A Pentagon Problem.

Let the side of the small pentagon be $1$ and the side of the large pentagon be $x$ . The similarity ratio will then be $x$ .

Since $BD' = AB = x$ and $E'D' = 1$ , $BE' = x - 1$ . By symmetry, $AE' = AD' = BE' = x - 1$ .

Since $\triangle AD'E' \sim \triangle BAD'$ , $\frac{AD'}{E'D'} = \frac{BA}{AD'}$ or $\frac{x - 1}{1} = \frac{x}{x - 1}$ . This rearranges to $x^2 - 3x + 1 = 0$ , which has a solution of $x = \frac{3 + \sqrt{5}}{2} \approx \boxed{2.618}$ for $x > 1$ .

Bonus: The answer is also $x = \phi^2$ , where $\phi = \frac{1 + \sqrt{5}}{2}$ , the golden ratio.

Let $|DE|=a$ and $|A'D|=b$ then $|A'E|=|B'E|=b$

$[A'E]$ is bisector of $∠DEB'$ hence $b \cdot b=a \cdot x⇒x=\dfrac{b^{2}}{a}$

If we apply Law of Cosines on △ $A'B'E$

$\left( \dfrac{b^{2}}{a}\right) ^{2}=2 \cdot b^{2}- 2 \cdot b^{2} \cdot cos36°$ , $b^{2}$ s simplifies : $\frac{b^{2}}{a^{2}}=2-2cos36°$

We want to find similarity ratio that is equal to $\dfrac{a}{x}$ and we know $x=\dfrac{b^{2}}{a}$ ,so $\dfrac{a}{x}=\dfrac{a}{\frac{b^{2}}{a}}=\dfrac{a^{2}}{b^{2}}$

We got : $\dfrac{b^{2}}{a^{2}}=2-2cos36°⇒\dfrac{a^{2}}{b^{2}}=\dfrac{1}{2-2cos36°}\approx \boxed{2.618}$

Let small pentagon side be =1, use bisecting theorem twice to get larger pentagon side =a by following two triangle bisecting Eqs:

x(2x+1)=a(1+x) and a=x^2

a=(1/4)(1+5^0.5)^2

Answer is a= 2.61803

$AB = x$

$E'D' = y$

$\angle C'E'D' = 108 \degree$ (Angles in a regular pentagon.)

$\angle BE'A = 108 \degree$ (Opposite angles.)

$\angle AE'D' = (360 - 2*108)/2 = 72 \degree$ (Angles in a circle sum to 360.)

$\angle E'AD' = 180 - 2*72 = 36 \degree$ (Angles in an isosceles triangle.)

Using the law of sines:

$\frac{x}{\sin108\degree} = \frac{AE'}{\sin36\degree}$

$AE' = \frac{x}{\sin108\degree} * \sin36\degree$

$\frac{AE'}{\sin72\degree} = \frac{y}{\sin36\degree}$

$y = \frac{AE'}{\sin72\degree} * \sin36\degree$

$\text{Ratio} = \frac{x}{y}$

$= \frac{x}{\frac{\frac{x}{\sin108\degree} * \sin36\degree}{\sin72\degree} * \sin36\degree}$

$= \frac{x}{\frac{\frac{x}{0.9511} * 0.5878}{0.9511} * 0.5878}$

$= \frac{x}{\frac{0.3633x}{0.951}}$

$= \frac{x}{0.382x}$

$= 1/0.382 \approx 2.618$

Let $AE' = 1$ . Then by symmetry, $BE' = BC' = AE' = 1. \newline\angle BAE' = \angle ABE' = \frac{180\degree - 108\degree}{2} = 36\degree \newline AB = AE'cos(36\degree) + BE'cos(36\degree) = 2cos(36\degree) = 2sin(54\degree)\quad\cdots \text{Eq. 1}\newline \angle C'BE' = 108\degree - 36\degree - 36\degree = 36\degree\newline E'C' = 2BE'sin(18\degree) = 2sin(18\degree)\quad\cdots\text{Eq. 2} \newline \large\frac{AB}{E'C'} = \frac{2sin(54\degree)}{2sin(18\degree)} = \frac{3sin(18\degree) - 4sin^3(18\degree)}{sin(18\degree)} = \small 3 - 4sin^2(18\degree) = 3 - 4\big(\frac{\sqrt{5} - 1}{4}\big)^2 = \frac{3+\sqrt{5}}{2} = 2.618$