How is the golden ratio related to these problems?

Question

I have created two problems which both have the golden ration integrated into the answer, and I would like an intuitive understanding as to why and how the golden ratio is linked to them.

Problem 1:

$\sin(108^\circ)/\sin(36^\circ) = \phi$

I would like to know a geometric understanding as to why this is (not that "sin of 108 equals whatever and sin of 36 equals whatever").

Problem 2:

Draw a square with corners a (top left),b (top right),c (bottom left),d (bottom right) with sides length x.

Now draw a line of length "l" starting from the a to the midpoint between the b and d. This line will have a length of $x\sqrt{5}/2$ through the Pythagorean theorem.

Then create a semicircle in line with the bottom (cd), with a corner on d, and then enlarge it such that it is just tangent with line l, like so:

This semicircle will have a radius of r, which is what we are solving for in terms of x.

To do this we look at the length l, from the tangent point to the midpoint of bd, which we will call $l_2$ (the other part of the line from a to the tangent point will be $l_1$).

Through the fact that tangents from the same point on a cirle are the same length, we see that $l_2$ is the same as $x/2$. Therefore $l_1$ is equal to $l-l_2$, which is $x((-1+\sqrt{5})/2)$ (or otherwise written as $x(\phi - 1)$)

To solve for r we first define the midpoint of the semicircle's diameter as the point R (capital). Then we create a triangle with corners a, c and R. The vertical side will be length x and the horizontal side will be length x - r, so the hypotenuse (aR) will be length $\sqrt{2x^2 -2xr + r^2}$.

Now construct a triangle with corners a, R and the tangent point of the semicircle with l. The hypotenuse of this triangle will be $\sqrt{(x(\phi-1))^2 + r^2}$

However, these belong to the same hypotenuse (aR) so we can set them equal to each other and get this system of equations:

$\sqrt{2x^2 -2xr + r^2}$ = $\sqrt{(x(\phi-1))^2 + r^2}$

You can then solve for r (I won't show the working for this) and manipulate your answer with some golden ratio identities ($\phi^2 = \phi + 1$, and $1/\phi = \phi - 1$), to get this;

$r = x\phi/2$

Is there a intuitive, visceral understanding for why the golden ratio pops up or is it just a coincidence (although I don't think it is).

Thanks for helping!

Answer 1

For the second problem, first consider the figure below:

In this figure, $FT$ is a diameter of a circle of diameter $1,$ the points $E,F,T$ are collinear, and the segment $ED$ has length $1$ and is tangent to the circle at $D.$

There is a theorem of geometry that says $$ EF \times ET = (ED)^2, \tag1$$ but since $ED=1,$ Equation $(1)$ implies that $$ EF = \frac1{ET},$$ but the collinear arrangement of $E,F,T$ then tells us that $ET = EF + 1$; therefore $$ ET = \frac1{ET} + 1. \tag2$$ There is only one positive solution of Equation $(2).$ It is $ET = \phi,$ where $\phi$ is the golden ratio. It follows that $EF = 1/\phi.$

Now consider the figure below:

This is your square (I wrote the names of the vertices with capital letters $A,B,C,D$ instead of lower-case letters) with a few names added to some points and with some additional construction added to it. Specifically, $M$ is the midpoint of $BD,$ $T$ is the tangent point of $AM$ with the circle about $R,$ the radius $RT$ is extended to the point $P$ on $AB,$ a circle is constructed on the diameter $BD$ with center at $M,$ another square is adjoined on the other side of $BD,$ the segment $AM$ is extended to vertex $E$ of the adjacent square, intersecting the circle around $M$ at $F,$ the tangent to that circle at $F$ intersects $DE$ at $G,$ and a circle is constructed with center $G$ and radius $FG.$

For simplicity, I have assumed for this figure that $x = 1$ and labeled the side of the square accordingly. Just multiply all lengths in the figure by $x$ in order to obtain the "real" lengths implied by your problem statement.

Examining tangents to circles we have $MD = MT = \frac12,$ and also $GD = GF,$ which implies that the circle around $M$ passes exactly through $T$ and also that the circles around $R$ and $F$ are mutually tangent at $D.$

Observe that a copy of the first figure is embedded in the second figure. This means that $ET = \phi$ and $EF = 1/\phi.$

Also observe that triangles $\triangle EFG,$ $\triangle EDM,$ and $\triangle ETR$ are similar. (Their sizes happen to be in the geometric progression $\frac1\phi : 1 : \phi,$ though that's just another incidental beautiful fact about this figure.) Also note that $DM = \frac12 DE$ and therefore $FG = \frac12 EF = \frac1{2\phi}$ and $TR = \frac12 ET = \frac12 \phi,$ which is what you wanted to show.

I included a number of unnecessary details in this proof just to show how this figure is teeming with golden ratios, much like the diagonals of a regular pentagon.

Answer 2

Just to expand on @LordSharktheUnknown's comment, let $ABCDE$ be a regular pentagon, with $AC$ respectively cutting $BE,\,BD$ at $F,\,G$. Drawing in all diagonals, the diagram is full of two kinds of isosceles triangle, up to similarity: those with apex $\frac{3\pi}{5}$ (e.g. with two sides as legs and a diagonal as base), and, and those with apex $\frac{\pi}{5}$ (e.g. $\triangle BFG$, with the apex at $B$). Define $\phi$ as the diagonal-to-side ratio, so by similar triangles $\frac{1}{\phi}=\frac{BF}{AB}=\frac{AF}{AB}=\phi-1$, and $\phi$ is clearly the golden ratio. By the sine rule in $\triangle ABC$, $\phi=\frac{\sin\frac{3\pi}{5}}{\sin\frac{\pi}{5}}$.