## Verify Conway’s construction for taking cube roots

Verify Conway’s construction for taking cube roots using a compass and a straightedge with a unit distance marked on it.

Suppose we have constructed a distance $k$, with $0 < k < 1$, with endpoints $A$ and $F$. Construct the line perpendicular to $AF$ through $F$, and construct the point $B$ on this line whose distance from $F$ is $\sqrt{1-k^2}$. (Note that, by the Pythagorean theorem, the distance from $A$ to $B$ is 1.) Now construct $E$ on $FB$ such that $F$ is between $E$ and $B$ and the distance from $E$ to $F$ is $\frac{1}{3}\sqrt{1-k^2}$. Extend the line through $A$ and $E$. Now, using the marked straightedge, construct the line passing through $B$ whose points of intersection with $AF$ and $AE$ are separated by a distance of 1. Call these points $C$ and $D$, and call the distances $BC$ and $AD$ $a$ and $b$, respectively, as in the following diagram.

Finally, construct the lines through $C$ perpendicular to $AF$ and $BE$, and call the distances from $C$ to these lines $y$ and $x$, respectively, as we have labeled in green in the following diagram.

Now $\triangle BFD$ and the triangle with vertices $C$ and $D$ and a side length $y$ are similar, so that (1) $\dfrac{y}{1} = \dfrac{\sqrt{1-k^2}}{1+a}$. Similarly, $\triangle BFD$ and the triangle with vertices $B$ and $C$ and side length $x$ are similar, so that (2) $\dfrac{x}{a} = \dfrac{b+k}{1+a}$. Likewise, $\triangle FAE$ and the triangle having vertices $A$ and $C$ and a side length $y$ are similar, so that (3) $\dfrac{y}{x-k} = \dfrac{\frac{1}{3}\sqrt{1-k^2}}{k} = \dfrac{\sqrt{1-k^2}}{3k}$.

Finally, using the Pythagorean theorem on $\triangle BDF$, we have (4) $(1-k^2) + (b+k)^2 = (1+a)^2$. We can expand (4) and simplify to get the equation (4′) $b^2+2bk = 2a+a^2$.

Solving (1) and (3) for $y/\sqrt{1-k^2}$, we see that $\dfrac{1}{1+a} = \dfrac{x-k}{3k}$. Solve for $x$ to get (5) $x = k + \dfrac{3k}{1+a}$. Equating with (2), we have $\dfrac{3k}{1+a} + k = \dfrac{a(b+k)}{1+a}$, which simplifies to (6) $ab = 4k$. Using (6) to eliminate $b$ in (4′), we have $a = 2k^{2/3}$. (Hint: clear denominators and note that $2+a$ is a factor.) Similarly, eliminating $a$ yields $b = 2k^{1/3}$, as desired.