geometry
I am trying to write a code to draw pentagon star and have stumbled upon a problem. In the attached diagram, what is the ratio of the diameter of outer (red) circle to the inner (blue) circle. The circles are concentric. I've read that it is ~2.61 but would like to develop the relationship myself.
(Converting comment to answer, as requested.)
If the side of each sub-square has length $2$, then the radius of the circle is $\sqrt{10} = \sqrt{2}\cdot\sqrt{5}$. Once you have $\sqrt{5}$ floating around, it's not too difficult to find $\phi$.
Edit. It seems you could use a geometry lesson to help put your "discoveries" into some perspective.
Let us first note this basic property of parallel lines:
Theorem. Three (or more) parallel lines cut any transversal into segments with a fixed proportion (such as $p:q$ in the figure).
And now, let us consider your circle-and-grid diagram, in which one readily shows that $a/b = \phi = 1.618\dots$. We'll call this the figure's primary instance of the golden ratio.
By the nature of $\phi$, an immediate (and thoroughly-unsurprising) corollary is that $b/c =\phi$, as well. We can write $$a:b:c \quad=\quad \phi : 1 : \frac{1}{\phi}$$
Adding-in some lines through the dividing-point of $b$ and $c$, parallel to the sides of the square, we can exploit the parallelism Theorem by effectively transferring the $a:b:c$ proportion to the edges of the square:
By symmetry, we get proportions all around the square:
It's important to note, though, that these additional instances of the golden ratio are unremarkable; each is merely an echo of the primary instance. Indeed, thanks to the parallelism Theorem, we can effortlessly construct countless more instances of the golden ratio; here are just a few (dozen):
Above, because a particular set of horizontal lines cut the sides of the square in the proportion $\phi:1$, those lines cut all those segments in the same way. We can get even more instances of $\phi$ by connecting other pairs of points, midpoints of those points, or whatever ... and let's not forget about the rotational and reflective symmetries in the figure. OMG! Golden ratios everywhere! And yet, as before, these instances of $\phi$ are unsurprising and unremarkable, either individually or collectively, because they're obvious follow-ons of obvious follow-ons of the original primary instance. That's not a symphony you hear; it's the accumulation of echoes of one single note.
You seem to be spending a lot of time and effort "discovering" various instances of $\phi$ by manually plotting points and measuring ratios in GeoGebra. Please do yourself (and the rest of us) a favor: pay attention to some of the dead-simple geometric principles at work here. You'll do far less measuring, and you'll get far fewer false-positives (and -negatives); most importantly, you'll learn to distinguish utter trivialities from whatever legitimately-interesting construction you might happen to come across, vastly improving the value of your questions here.
Oh ... As to whether your recent "green square", "red segment", and "orange circles" additions are "utter trivialities" or "legitimately-interesting constructions" ... well ... I believe the first two go directly into the "trivialities" box.
I'm not sure how you think the orange circles relate to each other. If you're suggesting that the smaller circle (with $2/3$ the diameter of the larger circle) is tangent to both the square and the bigger circle ... then we're going to need a third box: "bogus claims". The double-tangency requires that $$\frac{\text{smaller diameter}}{\text{larger diameter}} = \frac{1+2\phi^{-1}}{2\phi} = \frac{1}{2}(3 - \phi ) = 0.6909\dots \neq \frac{2}{3}$$
(I honestly don't understand how a Ph.D in physics lets such easily-caught computational errors slip by, but I guess that's just a failure of imagination on my part.)
Anyway ... I hope this discussion helps you in your journey. It's the last contribution I intend to make.
As shown in the diagram above, let $A$, $B$ and $C$ be the centers of the $3$ circles. Draw the lines $AC$ and $BC$ to join these centers. Since the circles are tangent to each other, these lines go through the points where they are tangent, so each length is the sum of the circle radii, i.e.,
$$|AC| = r_1 + r_3 \tag{1}\label{eq1B}$$
$$|BC| = r_2 + r_3 \tag{2}\label{eq2B}$$
Assign the point coordinates to be $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$. With \eqref{eq1B} and \eqref{eq2B}, using these coordinates in the squares of the lengths of $|AC|$ and $|BC|$ gives
$$\begin{equation}\begin{aligned} (x_1 - x_3)^2 + (y_1 - y_3)^2 & = (r_1 + r_3)^2 \\ x_1^2 - 2x_1x_3 + x_3^2 + y_1^2 - 2y_1y_3 + y_3^2 & = r_1^2 + 2r_1r_3 + r_3^2 \end{aligned}\end{equation}\tag{3}\label{eq3B}$$
$$\begin{equation}\begin{aligned} (x_2 - x_3)^2 + (y_2 - y_3)^2 & = (r_2 + r_3)^2 \\ x_2^2 - 2x_2x_3 + x_3^2 + y_2^2 - 2y_2y_3 + y_3^2 & = r_2^2 + 2r_2r_3 + r_3^2 \end{aligned}\end{equation}\tag{4}\label{eq4B}$$
Next, \eqref{eq3B} minus \eqref{eq4B} gives
$$\begin{equation}\begin{aligned} & x_1^2 - x_2^2 + 2(x_2 - x_1)x_3 + y_1^2 - y_2^2 + 2(y_2 - y_1)y_3 = r_1^2 - r_2^2 + 2(r_1 - r_2)r_3 \\ & 2(x_2 - x_1)x_3 + 2(y_2 - y_1)y_3 = r_1^2 - r_2^2 + 2(r_1 - r_2)r_3 - x_1^2 + x_2^2 - y_1^2 + y_2^2 \\ & c_1x_3 + c_2y_3 = c_3 \end{aligned}\end{equation}\tag{5}\label{eq5B}$$
where, to make the algebra easier to deal with, I have set
$$c_1 = 2(x_2 - x_1) \tag{6}\label{eq6B}$$
$$c_2 = 2(y_2 - y_1) \tag{7}\label{eq7B}$$
$$c_3 = r_1^2 - r_2^2 + 2(r_1 - r_2)r_3 - x_1^2 + x_2^2 - y_1^2 + y_2^2 \tag{8}\label{eq8B}$$
In \eqref{eq3B}, move the $x_1^2 + y_1^2$ terms to the right to get
$$-2x_1x_3 + x_3^2 - 2y_1y_3 + y_3^2 = c_4 \tag{9}\label{eq9B}$$
where
$$c_4 = r_1^2 + 2r_1r_3 + r_3^2 - x_1^2 - x_2^2 \tag{10}\label{eq10B}$$
Assuming $x_2 \neq x_1$, so $c_1 \neq 0$ in \eqref{eq6B}, you can get from \eqref{eq5B} that
$$x_3 = \frac{c_3 - c_2y_3}{c_1} \tag{11}\label{eq11B}$$
Substituting this into \eqref{eq9B} gives
$$\begin{equation}\begin{aligned} & -2x_1\left(\frac{c_3 - c_2y_3}{c_1}\right) + \left(\frac{c_3 - c_2y_3}{c_1}\right)^2 - 2y_1y_3 + y_3^2 = c_4 \\ & -\frac{2x_1c_3}{c_1} + \left(\frac{2x_1c_2}{c_1}\right)y_3 + \frac{c_3^2}{c_1^2} - \left(\frac{2c_2c_3}{c_1^2}\right)y_3 + \left(\frac{c_2^2}{c_1^2}\right)y_3^2 - 2y_1y_3 + y_3^2 = c_4 \\ & \left(\frac{c_2^2}{c_1^2} + 1\right)y_3^2 + \left(\frac{2x_1c_2}{c_1} - \frac{2c_2c_3}{c_1^2} - 2y_1\right)y_3 - \frac{2x_1c_3}{c_1} + \frac{c_3^2}{c_1^2} - c_4 = 0 \\ & c_5y_3^2 + c_6y_3 + c_7 = 0 \end{aligned}\end{equation}\tag{12}\label{eq12B}$$
where, once again, to keep the algebra somewhat simpler, I've set
$$c_5 = \frac{c_2^2}{c_1^2} + 1 \tag{13}\label{eq13B}$$
$$c_6 = \frac{2x_1c_2}{c_1} - \frac{2c_2c_3}{c_1^2} - 2y_1 \tag{14}\label{eq14B}$$
$$c_7 = -\frac{2x_1c_3}{c_1} + \frac{c_3^2}{c_1^2} - c_4 \tag{15}\label{eq15B}$$
Using the quadratic formula in \eqref{eq12B} gives
$$y_3 = \frac{-c_6 \pm \sqrt{c_6^2 - 4c_5c_7}}{2c_5} \tag{16}\label{eq16B}$$
There are $2$ roots because circle $C$ could be located either below the tangent point between $A$ and $B$ or above it. Assuming you want the one below, choose the smaller root in \eqref{eq16B}. Finally, you can substitute that value into \eqref{eq11B} to get $x_3$. You will then have the coordinates of the center of the blue circle.
Best Answer
@Robert Israel : Here is an alternate proof, involving only triangle trigonometry.
With notations displayed on the figure below, we have to find the ratio $\dfrac{OA}{OB}.$
by reasoning in right triangle $OCA \ $: $ \ \widehat{OAB}=90°-72°=18°.$
by reasoning in right triangle $OCB \ $: $ \ \widehat{OBC}=90°-72°/2=54°$, thus $\widehat{ABO}=180°-54°=126°.$
Thus, sine law in triangle $OAB$ gives:
$$\tag{1}\dfrac{OA}{OB}=\dfrac{\sin \widehat{ABO}}{\sin \widehat{OAB}}=\dfrac{\sin 126°}{\sin 18°}=\dfrac{\cos 36°}{\cos 72°}$$ (exact value); thus $\dfrac{OA}{OB}\approx 2.618.$
Knowing that $\cos 36°=\dfrac{1+\sqrt{5}}{4}$, it is not difficult to conclude that the exact value can also be written$1+\Phi$ (where $\Phi:=\dfrac{1+\sqrt{5}}{2}$ is the golden ratio).