Euler's theorem in Geometry is a great result that connects the distance between the circumcentre and incentre of a triangle with the circumradius and inradius of the triangle. A similar result, known as Fuss's theorem, holds for bi-centric quadrilaterals. It was surprising when I found a similar relation holds for three mutually tangent circular arcs.
Consider the three blue circles $A$, $B$ and $C$ (with radii $r_A$, $r_B$ and $r_C$ respectively) that are mutually tangent to each other. Let the points of tangency be $T_A$, $T_B$ and $T_C$ which becomes the vertices of our 'triangle of circular arcs'.
We can now draw two circles: a circumcirle (red) that passes through all the tangency points and an incircle (green) that is mutually internally tangent to all the blue circles. Let's position the configuration in the Argand plane such that center of the incircle is at the origin.
Let the centers of the blue circles be $z_A$, $z_B$ and $z_C$, and the center of the circumcircle be $z$.
Now if we let $R$ and $r$ be the circumradius and inradius of our triangle, then
$$\text{Eqn (1): }d^2=R^2-4Rr+r^2$$
where $d=\vert z \vert$ is the distance between the centers of the circumcircle and incircle.
To prove this, Consider $\triangle z_A z_B z_C$ whose side lengths are $r_A+r_B$, $r_B+r_C$ and $r_C+r_A$. It is easy to show that,
$\text{Eqn (2): }R^2=\displaystyle\frac{r_Ar_Br_C}{r_A+r_B+r_C}$ and $\text{Eqn (3): }r^{-1}={r_A}^{-1}+{r_B}^{-1}+{r_C}^{-1}+2R^{-1}$
$(2)$ is obtained by noting that the red circle is the incircle of $\triangle z_A z_B z_C$ and $(3)$ is a direct application of Descartes' theorem.
Now given the vertices of a complex triangle, we can express the incenter of the triangle using its vertices (using this). Now we have,
$\displaystyle z=\sum_{\text{cyc}}\frac{r_B+r_C}{2(r_A+r_B+r_C)}z_A \implies \text{Eqn (4): }2(r_A+r_B+r_C)z=\sum_{\text{cyc}}(r_B+r_C)z_A$
Multiplying $(4)$ with its complex conjugate, we have,
$\text{Eqn (5): }4(r_A+r_B+r_C)^2\vert z \vert^2=\displaystyle\sum_{\text{cyc}}(r_B+r_C)^2\vert z_A \vert^2+\sum_{\text{cyc}}(r_B+r_C)(r_C+r_A)(z_A\bar{z_B}+\bar{z_A}z_B)$
From the figure, we see that $\vert z_B-z_A \vert=r_A+r_B$. Using the fact that the product of a complex number and its conjugate is the square of its magnitude, we can see that,
$\text{Eqn (6): }z_A\bar{z_B}+\bar{z_A}z_B=\vert z_A \vert^2+\vert z_B \vert^2-\vert z_B-z_A\vert^2=\vert z_A \vert^2+\vert z_B \vert^2-(r_A+r_B)^2$
Using $(6)$ in $(5)$, and cancelling the common terms, we get,
$\text{Eqn (7): }2(r_A+r_B+r_C)\vert z \vert^2=\displaystyle\sum_{\text{cyc}}\vert z_A \vert^2(r_B+r_c)-\prod_{\text{cyc}}(r_A+r_B)$
From the figure above, we have $\vert z_x \vert=r+r_x$ for $x \in \{A,B,C\}$. Using this we get,
$\begin{align}
\displaystyle\text{Eqn (8): }\sum_{\text{cyc}}\vert z_A \vert^2(r_B+r_c)&=\sum_{\text{cyc}}(r^2+2rr_A+r_A^2)(r_B+r_C) \\
&= 2r^2(r_A+r_B+r_C)+4r(r_Ar_B+r_Br_C+r_Cr_A)+\sum_{\text{cyc}}r_A^2(r_B+r_C) \\
\end{align}$
Using $(8)$ in $(7)$, we get,
$\text{Eqn (9): }\displaystyle2(r_A+r_B+r_C)\vert z \vert^2=2r^2(r_A+r_B+r_C)+4r(r_Ar_B+r_Br_C+r_Cr_A)+\sum_{\text{cyc}}r_A^2(r_B+r_C)-\prod_{\text{cyc}}(r_A+r_B)$
Now the second term of $(9)$ can be simplified as
$\begin{align}
\text{Eqn (10): }r(r_Ar_B+r_Br_C+r_Cr_A)&=rr_Ar_Br_C(r_A^{-1}+r_B^{-1}+r_C^{-1})\\
&=rR^2(r_A+r_B+r_C)(r^{-1}-2R^{-1})\\
&=R(R-2r)(r_A+r_B+r_C)
\end{align}$
where we have used $(2)$ and $(3)$ in the second equality.
The last two terms of $(9)$ can be expanded and most of the terms gets cancelled giving,
$\displaystyle\text{Eqn (11): }\sum_{\text{cyc}}r_A^2(r_B+r_C)-\prod_{\text{cyc}}(r_A+r_B)=-2r_Ar_Br_C=-2R^2(r_A+r_B+r_C)$
where we have used $(2)$ in the last equality.
Using $(10)$ and $(11)$ in $(9)$ and dividing both sides by $2(r_A+r_B+r_C)$ , we finally have $(1)$.
From $(1)$, it follows that, $R\geq (2+\sqrt{3})r$ which holds with equality only when $r_A=r_B=r_C$.
This can probably be proved with geometry but I couldn't get my head around it.
Finally, before we finish this discussion, we also note that we can completely parametrize $(1)$ for non-co-prime integer solutions. For integer $a,b$ such that $\gcd(a,b)=1$, $3 \nmid b$ and $b \leq \sqrt{3}a$
(i) If $a\not\equiv b \pmod{2}$, then $R=3a^2+4ab+b^2$, $r=2ab$ and $d=3a^2-b^2$
(ii) If $a\equiv b \pmod{2}$, then $R=(3a^2+4ab+b^2)/2$, $r=ab$ and $d=(3a^2-b^2)/2$
Hope you found this interesting. See ya later.
References:
Tangencies: Three tangent Circles
Efficiently constructing tangent circles
Inversion and applications to Ptolemy and Euler
Until then
Yours Aye
Me
Consider the three blue circles $A$, $B$ and $C$ (with radii $r_A$, $r_B$ and $r_C$ respectively) that are mutually tangent to each other. Let the points of tangency be $T_A$, $T_B$ and $T_C$ which becomes the vertices of our 'triangle of circular arcs'.
We can now draw two circles: a circumcirle (red) that passes through all the tangency points and an incircle (green) that is mutually internally tangent to all the blue circles. Let's position the configuration in the Argand plane such that center of the incircle is at the origin.
Let the centers of the blue circles be $z_A$, $z_B$ and $z_C$, and the center of the circumcircle be $z$.
Now if we let $R$ and $r$ be the circumradius and inradius of our triangle, then
$$\text{Eqn (1): }d^2=R^2-4Rr+r^2$$
where $d=\vert z \vert$ is the distance between the centers of the circumcircle and incircle.
To prove this, Consider $\triangle z_A z_B z_C$ whose side lengths are $r_A+r_B$, $r_B+r_C$ and $r_C+r_A$. It is easy to show that,
$\text{Eqn (2): }R^2=\displaystyle\frac{r_Ar_Br_C}{r_A+r_B+r_C}$ and $\text{Eqn (3): }r^{-1}={r_A}^{-1}+{r_B}^{-1}+{r_C}^{-1}+2R^{-1}$
$(2)$ is obtained by noting that the red circle is the incircle of $\triangle z_A z_B z_C$ and $(3)$ is a direct application of Descartes' theorem.
Now given the vertices of a complex triangle, we can express the incenter of the triangle using its vertices (using this). Now we have,
$\displaystyle z=\sum_{\text{cyc}}\frac{r_B+r_C}{2(r_A+r_B+r_C)}z_A \implies \text{Eqn (4): }2(r_A+r_B+r_C)z=\sum_{\text{cyc}}(r_B+r_C)z_A$
Multiplying $(4)$ with its complex conjugate, we have,
$\text{Eqn (5): }4(r_A+r_B+r_C)^2\vert z \vert^2=\displaystyle\sum_{\text{cyc}}(r_B+r_C)^2\vert z_A \vert^2+\sum_{\text{cyc}}(r_B+r_C)(r_C+r_A)(z_A\bar{z_B}+\bar{z_A}z_B)$
From the figure, we see that $\vert z_B-z_A \vert=r_A+r_B$. Using the fact that the product of a complex number and its conjugate is the square of its magnitude, we can see that,
$\text{Eqn (6): }z_A\bar{z_B}+\bar{z_A}z_B=\vert z_A \vert^2+\vert z_B \vert^2-\vert z_B-z_A\vert^2=\vert z_A \vert^2+\vert z_B \vert^2-(r_A+r_B)^2$
Using $(6)$ in $(5)$, and cancelling the common terms, we get,
$\text{Eqn (7): }2(r_A+r_B+r_C)\vert z \vert^2=\displaystyle\sum_{\text{cyc}}\vert z_A \vert^2(r_B+r_c)-\prod_{\text{cyc}}(r_A+r_B)$
From the figure above, we have $\vert z_x \vert=r+r_x$ for $x \in \{A,B,C\}$. Using this we get,
$\begin{align}
\displaystyle\text{Eqn (8): }\sum_{\text{cyc}}\vert z_A \vert^2(r_B+r_c)&=\sum_{\text{cyc}}(r^2+2rr_A+r_A^2)(r_B+r_C) \\
&= 2r^2(r_A+r_B+r_C)+4r(r_Ar_B+r_Br_C+r_Cr_A)+\sum_{\text{cyc}}r_A^2(r_B+r_C) \\
\end{align}$
Using $(8)$ in $(7)$, we get,
$\text{Eqn (9): }\displaystyle2(r_A+r_B+r_C)\vert z \vert^2=2r^2(r_A+r_B+r_C)+4r(r_Ar_B+r_Br_C+r_Cr_A)+\sum_{\text{cyc}}r_A^2(r_B+r_C)-\prod_{\text{cyc}}(r_A+r_B)$
Now the second term of $(9)$ can be simplified as
$\begin{align}
\text{Eqn (10): }r(r_Ar_B+r_Br_C+r_Cr_A)&=rr_Ar_Br_C(r_A^{-1}+r_B^{-1}+r_C^{-1})\\
&=rR^2(r_A+r_B+r_C)(r^{-1}-2R^{-1})\\
&=R(R-2r)(r_A+r_B+r_C)
\end{align}$
where we have used $(2)$ and $(3)$ in the second equality.
The last two terms of $(9)$ can be expanded and most of the terms gets cancelled giving,
$\displaystyle\text{Eqn (11): }\sum_{\text{cyc}}r_A^2(r_B+r_C)-\prod_{\text{cyc}}(r_A+r_B)=-2r_Ar_Br_C=-2R^2(r_A+r_B+r_C)$
where we have used $(2)$ in the last equality.
Using $(10)$ and $(11)$ in $(9)$ and dividing both sides by $2(r_A+r_B+r_C)$ , we finally have $(1)$.
From $(1)$, it follows that, $R\geq (2+\sqrt{3})r$ which holds with equality only when $r_A=r_B=r_C$.
This can probably be proved with geometry but I couldn't get my head around it.
Finally, before we finish this discussion, we also note that we can completely parametrize $(1)$ for non-co-prime integer solutions. For integer $a,b$ such that $\gcd(a,b)=1$, $3 \nmid b$ and $b \leq \sqrt{3}a$
(i) If $a\not\equiv b \pmod{2}$, then $R=3a^2+4ab+b^2$, $r=2ab$ and $d=3a^2-b^2$
(ii) If $a\equiv b \pmod{2}$, then $R=(3a^2+4ab+b^2)/2$, $r=ab$ and $d=(3a^2-b^2)/2$
Hope you found this interesting. See ya later.
References:
Tangencies: Three tangent Circles
Efficiently constructing tangent circles
Inversion and applications to Ptolemy and Euler
Until then
Yours Aye
Me
