Two semicircles in a regular pentagon

Labelling the vertices and drawing a couple of radii to see the size of the semicircles:

The angles and the ratios between side lengths and diagonal lengths of a regular pentagon are well known, so:

angle ABC is $\pi/5$ and angle ADE is $2\pi/5$.

Obviously, angle FGB and angle HJD are

$\pi/2$.

If we say the side-length of the pentagon is $2$, then the lengths HD and FB are respectively

$1$ and $\phi=\frac{1+\sqrt{5}}{2}.$

So the radii of the pink and blue semicircles are respectively

length FG equals $\phi\sin(\pi/5)$ and length HJ equals $\sin(2\pi/5)=2\sin(\pi/5)\cos(\pi/5)$ (using double angle formulae).

Taking their ratio, we find

$$\frac{HJ}{FG}=\frac{2\sin(\pi/5)\cos(\pi/5)}{\phi\sin(\pi/5)}=\frac{\cos(\pi/5)}{\left(\frac{1+\sqrt{5}}{4}\right)}=1$$

so the answer is

they are both the same size!

I think the picture below speaks for itself, and is a very simple solution.

Originally (see below) I used an algebraic approach. Here I now consider something more geometric.

If a semicircle is tangent to $n-1$ sides of an $n$-sided polygon gland its diameter lies on the remaining side, the radius is twice the area of the polygon divided by the sum of the lengthsbof the $n-1$ tangent sides.

Let $s$ measure a side of the regular pentagon and $d$ measure a diagonal. Thus the blue semicircle is inscribed in a triangle with area

$\frac12sd\sin(72°)$

and the tangent sides are two diagonals, thus

$r_b=\frac12s\sin(72°)$

For the pink semicircle, the triangle has area

$\frac12s^2\sin(108°)$

and two tangent sides that are sides of the pentagon, so in this case

$r_p=\frac12s\sin(108°)$

But 72°and 108°are supplementary, so have the same sine, therefore the two semicircles have the same size.

Initial answer

This is fairly calculation-heavy also, but entirely algebraic, using the fact that the diagonal/side ratio is $\phi=(1+\sqrt5)/2$.

Assume the unit is half the side of the pentagon (putting the one-half here avoids lots of fractions later on).

The blue semicircle is inscribed in an isosceles triangle with legs measuring $2\phi$ and base measuring $2$. We bisect the apex angle and identify the radius $r_b$ as the height to the hypotenuse of either resulting right triangle. We then have

$\text{(Altitude)}^{-2}=\text{(Leg 1)}^{-2}+\text{(Leg 2)}^{-2}$

where the hypotenuse times its altitude must equal the product of the legs and we put that into the Pythagorean Theorem.

Then

$r_b^{-2}=1+\dfrac{1}{4\phi^2-1}=\dfrac{4+4\phi}{3+4\phi},$

where we plug in $\phi^2=1+\phi$.

For the pink semicircle the circumscribed isosceles triangle has legs measuring $2$ and base measuring $2\phi$, and calculations similar to the one above give

$r_p^{-2}=\dfrac1{\phi^2}+\dfrac1{4-\phi^2}=\dfrac4{(1+\phi)(3-\phi)}=\dfrac4{2+\phi}$

To compare the two radii we rationalize the denominators in the above expression using the conjugate multiplication

$(a+b\phi)[(a+b)-b\phi]=a^2+ab-b^2$

Thus

$r_b^{-2}=\dfrac{4+4\phi}{3+4\phi}=\dfrac{(4+4\phi)(7-4\phi)}5=\color{blue}{\dfrac{12-4\phi}5}$

$r_p^{-2}=\dfrac4{2+\phi}=\dfrac{4(3-\phi)}5=\color{blue}{\dfrac{12-4\phi}5}$

The two circle radii are the same.

I'd like to propose the following.

Let's draw a regular semi-decagon around the pink semi-circle.

Let's take the side of the decagon to be 1, ED is 1.

Then remember that the proportions of the sides of triangle 36-36-108 is ϕ. Since EE' is 1, AE is 1/ϕ and since DG is 1, DB is ϕ.

We have ϕ = 1 + 1/ϕ. That means BD = DE + EA = DA. D is exactly midway between A and B.

Since DG is parallel to AF, G is exactly midway between B and F.

Since BCF is isoceles, GC cuts BF at a right angle. GC is vertical, so is G'C'.

So you can translate the blue circle up with its defining lines. C becomes G, C' becomes G', CF overlaps GD. The result is that the construction of the pink circle is the construction of the blue circle translated.

This means both semicircles have the same size.

PS: You can make it easier if you reframe the question like this. Which circle is larger?