Three Equal Circles in a Semicircle

If the small circles each have a radius of 1, what is the radius of the semi-circle?

Papa showed me this really cool problem courtesy of STEPMaths. At the time I was doing a really frustrating chemistry paper and by the time I got to starting to work on it Papa showed me the solution... So I am going to present it here to (hopefully) learn it.

SOLUTION:

Join up the midpoint of two of the circles as well as the midpoint of semi-circle. Extend the hypotenuse of the right-angled triangle formed to extend to touch another point on the semi-circle.

By definition, this extended line is the radius of the semi-circle. Let this be R
Therefore the hypotenuse of the triangle is (R - 1)
Using Pythagoras' Theorem on the right-angled triangle we can show that:

(R-1)² = 1² + 2²

(R-1)² = 5

R - 1 = ±√5

R = 1 ±√5

As R ⟩ 0, R = 1 + √5

Now let's pretend we have no given values and look at the relationship between the radius of the semi-circle and that of the small circles.

We will keep the radius of the semi-circle as R and let the radius of the smaller circles be r

(R-r)² = r² + (2r)²

(R-r)² = 5r²

R - r = ± r√5

R = r ± r√5

As R ⟩ 0, R = r + r√5 = r(1 + √5)

This is interesting because we can see a link between the radius of the semi-circle and the Golden Ratio. The radius of the semi-circle is equal to radius of the small circle times twice the Golden Ratio.

In other words: the ratio of the radius of the semi-circle to the diameter of the small circles is the golden ratio.

I need to do some more research on the Golden Ratio because right now all I know is what Emily tried discussing with some very confused year sevens.