Given the graph $y=x^4$, can we construct the $y$-axis using only a straightedge and a compass?

Question

This is a question my friend raised and we have had difficulty solving it.

Suppose that the graph of the polynomial function $f(x)=x^4$ is drawn on a plane. Can we construct the $y$-axis of this plane using only a straightedge and a compass?

In the case where the function is replaced by $f(x)=x^3$, the problem can be solved as follows:

1. Draw two parallel lines $n$ and $m$ so that each of them intersect the given curve exactly three times.
2. Let $a_1, a_2, a_3$ be the intersection points of the curve and $n$, and $b_1, b_2, b_3$ be those of the curve and $m$.
3. By Vieta's formulas, the centroids of $a_1, a_2, a_3$ and $b_1, b_2, b_3$ both lie on the $y$-axis. Hence, the unique line passing through these two points is the $y$-axis.

We cannot replace $x^3$ with $x^2$ in this approach; however, in that case the line $l$ drawn in step $(3)$ is parallel to the $y$-axis and we can obtain $y$-axis using $l$ (see the image below).
This approach does not work for the original problem because we cannot draw a line that intersects the graph exactly four times. Can we still draw the $y$-axis? If not, how can we prove it?

Answer 1

Too long for comment

EDIT: Thank you @Bart_Michels for pointing ou that it is possible to construct the cubic with only two 'free variables'. What this means for this problem I'm not sure but I'm leaning towards it being impossible for the quartic.

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An interesting observation I made after far too much thought into this problem: The quadratic can be solved using only two 'free variables' while the cubic seems to require three 'free variables'. What do I mean by free variable? I mean an arbitrary coordinate that cannot be set by the constructor.

In the case of the quadratic, you need two $x$ coordinates (and then the corresponding $y=x^2$ $y$ coordinates which are not free). For the quadratic below, the blue diamonds are the free variables (the green is the function and the red is the $y$-axis). In Geogebra, you can move these points around and the construction will always output the $y$-axis.

In the case of the cubic (see below), you need three $x$ coordinates, chosen such that the first two define a line with three intersections with the cubic and the third point will define a second line with the same slope as the first. Note that the quadratic above shows the full construction while I took some of Geogebras shortcuts (perpendicular and parallel line through points and midpoints) to cut down on the visual clutter.

What does this mean for the quartic? It is evidence that any potential quartic construction will require four free variables but what that looks like I don't know. It could be two arbitrary lines, it could be three lines that share an intersection point with the quartic, or it could be two points on the quartic and some midpoint that defines a circle (the midpoint would have two free variables, both $x$ and $y$ coordinates).

Answer 2

One thing occurred to me; a related quartic problem for $ y = x^4 - a x^2 + b$ can be solved by the "centroid" method.

It took me a while to conclude that the centroid of $n$ points on a line can be constructed with compass and straightedge. The trick is that we may provide a segment of length defined to be $1.$ Along the line in question, we draw a point labelled $0$ and a point labelled $1.$ Given this number line, we can construct the sum of four signed distances to the "right" of $0,$ then find one fourth of this, still relative to the $0.$

Below I display $y = x^4 - 5 x^2 + 4,$ which is symmetric around the $y$ axis, and $y = 4 - 2x$ with evident intersections at $(0,4)$ and $(2,0).$ The other intersections have $x = -1 + \sqrt 2$ and $x = -1 - \sqrt 2.$ The centroid has $x$ value $$\frac{1}{4} \left( (-1-\sqrt 2) + 0 +(-1 + \sqrt 2) + 2 \right) = \frac{0}{4} = 0. $$

Answer 3

This is true for any $y=x^n$, $n\geq2$.

Let $S$ be a set of at least two points in the plane, which we can identify with $\mathbb C$, and $0_S$, $1_S$ be distinct points of $S$. Then it is elementary to prove that the set of points constructible from $S$ is of the form $0_S + (1_S-0_S)\cdot F$ where $F$ is a subfield of $\mathbb C$.

In fact, $F$ can be characterized as follows, but we will not require this: It is the smallest subfield containing $(S-0_S)/(1_S-0_S)$ that is closed under square roots and complex conjugation.

Take some points $S$ anywhere in the plane. The polynomial curve $y=x^n$ can be expressed using a degree $n$ polynomial equation $P(x_S,y_S)=0$ in the coordinate system of $F$ with basis $(1_S,i_S)$. In fact it is simply of the form $(a+bx_S+cy_S)^n - (d-cx_S + by_S) = 0$ with $a,b,c,d$ real numbers.

Now consider the field extension $E$ of $F$ that is closed under constructions and under taking intersections of lines with the curve $y=x^n$ (we don't even require intersections with the curve; it's enough to be able to sample some points on the curve).

$0_S + (1_S-0_S)\cdot E$ is the set of points that we can construct.

We have $\mathbb R[x,y]/P(x,y)\equiv \mathbb R[x]$ because this is true in the original coordinates, and a polynomial of degree $m$ has degree at most $mn$ in the quotient (again because $y$ maps to $x^n$ in the original coordinates).

In particular, the zero set of any polynomial of degree $n$ different from multiples of $P$ intersects the curve in at most $n^2$ points because it has degree at most $n^2$ in the quotient.

Now take any $n^2+1$ points on the curve. Evaluating polynomials $Q(x_S,y_S)$ of degree $n$ at these points gives linear system in $Q$ over $E$ whose only solutions are multiples of $P$. Therefore there exists $k\in \mathbb R$ such that $kP\in E[x_S,y_S]$.

In particular $b^2/c^2\in E$. because there are monomials of degree $2$ with coefficients $b^2a^{n-2}$ and $c^2a^{n-2}$ up to rational numbers.

Therefore $b/c\in E$, which means we can construct a line parallel to the $y$ axis, and we are done.

In practice it suffies to choose $n^2+1$ points on the curve and find a full rank submatrix with $\binom{n+2}2 -1$ rows (by trial and error, by computing determinants and checking if they are distinguishable from $0_S$). So to try this in Geogebra for $n=4$ you can take any $14$ points and this will work for a generic choice of points, and there is a deterministic procedure to proceed if the choice is unlucky (or if you are not sure whether some point coincides with $0_S$).