Pentagon in a Pentagon

Let the side of the blue pentagon be $1$ and that of the large pentagon be $a$ . Then we have, through similiar isoceles triangles

$\dfrac{a-1}{1}=\dfrac{a}{a-1}$

which yields $a=\dfrac{1}{2}(3+\sqrt{5})=\phi ^2$ . Thus the ratio of the areas is $1:\phi^4$

Triangles $\Delta ABC$ and $\Delta BCD$ are isosceles. This means $AC=BC=BD=a-1$ . Furthermore, because angles $\angle ACB=\angle ADC$ *, isoceles triangles $\Delta ABC$ and $\Delta ADC$ are similiar, so that the equation to solve for $a$ then follows.
*Note: In a regular polygon, the angle between any two consecutive diagonals are the same. Hence, $\angle ACB =\angle BCD=\angle CDB$ so that $\angle ABC = \angle BAC = \angle ACB + \angle BCD$ , thus establishing that triangle $\Delta ACD$ is isosceles.

Only to get the correct answer, it might be easier to estimate..

Since n is an integer, make a list of $\phi$ 's powers:

$\phi \approx 1.618 \\ \phi^2 \approx 2.618 \\ \phi^3 \approx 4.236 \\ \phi^4 \approx 6.854 \\ \phi^5 \approx 11.090 \\ \vdots$

so the answer may be $3$ or $\boxed{4}$

To begin with, it is known that the area of a reguliar polygon is proportional to the square of th lenght of its side. Be $A$ the area of the bigger pentagon and $A'$ of the smaller one, and respectively $a$ and $a'$ the lenght of their sides.

$A=\alpha.a^2$ and $A'=\alpha.a^2$ **

Meaning : $\phi^n=\frac{A}{A'}=(\frac{a}{a'})^2$

From here, we can use Thales' theorem to write :

$\frac{a}{a'}=\frac{B}{b}$ (see the figure for B and b)

Now, let's find B and b expressions.

Using Pythagore's theorem in red and green triangles, and knowing that a pentagon's internal angle is 108° (purple filled arc on figure), we get :

$b=\frac{a}{2.cos(36)}$ (108/3 = 36, orange arc on figure)

$B=2.a.sin(54)$ (108/2 = 54, blue arc on figure)

Hence : $\phi^n=(4.sin(54).sin(36))^2$

$n.log(\phi)=log((4.sin(54).sin(36))^2)$

Which is equal to 4.

**out of curiosity, I've checked on wikipedia the value of the ratio : $\alpha \approx 1,72$

We know the interior angle of pentagon is $108^\circ$

Thus, $\angle CBA$ = $72^\circ$ and $\angle BCD$ = $36\circ$

So, the side ratio $\frac{CD}{AB}$ = $\frac{CD/2}{AB/2}$ = $\frac{cos36}{cos72}$ = $\frac{1}{2}$ (3+ $\sqrt {5}$ ) = $\phi ^2$ ; you may check the rational form of cos $36^\circ$ and cos $72^\circ$ to get the value

And area ratio is $\phi ^4$

The small pentagon is also regular. Its side length is

  x=a/4(cos 36°)^2  (using sine law)
 (a= side length of large pentagon)

The area of the large pentagon is

                na^2/4(cot π/n)

The area of the small penta gon is

               nx^2/4(cot π/n)

Therefore the ratio is (a/x)^2

        =[4(cos36°)^2]^2


        =[4×{(√5+1)/4}^2]^2


        =[(√5+1)/2]^4


      Answer=<<<<<4>>>>>😃

In reference to Michael Mendrin's drawing, we have that angle at C is 108deg (=540/5) and angle at BDC is 36deg (because the triangle with top at C and horizontal opposite side is isocele). This means that ACB is 36deg (=108-2 36). From that we can infer - Length of segment from C to mid[A,B] = sin(BDC)=sin36 with unit side of large pentagon - Length of segment CB = (sin36)/cos(ACB/2)=cos18 - Length of blue pentagon side (segment AB) = 2 (sin36/cos18) sin(ACB/2) = 2 (sin36/cos18) sin18 = 4 (sin18)^2 = 0.3820 (by sin(2*18) formula) By testing, 0.3820 = 1/phi^2. Thus, ratio of large/small pentagon side is phi^2. The ratio of area is the square of it, i.e. phi^4.

A solution based on guesswork: Call the side of the small pentagon 1. This is the base of a small isosceles triangle. I'm guessing the other two sides are length $\phi$ . One of these sides is the base of a bigger isosceles triangle whose other sides are therefore (another guess) $\phi$ squared. One of these 'other sides' is the length of the big pentagon. So if the proportion of the lengths is $\phi$ squared, the proportion of the area is $\phi$ to the power of 4.