Geometric picture of 3-torsion points on an elliptic curve

Question

I'm faced with what seems a paradox. If we have an elliptic curve $E/\Bbb C$ in Weierstrass form so that $\mathcal O_E$ is at infinity, then the addition law is quite easy to picture geometrically. In particular, the 2-torsion points are exactly the ones whose tangent is the vertical line $y=c$.

The 3-torsion points must therefore not have a vertical tangent line (except for $\mathcal O_E$), but instead they must satisfy $[2]P=-P$. Denoting $P=(P_x,P_y)$, this means that $-P=(P_x,-P_y)$. Now my problem is, if the $x$-coordinate of $[2]P$ is going to be different than $P$'s $x$-coordinate ($P_x\neq [2]P_x$) because the tangent line at $P$ is not vertical, how can the secant joining $[2]P$ and $P$ meet $\mathcal O_E$? The last condition implies that $P_x=[2]P_x$, right?

Answer 1

Look at the curve $y^2=x^3+1$. The point $P=(0,1)$ has order three. Draw the curve, find $2P$, and everything should be a lot clearer.

Answer 2

There is exactly one way in which the tangent line at $P$ can meet $E$ at a point $[-2]P$ with the same $x$-coordinate as $P$ which I missed. This is so when $P$ is in fact a flex. That totally makes sense because there are 9 flexes on an elliptic curve and $|E(3)|=9$.