A) You are given the points of the curve $y = x^2$ on the 2D plane. Using only straightedge and compass, can you locate the origin?
B) What if the curve was $y = x^3$?
You are not given the co-ordinate axes obviously. Just the points on the curve and a vast empty plane...
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A) $y = x^2$
Consider a chord joining $(a,a^2)$ to $(b,b^2)$. The x-coordinate of the midpoint of this chord is $\frac{a+b}{2}$ and the slope is $a+b$. Thus if two chords are parallel, their midpoints will have the same x co-ordinate!
This observation now makes it easy.
Get two parallel chords and locate their midpoints. Draw the line L joining the midpoints. L is parallel to the y-axis. Now draw perpendicular chords to L, though two points on the parabola. The midpoints of those lie on the y-axis. Draw the y-axis and intersect with the parabola to get $(0,0)$.
B) $y = x^3$
Take two points $A,B$ on the curve and draw the line joining them. This line cannot be parallel to the y-axis (this is easy to see) and must intersect the curve at a third point $C$.
Because: If the line is $y = mx + c$, then the intersection points are $(\alpha_i, \alpha_i^3)$ where $\alpha_i$ are the roots of $x^3 = mx + c$. If this has $2$ real roots, it must have $3$ real roots as complex roots come in conjugate pairs. Note that if there is a repeated root (and so we only have 2 points of intersection), we can perturb B a little and then $3$ points will be guaranteed.
Also notice $\sum \alpha_i = 0$, because the coefficient of $x^2$ is $0$. Thus the centroid of $A,B,C$ lies on the y-axis!
wlog, assume $B$ lies between $A$ and $C$.
Now pick a point $B'$ on the curve and draw the line $A, B'$, to get a new line $A, B', C'$. These two lines intersect at $A$ and thus $(A,B,C)$ and $(A,B',C')$ cannot have the same centroid, giving us a second point on the y-axis.
Construct the line joining these two centroids, and intersect with the curve to get $(0,0)$.
