Exercise 10.3.5

Proof. As the discriminant of $f(x)=x^3+2x+1(p=2,q=1)$ is ${\mathrm{\Delta }}_f=-4p^3-27q^2=-59<0$, there is a unique real root. First we compute this real root $\alpha$ with Cardano’s method. There exists a pair of complex numbers $u,v$ such that

(the roots of $y^2-\mathit{\alpha y}-\frac{2}{3}$).

Since $2+3\mathit{uv}=0$,

So $u^3,v^3$ satisfy

Therefore $u^3,v^3$ are the roots of $y^2+y-\frac{8}{27}$:

Thus

which is equal to

and the focus and directrix of $P_2:y=\frac{x^2}{2}$ are

As all the points have rational coordinates, and the lines rational coefficients, they are constructible by origami. More precisely, take two arbitrary points on the $x$-axis to represent $(0,0)$ and $(1,0)$. To respect the instructions, the directrix $D_1:x=-1∕2$ must be the left side of the sheet, so the $y$-axis is placed at $1∕8$ of the length of the sheet, and the unit is $1∕4$ of this length, so we obtain the $y$-axis by three vertical folds (see figure).

To obtain the two foci $F_1,F_2$, and the directrix $D_2$, we divide in eight equal parts the sheet by horizontal folds. (b) As seen in the preliminaries, the discriminant of $x^3+2x+1$ is ${\mathrm{\Delta }}_f=-59<0$, the equation (10.12) $m^3+\mathit{am}+b=0$, where $a=2,b=1$, has a unique real solution $\alpha$. As the slope of a common tangent to the parabolas $P_1,P_2$ is a root of this equation, the slope of such a tangent is $m=\alpha$. Since the intersection point $M_1(x_1,y_1)$ of the tangent with the parabola $P_1$ satisfies (10.10)

this intersection point is determined by $m=\alpha$, and similarly the tangent point $M_2(x_2,y_2)$ with $P_2$ is uniquely determined by

so there is at most one common tangent to the two parabolas, and at most one way to take the focus of each parabola to a point on the corresponding directrix.

Conversely, we must prove the existence of a common tangent.

Let $D$ the line with slope $\alpha$ passing by $M_2(x_2,y_2)$, where $(x_2,y_2)=(\alpha ,\frac{{\alpha }^2}{2})$. The equation of $D$ is given by $y-\frac{{\alpha }^2}{2}=\alpha (x-\alpha )$, so

$D$ contains $M_1(x_1,y_1)$, where $(x_1,y_1)=(\frac{1}{2{\alpha }^2},1+\frac{1}{\alpha })$, since

By Example 10.3.2 and Exercise 4, the slope of the tangent $T_1$ to $P_1$ at $M_1$ is $m=\frac{b}{y_1-\frac{1}{2}a}=\alpha$, and the slope of the tangent $T_2$ to $P_2$ at $M_2$ is $m=x_2=\alpha$, so $D=T_1=T_2$, therefore $D$ is tangent to $P_1$ and $P_2$. This proves the existence of one common tangent $D$ to the two parabolas, with equation

Conclusion: there is exactly one way to take the focus of each parabola to a point on the corresponding directrix.

To obtain the corresponding line (the dotted line of the figure) by an origami fold, we take $F_1$ to $D_1$ (the left side of the sheet), and simultaneously $F_2$ to $D_2$. The slope $m$ of this line is the root $\alpha$ of $x^3+2x+1$. (c) Let $F_2^{\prime }$ be the reflection of the focus $F_2$ of $P_2$ about the common tangent $D$, and $M_2$ the point of tangency. By Exercise 3, the line $(M_2{F}_2^{\prime })$ is parallel to the $y$-axis, so by (10.11) the $x$-coordinate of $F_2^{\prime }$ is

The intersection of the vertical fold passing by $F_2^{\prime }$ with the $x$-axis gives the point $(m,0)$, with $m\approx -0.45$.

We have resolved the equation $x^3+2x+1=0$ by origami ! □