It's always interesting anytime we get to visit one of the three classical problems in Geometry. As the title of this post says, we are going to talk about Doubling the cube.
It is well known that doubling the cube is impossible with only a compass and straightedge but is tractable with a neusis ruler. One of the simplest such constructions is given Wikipedia. We use a slightly modified version in this post.
Construct an equilateral triangle $ABC$. Draw a line through $B$ perpendicular to $BC$. We now use the neusis to find point $D$ (on line $AB$) and $E$ such that $AB=ED$. Then, $CE=\sqrt[3]{2}AB$.
The wiki article quoted above does not prove this and I decided to try it on my own. To begin with, because $\angle CBE$ is right angled, we see that $BE=\sqrt{CE^2-CB^2}$.
Now we construct a point $E'$ on $BD$ such that $EE'$ is parallel to $CB$.
Because $\triangle DEE' \sim \triangle DCB$, their sides are in proportion. Therefore,
$\displaystyle \frac{EE'}{CB} = \frac{DE}{DC} \implies EE'=BC \cdot \frac{BC}{BC+CE}$.
As $\triangle BE'E$ is a $30-60-90$ triangle, $BE=\sqrt{3}EE'$. Therefore,
$\displaystyle \sqrt{3}\frac{BC^2}{BC+CE}=\sqrt{CE^2-BC^2} \implies (CE^2-BC^2)(CE+BC)^2=3BC^4$.
If we let $CE=xBC$, then the above expression reduces to $(x^2-1)(x+1)^2=3$.
By ratio test, we can guess that $-2$ is a root of the equation. Taking that factor out, we can see that $x=\sqrt[3]{2}$ thus proving that $CE=\sqrt[3]{2}BC$.
This last part of the proof is the one I find a little unsatisfactory. For one of the classical problems in Geometry, this proof is more algebraic than geometric.
I went in search of a more geometric proof myself. I couldn't find one and I went on an internet search and to my surprise, there aren't a lot. Finally, I found Nicomedes' proof in this page which gave me what I was looking for. I also get to know about Crockett Johnson's painting on Doubling the cube.
Luckily, I was able to modify (and simplify) the argument to get something that looks elegant to my eyes. This proof starts by constructing a line through $D$ perpendicular to $CB$. Let this line meet (the extension of) $CB$ at $J$.
As $\triangle CBE \sim \triangle CJD$,
$\displaystyle \frac{CB}{BJ}=\frac{CE}{ED} \implies CE \cdot BJ=CB \cdot ED=CB \cdot AB$
Note that $\triangle DBJ$ is a $30-60-90$ triangle. Therefore, $BD=2BJ$. This finally shows us that
$CE \cdot BD=2CB \cdot AB \tag{1} \label{reqn}$
Now comes the second part of our construction. Extend $AC$ and mark a point $F$ on it such that $CF=CD$. Join $D$ and $F$. Draw a line through $B$ perpendicular to $AD$ and mark its intersection with $AF$ as $G$. Draw a line through $G$ parallel to $AB$ and let it meet $DF$ at $H$. Join $H$ and $B$.
$\triangle GAB$ is a $30-60-90$ triangle. Therefore, $GA=2AB=2CB$.
Because $CD=CF$ by construction, $CE=GF$. Using this, we recast $\eqref{reqn}$ to
$GF \cdot BD=GA \cdot AB \tag{2} \label{rreqn}$
Our plan now is to prove that $GH=AB$. As $\triangle FGH \sim \triangle FAD$, we have
$\displaystyle \frac{GH}{GF}=\frac{AD}{AF} \implies GH \cdot AF=GF \cdot ( AB+BD)=GF\cdot AB+GF\cdot BD$
Using $\eqref{rreqn}$ now, we have,
$\displaystyle GH \cdot AF=GF\cdot AB+GA\cdot AB=AB\cdot (GF+GA)=AB \cdot AF \implies GH = AF$
This also shows that $BH$ is parallel to $AF$. Therefore,
$\displaystyle\frac{GH}{GF}=\frac{AD}{AF}=\frac{BD}{BH} \tag{3} \label{feqn}$
because $\triangle FGH \sim \triangle FAD \sim \triangle HBD$.
We now construct a point $C'$ on $AB$ such that $CC'$ is perpendicular to $AB$. Note that $C'$ will also be the midpoint of $AB$. Now,
$DB\cdot DA=(DC'-C'B)(DC'+C'B)=DC'^2-C'B^2$
Now $CC'^2=CD^2-C'D^2=CB^2-C'B^2$ both by Pythagoras theorem on respective $\triangle CC'D$ and $\triangle CC'B$. Using this,
$DB \cdot DA=CD^2-CB^2=CF^2-CA^2=(CF-CA)(CF+CA)=GF\cdot AF$
Therefore,
$\displaystyle \frac{GF}{DB}=\frac{DA}{AF}$
(Alternately, we can construct a circle centered at $C$ with $CA$ as radius. Because both $D$ and $F$ are equidistant from $C$, tangents from $D$ and $F$ to the circle will be of the same length. In other words, both points have the same power w.r.t the circle. But it is apparent from the configuration that $\mathcal{P}(C,D)=DB\cdot DA$ and $\mathcal{P}(C,F)=FG \cdot FA$)
Using $\eqref{feqn}$, we finally have $\displaystyle \frac{GH}{GF}=\frac{GF}{BD}=\frac{BD}{BH}$
Because $ABHG$ is a parallelogram, $BH=AG=2GH$. Therefore,
$\displaystyle \left(\frac{GF}{GH}\right)^3=\frac{GF}{GH}\frac{GF}{GH}\frac{GF}{GH}=\frac{GF}{GH}\frac{BD}{GF}\frac{BH}{BD}=\frac{BH}{GH}=2$
Using the fact that $GF=CE$ and $GH=AB$, we finally have
$CE^3=2AB^3$ (or) $CE=\sqrt[3]{2}AB$
Until then
Yours Aye
Me
