In the figure, construct an equilateral triangle inside regular pentagon. The line connecting orthocenter and the second Isodynamic point of green triangle divides the side of pentagon in red and blue segments.
My question. How to prove that $$\frac{\rm{\color{red}{red}}}{\rm{ \color{blue}{blue}}}=\frac{\sqrt{5}+1}{2}?$$
On
Here's a thoroughly-unsatisfying approach that makes absolutely no attempt to exploit the specific geometry of the situation. It's just an algebraic slog.
First, we define $\theta := 6^\circ$, $c:=\cos\theta$, $s := \sin\theta$, noting that the minimal polynomial for $s$ is $$\mu(s) = 1- 8s-16s^2+8s^3+16s^4 \tag{1}$$
Now, consider an origin-centered pentagon of circumradius $r$, with vertices $A$, $B$, $U$, $V$ as shown:
Point $C$ is the apex of the equilateral triangle erected upon the side opposite $A$. A little angle-chasing gives the various measures as multiples of $\theta$. Note that $$\frac{|OC|}{\sin\theta} = \frac{r}{\sin 5\theta} \quad\to\quad |OC| = 2 r \sin\theta = 2 r s \tag{2}$$ Abusing notation and writing $\operatorname{cis}\psi = (\cos\psi ,\sin\psi)$, we can coordinatize thusly: $$A = r \operatorname{cis}0 \qquad B = r \operatorname{cis} 12\theta \qquad C = (2r s,0)$$
$$U= r\operatorname{cis}(-24\theta) \qquad V = r\operatorname{cis}(-12\theta) \tag{3}$$
With the help of a computer algebra system, we expand the various trig functions, re-write as much as possible in terms of $s=\sin\theta$, and reduce modulo $\mu(s)$ to get $$\begin{align} A &= r (1, 0) \\[4pt] B &= r \left(\;s \left(3 - 4 s^2\right), c \left(1 - 4 s^2\right)\;\right) \\[4pt] C &= r ( 2 s, 0 ) \\[8pt] U &= r \left(\;\tfrac12\left(-1 - 6 s + 8 s^3\right), -c \left( 1 - 2 s\right) \left( 1 - 2 s - 4 s^2\right)\;\right) \\[4pt] V &= r\left(\; s \left( 3 - 4 s^2 \right), -c \left( 1 - 4 s^2\right)\;\right) \end{align}\tag{4}$$
Orthocenter $P$ has barycentric coordinates $$\alpha:\beta:\gamma \;=\;\sin A \cos B \cos C : \cos A \sin B \cos C : \cos A \cos B \sin C \tag{5}$$ so that, after the same kind of reduction as above, we get $$P = \frac{\alpha A + \beta B + \gamma C}{\alpha+\beta+\gamma} = \left(\; \frac{r ( 5 + 14 s - 4 s^2 - 16 s^3)}{ 2 (9 + 14 s - 8 s^2 - 16 s^3)}\;,\; \frac{ c r (1-2s)^2 (1+2s)}{9 + 14 s - 8 s^2 - 16 s^3}\;\right) \tag{6}$$
Second-Isodynamic Point $Q$ has barycentric coordinates $$\alpha:\beta:\gamma = \sin A \sin(A-60^\circ) : \sin B \sin(B-60^\circ) : \sin C \sin(C-60^\circ) \tag{7}$$ so that $$Q = \left(\;-\frac{r ( 11 + 8 s - 12 s^2 - 8 s^3)}{ 2 (9 + 14 s - 8 s^2 - 16 s^3)}\;,\; -\frac{2 c r (8 + 11 s - 8 s^2 - 12 s^3)}{ 9 + 14 s - 8 s^2 - 16 s^3}\;\right) \tag{8}$$ And, noting that we can write the Golden Ratio as $\phi = 2 \cos 6\theta$, and taking $W$ as the point that separates $\overline{UV}$ in that ratio, we have $$W = \frac{U + \phi V}{1+\phi} = \left(\; -\frac{r s (3 - 4 s^2)}{2 (1 - s) (1 + 2 s)^2}\;,\; -\frac{c r (1 - 2 s)^2 (3 + 4 s)}{2 (1 - s) (1 + 2 s)^2} \;\right) \tag{9}$$
Now, we "simply" determine the equation of the line through $P$ and $Q$, which "reduces" to
$$\begin{align} 2 c x &( 171 + 254 s - 180 s^2 - 264 s^3) \\ - \phantom{2x}y &( 163 + 254 s - 164 s^2 - 272 s^3) \\ = 2 c r &( 47 + 68 s - 48 s^2 - 72 s^3) \end{align}\tag{10}$$
Finally, we substitute $W$ into this equation, reduce modulo $\mu(s)$, and find that everything vanishes, verifying the claim. $\square$
While this approach offers no hint as to what magic lurks in the result, knowing that the proposition is true assures us that the quest for a magic-revealing proof is not inherently doomed.
Addendum. As one might expect, a few sign changes in the above analysis would work themselves out to validate a related result. Specifically, if we take $C$ to be apex of an equilateral triangle erected on the outside of the pentagon, and we take $Q$ to be the first isodynamic point of $\triangle ABC$, then the line through $Q$ and orthocenter $P$ meets $\overline{UV}$ in the same $W$, dividing the segment in the golden ratio.
I'll forego rewriting all the equations here, as I'm still seeking an appropriately-magical approach to understanding this construction. However, I'll give the following unifying notes:
$$C = r \left(\; -2 \sin(9\theta \mp 10\theta), 0\;\right) = r\left(\;-2\sin(9\theta\mp 60^\circ), 0\;\right)$$ $$\angle B = 12\theta \mp 5\theta = 12\theta\mp 30^\circ \qquad \angle C = 9\theta \pm 5 \theta = 9\theta\pm 30^\circ$$ $$Q \;\stackrel{\text{bary}}{=}\; \sin A \sin(A\mp 60^\circ):\cdots$$ where, in each "$\pm$" and "$\mp$", the top sign corresponds to OP's construction, and the bottom corresponds to mine. Seeing that the angles vary by multiples of $60^\circ$ is nicely compatible with the change in the isodynamic point's coordinates.
Addendum 2. What works for pentagons often works for pentagrams, and that's the case here ... mostly. Curiously, the Second Isodynamic Point appears in both constructions; the First Isodynamic Point may also lead to another golden ratio somewhere, but I haven't looked very hard for it.
The figure shows a composite of two constructions with an equilateral triangle (apexes $C$ and $C'$) erected upon the edge of a pentagram (ie, a diagonal of the pentagon). Points $P$ (not shown) and $P'$ are the respective orthocenters, and $Q$ and $Q'$ the respective Second Isodynamic Points. Lines $\overleftrightarrow{PQ}$ and $\overleftrightarrow{P'Q'}$ concur with edge $\overleftrightarrow{UV}$ at $W$, such that $|VW|/|UW|=|UW|/|UV|=\phi$.
Interestingly, $\overleftrightarrow{UV}$ bisects $\angle QWQ'$, and $\overline{QW}\cong\overline{Q'W}$. This seems like it should be significant, but I don't know how.
For comparison, here's a composite image of the two pentagon constructions, showing the corresponding possibly-significant relations among $W$, $Q$, and $Q'$.
There's something going on here, but I just can't see it (yet).
Here's a much better argument than my previous attempt. I'm leaving out a few details due to time constraints, but will update later.
Let pentagon $ABCDE$ have side-length $s$, and let $F$ be the apex of an equilateral triangle erected internally upon $\overline{CD}$. The orthocenter of $\triangle ABF$ is the intersection of $\overline{BE}$ with the perpendicular $\overline{FF'}$ from $F$ to $\overline{AB}$. Let $X$ be the point that separates $\overline{DE}$ in the golden ratio: $|DX|/|XE|=\phi$. Here's the key insight (proof left to the reader for now):
We establish the result by showing the collinearity of $P$, $Q$, $X$ via Menelaus' Theorem applied to $\triangle BEM$, where $M$ is the midpoint of $\overline{DE}$. A little angle chasing gives the various angle measures; we also use the facts $|BE|=s\phi$ and $\phi-1=\phi^{-1}$. Calculating the appropriate signed ratios ...
$$\begin{align} \frac{|EX|}{|XM|}&=\phantom{-}\frac{s/(1+\phi)}{s/2-s/(1+\phi)}=\frac{2}{\phi-1} = 2\phi \\[4pt] \frac{|MQ|}{|QB|}&= -\frac{s\sin 60^\circ}{|EB|\sin\angle BEQ/\sin\angle BQE}= -\frac{s\sin 60^\circ}{s\phi\sin 132^\circ/\sin 30^\circ}= -\frac{s\sin 60^\circ}{2\phi\sin 48^\circ} \\[8pt] \frac{|BP|}{|PE|} &= \phantom{-}\frac{|BF|\sin 48^\circ/\sin\angle FPB}{|EF|\sin 120^\circ/\sin\angle FPE} = \frac{\sin 48^\circ}{\sin 60^\circ} \end{align}$$ ... we see that their product is $-1$. Thus, Menelaus holds, and collinearity is verified. $\square$
Note: When $F$ is the apex of the triangle erected externally on $\overline{CD}$, then the First Isodynamic Point of $\triangle ABF$ is the apex of the triangle erected internally on $\overline{DE}$, and a similar argument holds to give the companion construction I described in my previous answer. The pentagram constructions follow similarly.