Exercise 10.3.15

Proof. (a) We suppose that $k\ge 0$. Let $I=M$ be the midpoint of $\mathit{AB}$. As $△\mathit{ABC}$ is isosceles, by Pythagoras’ Theorem, $C{I}^2+{(k∕8)}^2=A{C}^2=1$, therefore $0\le k∕8\le 1$, so $k\le 8$. If $k>8$, there exists no point $C$ with the conditions of the text. If $k=8$, $C$ is the midpoint of $\mathit{AB}$ and $l_1$ = $l_2$, and if $k=0$, $A=B$ and $l_2$ is not defined. The restriction $0<k<8$ is necessary. (b) 2 solutions.

Solution 1.

Personal solution with analytic geometry.

Take the origin in $O=M$, where $M$ is the midpoint of $\mathit{AB}$, and take $\mathit{MB},\mathit{MC}$ as the $x$ and $y$-axis. Write for simplicity $\lambda =\mathit{MB}=k∕8,\mu =\mathit{MC}$, so ${\lambda }^2+{\mu }^2=1$. The unknown is $x=\overline{\mathit{BR}}$.

The coordinates of $A,B,C,D,R$ are

$$A|\begin{array}{c}\hfill -\lambda \hfill \\ \hfill 0\hfill \end{array},B|\begin{array}{c}\hfill \lambda \hfill \\ \hfill 0\hfill \end{array},C|\begin{array}{c}\hfill 0\hfill \\ \hfill \mu \hfill \end{array},D|\begin{array}{c}\hfill -2\lambda \hfill \\ \hfill -\mu \hfill \end{array},R|\begin{array}{c}\hfill \lambda +x\hfill \\ \hfill 0\hfill \end{array}.$$

Let $(X,Y)$ be any point in the plane.

$$\begin{array}{llll}\hfill (X,Y)\in l_1=(\mathit{DB})& ⟺0=\left|\begin{array}{cc}\hfill X-x_B\hfill & \hfill x_B-x_D\hfill \\ \hfill Y-y_B\hfill & \hfill y_B-y_D\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill X-\lambda \hfill & \hfill 3\lambda \hfill \\ \hfill Y\hfill & \hfill \mu \hfill \end{array}\right|& \hfill & \\ \hfill & ⟺\mu (X-\lambda )-3\mathit{\lambda Y}=0.& \hfill & \\ \hfill & & \hfill \end{array}$$

So the equation of $(\mathit{DB})$ is

$$(\mathit{DB}):\mathit{\mu X}-3\mathit{\lambda Y}=\mathit{\lambda \mu }.$$

Similarly,

$$\begin{array}{llll}\hfill (X,Y)\in (\mathit{CR})& ⟺0=\left|\begin{array}{cc}\hfill X-x_C\hfill & \hfill -(x_R-x_C)\hfill \\ \hfill Y-y_C\hfill & \hfill -(y_R-y_C)\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill X\hfill & \hfill -(\lambda +x)\hfill \\ \hfill Y-\mu \hfill & \hfill \mu \hfill \end{array}\right|& \hfill & \\ \hfill & ⟺\mathit{\mu X}+(\lambda +x)Y=\mu (\lambda +x).& \hfill & \\ \hfill & & \hfill \end{array}$$

So the equation of $(\mathit{CR})$ is

$$(\mathit{CR}):\mathit{\mu X}+(\lambda +x)Y=\mu (\lambda +x).$$

The coordinates $(x_Q,y_Q)$ are given by the system

$$\begin{array}{llll}\hfill \mathit{\mu X}-3\mathit{\lambda Y}& =\mathit{\lambda \mu },& \hfill & \\ \hfill \mathit{\mu X}+(\lambda +x)Y& =\mu (\lambda +x).& \hfill & \end{array}$$

Since $\mu \ne 0$, this is equivalent to

$$\begin{array}{llll}\hfill (4\lambda +x)X& =4\lambda (\lambda +x),& \hfill & \\ \hfill (4\lambda +x)Y& =\mathit{\mu x},& \hfill & \end{array}$$

so

$$Q|\begin{array}{c}\hfill \frac{4\lambda (\lambda +x)}{4\lambda +x}\hfill \\ \hfill \frac{\mathit{\mu x}}{4\lambda +x}\hfill \end{array},R|\begin{array}{c}\hfill \lambda +x\hfill \\ \hfill 0\hfill \end{array}.$$

Therefore

$$\begin{array}{llll}\hfill \mathit{QR}=1& ⟺1={\left[\frac{4\lambda (\lambda +x)}{4\lambda +x}-(\lambda +x)\right]}^2+{\left[\frac{\mathit{\mu x}}{4\lambda +x}\right]}^2& \hfill & \\ \hfill & ⟺{(4\lambda +x)}^2={(\lambda +x)}^2{x}^2+{\mu }^2{x}^2& \hfill & \\ \hfill & =x^4+2\lambda x^3+({\lambda }^2+{\mu }^2)x^2& \hfill & \\ \hfill & =x^4+2\lambda x^3+x^2& \hfill & \\ \hfill & ⟺x^4+2\lambda x^3-8\mathit{\lambda x}-16{\lambda }^2=0& \hfill & \\ \hfill & ⟺(x+2\lambda )(x^3-8\lambda )=0& \hfill & \\ \hfill & ⟺\left(x+\frac{k}{4}\right)(x^3-k)=0& \hfill & \end{array}$$

As $Q\ne D,R\ne A$, so $x\ne -k∕4$. Therefore

$$\mathit{QR}=1⟺x=\sqrt[3]{k}.$$

Note: the factorization is easy because $x=-2\lambda =-k∕4$ gives the particular solution $R=A$.

The marked-ruler constructions of Pappus (for the trisection) and Nicomedes (for the cube root) give both a fourth degree equation with a known obvious solution.

Solution2.

From [15] (G.E. Martin, Geometric Constructions):

“ Let the parallel to $(\mathit{AB})$ that passes through $C$ intersect $(\mathit{DB})$ at $E$. So $△\mathit{ABD}$ and $△\mathit{CDE}$ are similar. Then, since $A$ bisects $\mathit{CD}$, we have $\mathit{CE}=2\mathit{AB}=k∕2$. Also, since $△\mathit{QBR}$ and $△\mathit{QCE}$ are similar, we have $(k∕2)∕\mathit{CQ}=\mathit{BR}∕1$. With $x=\mathit{BR}$, then $\mathit{CQ}=k∕(2x)$. With $M$ the midpoint of $A$ and $B$, by two applications of the Pythagorean Theorem, we now have

$$\begin{array}{llll}\hfill {\left(1+\frac{k}{2x}\right)}^2& =C{R}^2=C{M}^2+M{R}^2=(C{A}^2-A{M}^2)+M{R}^2& \hfill & \\ \hfill & =1^2-{\left(\frac{k}{8}\right)}^2+{\left(x+\frac{k}{8}\right)}^2& \hfill & \end{array}$$

So, multiplying the two members by $8^2{x}^2$,

$$\begin{array}{llll}\hfill 16{(2x+k)}^2& =8^2{x}^2-k^2{x}^2+x^2{(8x+k)}^2,& \hfill & \\ \hfill 64x^2+64\mathit{kx}+16k^2& =64x^2-k^2{x}^2+64x^4+16k{x}^3+k^2{x}^2,& \hfill & \\ \hfill 0& =64x^4+16k{x}^3-64\mathit{kx}-16k^2,& \hfill & \\ \hfill 0& =4x^4+k{x}^3-4\mathit{kx}-k^2.& \hfill & \end{array}$$

Fortunately, this quartic easily factors as $(4x+k)(x^3-k)=0$. Since $4x+k>0$, then we must have $x^3-k=0$. Therefore, $x$ is the real cube root of $k$, as desired."

(c) If $k$ is any positive number, let $s$ an integer such that $0<k<2^{3s+3}$, so $0<K<8$, where $K=\frac{k}{2^{3s}}$. The Nicomedes’ construction applied to $K$ gives $\sqrt[3]{K}$, thus $\sqrt[3]{k}=2^s\sqrt[3]{K}$ is constructible by marked-ruler. □