I came across a weird property of the function $f(x)=\dfrac{1}{(1-x)}$

Observe the following:

$$f(x) = \frac{1}{(1-x)}, \quad\quad f^2(x) = f(f(x)) = \frac{(x-1)}{x}, \quad\quad f^3(x) = f(f(f(x))) = x$$ ultimately implying that $f^2(x)=f^{-1}(x)$.

(Mini question: Do you know of any other functions $g(x)$ where $g \circ (g \circ g(x)) = g^3(x)=x$ aside from $f(x)$ and aside from the trivial case where $g(x)=x$? I was pretty shocked when I noticed this pattern with $f(x)$.)

Anyway, notice for every $x$, there is a set of triplets generated by repeatedly applying the function $f(x)$.

Specifically $\langle x\rangle =\{x,f(x),f^{-1}(x)\}=\{x,\frac{1}{(1-x)},\frac{(x-1)}{x}\}$

For an illustrative example let $x=2$, so then $\langle 2\rangle=\{2, -1, \frac{1}{2}\}$. See now that this can be thought of as 3 points on the graph of the function $f(x)$, where

Point $A$: $x \mapsto f(x)$

Point $B$: $f(x) \mapsto f^{2}(x)=f^{-1}(x)$

Point $C$: $f^{-1}(x) \mapsto x$

Explicitly, still using $x=2$ as the example:

Point $A$: $(x, f(x)) = (2,-1)$

Point $B$: $(f(x), f^{-1}(x)) = (-1,\frac{1}{2})$

Point $C$: $(f^{-1}(x),x) = (\frac{1}{2},2)$

OK so now my question!

Since 3 points uniquely define a circle, I'd like to know if we can derive a closed-form function $r(x)$ that calculates the radius of circle $R$, where circle $R$ is the circle uniquely defined by the 3 points $A$, $B$ and $C$ generated by $\langle x\rangle$.

Continuing the example where $x=2$, circle $R$ has center at Point $R=(\frac{3}{4},\frac{1}{4})$ (i.e. the circumcenter of points $A$, $B$ and $C$). The radius of circle $R$ is then simply:

$$|\overline{AR}|=\sqrt{{\left(2-\frac{3}{4}\right)}^2+{\left(-1-\frac{1}{4}\right)}^2}= \frac{5\sqrt{2}}{4}.$$

So evaluating $r(x)$ at $x=2$ gives us $r(2)=\dfrac{5\sqrt{2}}{4}\approx1.76777$.

Another cool example to consider is $x=\phi$, where $\phi=\dfrac{1+\sqrt{5}}{2}\approx1.61803$ (the Golden Ratio). Some cool characteristics that make $\phi$ unique among all numbers are:

$$\phi-1=\frac{1}{\phi}\quad\text{and}\quad \phi+1=\phi^2$$

You can calculate this on your own, but applying $f(x)$ on $x=\phi$ repeatedly results in $\langle\phi\rangle=\{\phi,-\phi,\frac{1}{\phi^{2}}\}$.

With the help of Wolfram Alpha, I was able to calculate $r(\phi)\approx1.93649$

(Circumcenter: https://tinyurl.com/y59trfn5 | Radius: https://tinyurl.com/y6jxs9sn)

Calculating the circumcenter seems to be the biggest issue, but maybe there's a cleaner way with the help of linear algebra? I was reading that there's a way to calculate the formula of a circle using matrices and determinants, but that seemed too complex for this. Maybe circles and triangles aren't the way to approach this at all -- I'd be happy to take suggestions and hear your thoughts!

Just some last conceptual thoughts...

1) $r(x)$ should always be positive (i.e. there is no $x$ where $r(x)$ is $0$ or negative), and therefore somewhere hit some positive minimum value for $r(x)$ (assuming/implying that $r(x)$ is smooth and differentiable on the interval $x \in (-\infty,1)\cup(1,+\infty)$).

2) $\lim\limits_{x \to 1^-}r(x)=+\infty$ and $\lim\limits_{x \to 1^+}r(x)=+\infty$

3) $\lim\limits_{x \to -\infty}r(x)=+\infty$ and $\lim\limits_{x \to +\infty}r(x)=+\infty$

4) $r(x)$ is NOT symmetric around $x=1$. Just as a quick check, $r(3)\approx2.12459$ and $r(-1)\approx1.76777$

5) $r(x)$ is actual VERY NOISY as a function since for any 1 value of $r(x)$, there are at least 3 unique variables that result in that value (i.e. all $x \in \langle x\rangle$)(e.g. $r(2)=r(-1)=r(\frac{1}{2})\approx1.76777$)

That last point makes me feel there's no true closed-form function for $r(x)$. Regardless, I'd be really curious to find out what the minimum radius is... (placing \$1 on $r(x)$ for $x \in \langle\frac{\pi^2}{4}\rangle$!)

  • $\begingroup$ "if we can derive a closed-form function r(x) that calculates the radius of circle R": of course. You even did it for your example points. $\endgroup$ – Yves Daoust May 8 at 11:58
  • $\begingroup$ About the "weird property" of function $f(x):=\dfrac{1}{1-x}$, see the group property I have explained in my answer here : math.stackexchange.com/q/2265961 $\endgroup$ – Jean Marie May 9 at 20:43

The equation of a circle through $A$, $B$, $C$ is given by: $$\left|\begin{array}{cccc} x^2+y^2 & x & y & 1 \\ A_x^2+A_y^2 & A_x & A_y & 1 \\ B_x^2+B_y^2 & B_x & B_y & 1 \\ C_x^2+C_y^2 & C_x & C_y & 1 \end{array}\right| = 0 \tag{1}$$ For $$A = \left(p,\frac{1}{1-p}\right)\qquad B = \left(\frac{1}{1-p},\frac{p-1}{p}\right) \qquad C = \left(\frac{p-1}{p},p\right)$$ this becomes (with the help of a computer algebra system) $$\left(x^2+y^2\right)p(p-1) - x\left(1- 3 p + p^3\right) -y\left(1 - 2 p - p^2 + p^3\right) = -1 - p + 4 p^2 - p^3 \tag{2}$$ Completing the square yields $$\left(x -\frac{p^3-3p+1}{2p(p-1)} \right)^2+\left(y - \frac{p^3-p^2-2p+1}{2p(p-1)} \right)^2 = \frac{\left(p^2 + 1\right) \left(p^2-2p+2\right) \left(2p^2-2p+1\right)}{4 p^2(p-1)^2} \tag{$\star$}$$

Thus, the circumcenter $K$ and radius $r$ are

$$\begin{align} K &= \left(\frac{p^3-3p+1}{2p(p-1)}, \frac{p^3-p^2-2p+1}{2p(p-1)}\right)\\[6pt] r &= \frac{\sqrt{\left(p^2+1\right) \left(\,(p-1)^2+1\,\right) \left(\,p^2+(p-1)^2\,\right)\;}}{2 |p(p-1)|} \end{align}$$

Interestingly, we can rewrite $r$ as $$r = \frac12 \;\sqrt{\left(\frac{p}{1} + \frac{1}{p}\right) \left(\frac{p - 1}{1} + \frac{1}{p - 1}\right) \left(\frac{p}{p - 1} + \frac{p - 1}{p}\right)}$$

Animation courtesy of @Jyrki:

enter image description here

By way of minimization ... We have $$\frac{d(r^2)}{dp} = \frac{(p-2) (p+1) (2p-1) (p^2-p+1)^2}{2 p^3(p-1)^3}$$ so that the non-extraneous critical points occur at $p=2,-1,\frac12$, which correspond exactly to OP's $\langle 2\rangle$ triangle, as can be seen in the animation.

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    $\begingroup$ Good job! As a way of confirmation of your solution I took the liberty of adding a hopefully self-explanatory animation. Feel free to remove it, if you don't like it. In the animation $p$ goes from $-4.98$ to $-0.02$ in increments of $0.04$. $\endgroup$ – Jyrki Lahtonen May 8 at 13:50
  • $\begingroup$ @JyrkiLahtonen: Thanks! :) $\endgroup$ – Blue May 8 at 13:51
  • $\begingroup$ This is so beautiful thank you for this T___T Can I DM you about the software and how you were able to model the very smooth animation? so satisfying... And for the mini-question: did I really just stumble upon a random function that has this property or do you think there are (infinitely) many others that can be constructed? $\endgroup$ – Andrew May 8 at 13:54
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    $\begingroup$ @Andrew About the miniquestion: Look up fractional linear transformations. Your $f$ corresponds to the matrix $$A=\pmatrix{0&1\cr -1&1\cr},$$ and you can easily verify that $A^3$ is the identity matrix. You can get, for example, an order five fractional linear transformation by using a matrix such that its fifth power is the identity. The rotation matrix $$R=\pmatrix{\cos(2\pi/5)&\sin(2\pi/5)\cr-\sin(2\pi/5)&\cos(2\pi/5)\cr}$$ springs to mind. $\endgroup$ – Jyrki Lahtonen May 8 at 13:58
  • $\begingroup$ I feel so full now; thank you all so much for your amazing insight. @JyrkiLahtonen do you mind sharing a graph of $r(p)$ and $r'(p)$ as well? $\endgroup$ – Andrew May 8 at 14:04

Given a point $x\in\Bbb{R}$, with $x\notin\{0,1\}$, the corresponding three points are $$P_1(x)=\left(x,\tfrac{1}{1-x}\right), \qquad P_2(x)=\left(\tfrac{1}{1-x},\tfrac{x-1}{x}\right), \qquad P_3(x)=\left(\tfrac{x-1}{x},x\right).$$ Then we want to find the unique circle passing through these three points. Note that these three points are not collinear for any choice of $x\in\Bbb{R}$ with $x\notin\{0,1\}$, so such a circle does indeed exist. Now we want to find $a,b,r\in\Bbb{R}$ such that $$||P_i(x)-(a,b)||^2=r^2,$$ for $i\in\{1,2,3\}$. This yields the following three equations: \begin{eqnarray*} \left(x-a\right)^2+\left(\tfrac{1}{1-x}-b\right)^2&=&r^2,\\ \left(\tfrac{1}{1-x}-a\right)^2+\left(\tfrac{x-1}{x}-b\right)^2&=&r^2,\\ \left(\tfrac{x-1}{x}-a\right)^2+\left(x-b\right)^2&=&r^2. \end{eqnarray*} Clearing all denominators then yields the following system of equations: \begin{eqnarray*} (1-x)^2(x-a)^2+(1-(1-x)b)^2&=&(1-x)^2r^2,\\ x^2(1-(1-x)a)^2+(1-x)^2((x-1)-xb)^2&=&x^2(1-x)^2r^2,\\ ((x-1)-xa)^2+x^2(x-b)^2&=&x^2r^2. \end{eqnarray*} We can rearrange these to make them look more like polynomials in $a$, $b$ and $r$: \begin{eqnarray*} (1-x)^2&a^2-2x(1-x)^2&a+(1-x)^2&b^2-2(1-x)&b+x^2(1-x)^2+1&=(1-x)^2&r^2,\\ x^2(1-x)^2&a^2-2x^2(1-x)&a+x^2(1-x)^2&b^2+2x(1-x)^3&b+(1-x)^4+x^2&=x^2(1-x)^2&r^2,\\ x^2&a^2+2x(1-x)&a+x^2&b^2-2x^3&b+x^4+(1-x)^2&=x^2&r^2. \end{eqnarray*} Multiplying the first equation by $x^2$ and subtracting it from the second shows that $$-2x^2(1-x)(x^2-x+1)a+2x(1-x)(x^2-x+1)b-(x-1)^2(x^4-x^2+2x-1)=0.$$ Note that this equation is linear in both $a$ and $b$. In the same way we can multiply the third equation by $(1-x)^2$ and subtract it from the second to find that $$-2x(1-x)(x^2-x+1)a-2x(1-x)^2(x^2-x+1)b-x^2(x^4-2x^3+x^2-1)=0,$$ which is again linear in $a$ and $b$. Now we have a system of two linear equations in two variables, which we can solve by linear algebra. Dividing by $x\neq0$ and $1-x\neq0$ yields the system: \begin{eqnarray*} -2x^2(x^2-x+1)&a&+2x(x^2-x+1)&b&=(x-1)(x^4-x^2+2x-1),\\ -2(1-x)(x^2-x+1)&a&-2(1-x)^2(x^2-x+1)&b&=x(x^4-2x^3+x^2-1). \end{eqnarray*} Because also $x^2-x+1\neq0$ for all real numbers $x$, we can also write this as \begin{eqnarray*} -x&a&+&b&=\frac{x-1}{x}\frac{x^4-x^2+2x-1}{2(x^2-x+1)},\\ -&a&-(1-x)&b&=\frac{x}{1-x}\frac{x^4-2x^3+x^2-1}{2(x^2-x+1)}. \end{eqnarray*} It follows that \begin{eqnarray*} a&=&\frac{x^5-x^4-2x^3+4x^2-4x+1}{2x(x-1)(x^2-x+1)}=\frac{x^3-3x+1}{2x(x-1)},\\ b&=&\frac{x^5-2x^4+2x^2-3x-1}{2x(x-1)(x^2-x+1)}=\frac{x^3-x^2-2x+1}{2x(x-1)}, \end{eqnarray*} and plugging this back into the original equations shows that $$r^2=\frac{2x^6-6x^5+11x^4-12x^3+11x^2-6x+2}{4x^2(x-1)^2} =\frac{(x^2+1)(x^2-2x+2)(2x^2-2x+1)}{(2x(x-1))^2}.$$ It is interesting to note that the square of the radius can also be expressed as \begin{eqnarray*} r(x)^2&=&\frac14\left|f^{0}(x)+\frac{1}{f^{0}(x)}\right| \left|f^{1}(x)+\frac{1}{f^{1}(x)}\right| \left|f^{2}(x)+\frac{1}{f^{2}(x)}\right|,\\ &=&\frac{1}{4|f^{0}(x)f^{1}(x)f^{2}(x)|} \left(f^0(x)^2+1\right) \left(f^1(x)^2+1\right) \left(f^2(x)^2+1\right)\\ &=&\frac14\prod_{i=0}^2||(f^i(x),1)||^2, \end{eqnarray*} and hence we see that $$r(x)=\frac12\prod_{i=0}^2\left|\left|(f^i(x),1)\right|\right|.$$

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    $\begingroup$ so awesome; I think we're a few numbers off here and there but I'll try to work through your calculations on my lunch break! Seems very promising and I like the approach $\endgroup$ – Andrew May 8 at 12:46
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    $\begingroup$ I redid the calculations for $a(x)$ and $b(x)$. Note your only mistake was a sign error on $a(x)$, which then messed up the whole polynomial for $b(x)$. They should be as follows: $a(x)=\dfrac{x^5-x^4-2x^3+4x^2-4x+1}{2x(x-1)(x^2-x+1)}$ and then you use that to derive $b(x)=\dfrac{x^6-3x^5+2x^4+2x^3-5x^2+4x-1}{2x(x-1)^2(x^2-x+1)}$. This checks out for all the examples I provided above (e.g. $<2>, <3>, <\phi>$). Now going to try to calculate $r(x)$ from here... will report back soon. $\endgroup$ – Andrew May 8 at 13:25
  • $\begingroup$ @Andrew: Your corrections for $a(x)$ and $b(x)$ factor, leaving the expressions given in my answer. $\endgroup$ – Blue May 8 at 13:31
  • $\begingroup$ Thank you for the corrections; I have updated my answer, and included a corrected expression for $r^2$. These now all agree with the more elegant answer given by @Blue. $\endgroup$ – Servaes May 8 at 13:36
  • $\begingroup$ @Andrew With the corrected expression for $r(x)$, I have also added a simpler closed form for $r(x)$ that has a more geometrical intepretation. $\endgroup$ – Servaes May 8 at 13:55


Let us derive the formula for the circumradius of three points.

WLOG, one of the points is the origin (otherwise, translate all three points). The equation of the circle must be of the form


When we plug the coordinates of the two other points, we get a system of two equations in two unknowns giving the center. Finally, as the radius is the distance of the center to the origin,

$$r=\frac{\sqrt{\begin{vmatrix}x_1^2+y_1^2&2y_1\\x_2^2+y_2^2&2y_2\end{vmatrix}^2+\begin{vmatrix}2x_1&x_1^2+y_1^2\\2x_2&x_2^2+y_2^2\end{vmatrix}^2}} {|\begin{vmatrix}2x_1&2y_1\\2x_2&2y_2\end{vmatrix}|}.$$

Now you can plug the coordinates of the three points, not forgetting the translation.

  • $\begingroup$ I tried this for the example of $<2>$ which gives the points A(2,-1), B(-1,0.5), and C(0.5,2). Assuming A(2,-1) becomes the origin WLOG, the new points are A(0,0), B(-3,1.5), and C(-1.5,3). According to your formula for r (calculating the determinants), I got an answer of $r=3\sqrt{2}\approx4,24264\neq1.76777$ which was calculated above. Maybe I'm misunderstanding but let me know what you think. $\endgroup$ – Andrew May 8 at 12:34
  • $\begingroup$ @Andrew: ouch, I dropped the factor $2$ ! $\endgroup$ – Yves Daoust May 8 at 12:36

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