Q}_a = \frac{1}{\sqrt{2}}(1,1,1) \quad ; \quad \mathbf{Q}_1 = (0,1,1) $$ This will give us the 3D curve $$ \mathbf{Q}(t) = \Big( 1 + (\sqrt{2} - 2)t + (1 - \sqrt{2})t^2, \sqrt{2}t + (1 - \sqrt{2})t^2, 1 + (\sqrt{2} - 2)t + (2 - \sqrt{2})t^2 \Big) $$ which is a parabola. The conical projection of this parabola onto the plane $z=1$ is $$ \mathbf{P}(t) = \left( \frac{1 + (\sqrt{2} - 2)t + (1 - \sqrt{2})t^2}{1 + (\sqrt{2} - 2)t + (2 - \sqrt{2})t^2}, \frac{\sqrt{2}t + (1 - \sqrt{2})t^2} {1 + (\sqrt{2} - 2)t + (2 - \sqrt{2})t^2} \right) $$ On this curve, you can check that $x^2+y^2=1$, so it's a portion of the unit circle.


An interesting question taken from Geometry by H. Jacobs goes as follows: If a line segment slides without slipping along the X-Y axis what is the curve generated by the bisector of the segment? The surprising answer is a circle whose radius is half the radius of the original segment. link: http://www.youtube.com/watch?v=_Jb0MNCmkY8


The OP's curve is (a portion of) the parabola with $(1/2,1/2)$ for its focus and $x+y=0$ for its directrix. With a slight modification (see below), the lines in the OP's drawing are tangent lines to the parabola, which can be thought of as (origami) crease lines created when the focus is "folded" to lie atop various points along the directrix.

The slight modification is this: The tangent lines for the parabola run from $(0,t)$ to $(1-t,0)$, but it appears from the drawing that the OP is connecting $(0,{k\over16})$ to $({17-k\over16},0)$ rather than $(1-{k\over16},0)$. In the limit, this doesn't matter, but it does make a small difference along the way.


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