Given a semicircle with an inscribed square of side S.

Let a be the length of the diameter on each side of the square. Find S / a.

First, I made a Geometer's Sketchpad script tool for the desired construction. If you need it, it's located inside the Assignment 5 script tools, titled InscribeFromSideRAM.gsp. You will define two points, A and B, which will form the bottom side of the square. It then makes the square. Then it forms a circle from the center of the bottom side, the midpoint of the circle, with radius equal to the distance to a far corner. By extending the bottom side into a line, we have the base for the diameter. The two intersections of the circle with the bottom side extended form the bottom edges of the semicircle. Finally, an arc is drawn from one bottom edge, around the two top corners of the square, to the other bottom edge. The construction is shown below, with S the side length.

Because the heights of the two upward sides are the same, they're the same distance from the center of the semicircle. This means that along the bottom of the semicircle, the distance from A to the center must be the same as the distance from B to the center, \frac{ S }{ 2 } units. We then draw a radius from the center to the intersection of the square and semicircle, forming a right triangle.

By the Pythagorean Theorem, we know the radius r of the semicircle satisfies the following:

r^2 = S^2 + \left( \frac{ S }{ 2 } \right)^2

We can also read the radius along the bottom of the semicircle, where we have a and half of S. This means r = a + \frac{ S }{ 2 } . Since we have two formulas for r, let's equate the two formulas for r^{ 2 } and then perform some algebra.

r^2 = S^2 + \left( \frac{ S }{ 2 } \right)^2 = \left( a + \frac{ S }{ 2 } \right)^2

S^2 + \left( \frac{ S }{ 2 } \right)^2 = a^2 + 2 ~ \left( \frac{ S }{ 2 } \right) ~ a + \left( \frac{ S }{ 2 } \right)^2

S^2 = a^2 + S ~ a

S^2 - a ~ S - a^2 = 0

We have a quadratic form, so we can solve for S via the quadratic formula.

S = \frac{ -(-a) \pm \sqrt{a^2 - 4 (1) (-a^2) } } {2} = \frac{a \pm \sqrt{5 a^2} }{2} = a ~ \left( \frac{ 1 \pm \sqrt{ 5 } }{ 2 } \right)

We need to be careful, because there are two possible roots. If we choose the negative root, \frac{ 1 - \sqrt{ 5 } }{ 2 } is negative. Distances can't be negative, so that's wrong. We want the positive root. Dividing by a gives us the desired ratio result.

\frac{S}{a} = \frac{ 1 + \sqrt{ 5 } }{ 2 }

If you continue on to Part 3 of the final, you'll see a story about rabbits. How might they be connected? (Cue the foreshadowing music.)