I remember being told in passing in a talk once the following theorem:

Let $y=x^3$, and let $x_1,x_2,x_3$ be the abscissae ($x$ co-ordinates) of three distinct points on this cubic. Then $x_1+x_2+x_3=0$ if and only if the three points are colinear.

So eventually I sat down and proved this, in the more general form

Let $y=ax^3+bx^2+cx+d$, and let $x_1,x_2,x_3$ be the abscissae ($x$ co-ordinates) of three distinct points on this cubic. Then $a(x_1+x_2+x_3)+b=0$ if and only if the points are colinear.

The proof I came up with was fairly uninspired, and went as follows: the points are colinear if and only if the line joining $(x_1,y_1)$ and $(x_2,y_2)$ is parallel to the line joining $(x_1,y_1)$ and $(x_3,y_3)$. (Trivial, since the lines intersect in one point, and are then the same line if and only if they are parallel.). Similarly with the line through $(x_2,y_2)$ and $(x_3,y_3)$.

Then we have the gradients of the first two lines, given by the expression $$ \frac{y_1-y_2}{x_1-x_2} = \frac{y_1-y_3}{x_1-x_3}, $$ or $$ \frac{(ax_1^3+bx_1^2+cx_1+d)-(ax_2^3+bx_2^2+cx_2+d)}{x_1-x_2} = \frac{(ax_1^3+bx_1^2+cx_1+d)-(ax_3^3+bx_3^2+cx_3+d)}{x_1-x_3}, $$ which simplifies to $$ \frac{a(x_1^3-x_2^3)+b(x_1^2-x_2^2)+c(x_1-x_2)}{x_1-x_2} = \frac{a(x_1^3-x_3^3)+b(x_1^2-x_3^2)+c(x_1-x_3)}{x_1-x_3}, $$ and since the denominators are both not zero, we can cancel to find $$ a(x_1^2+x_1 x_2 + x_2^2)+b(x_1+x_2)+c = a(x_1^2+x_1 x_3 + x_3^2)+b(x_1+x_3)+c. $$ This factorises to $$ (x_2-x_3)(a(x_1+x_2+x_3)+b)=0. $$ Doing the same thing with the labels on the points cyclicly permuted gives $$ (x_1-x_2)(a(x_1+x_2+x_3)+b)=0\\ (x_3-x_2)(a(x_1+x_2+x_3)+b)=0, $$ and hence $a(x_1+x_2+x_3)+b=0$, since the points are distinct. Further, all of the above steps can be done backwards with no loss of generality, so the converse also holds.

Now, the actual questions:

  1. Is there a simpler proof than this rather unpleasant algebraic proof, where we muddled through to the answer by doing the obvious thing? I'm thinking here of simplification beyond obvious things like the linear change of variable $x=a^{-1/3}z-a/(3b)$ to get rid of the quadratic term and make the polynomial monic.
  2. Is there a way to include the tangent case easily? (i.e. the tangent to a general cubic as in the second theorem intersects again at $-2x-b/a$)
  3. Is there any interesting generalisation of this? (Presumably the group law on elliptic curves does something like this.)
  4. The always dodgy "Does this result actually mean anything geometrically?"

For the sum $\implies$ line part of the proof, let $P(x) = ax^3 + bx^2 + cx + d$ be an arbitrary polynomial, and let $x_1$, $x_2$, and $x_3$ be any three values such that $a(x_1 + x_2 + x_3) + b = 0.$ Consider once again the polynomial $$ Q(x) = a(x-x_1)(x-x_2)(x-x_3). $$ The equation $Q(x)=0$ clearly has roots at $x_1,$ $x_2,$ and $x_3.$ But we can rewrite $Q(x)$ as $$Q(x) = ax^3 - a(x_1 + x_2 + x_3)x^2 + px + q = ax^3 + bx^2 + px + q$$ for some constants $p$ and $q$.

Now let $$\begin{eqnarray} R(x) &=& P(x) - Q(x) \\ &=& ax^3 + bx^2 + cx + d - ax^3 + bx^2 + px + q\\ &=& (c-p)x + (d-q). \end{eqnarray}$$ Then $y = R(x)$ is the equation of a line.

Since $Q(x_1) = Q(x_2) = Q(x_3) = 0,$ it follows that $R(x_1) = P(x_1),$ $R(x_2) = P(x_2),$ and $R(x_3) = P(x_3).$ That is, the points $(x_1,P(x_1)),$ $(x_2,P(x_2)),$ and $(x_3,P(x_3))$ all lie along the line $y = R(x).$

  • $\begingroup$ Ah yes, very nice. I still wonder if there's a better way to do the whole argument using some sort of mildly sophisticated algebraic geometry. $\endgroup$ – Chappers Apr 20 '15 at 13:30
  • $\begingroup$ I have a hunch there is something much more general about the intersections of polynomial curves. The little I know about that is kind of rusty. Maybe someone will enlighten both of us on this. $\endgroup$ – David K Apr 20 '15 at 16:57

An option that occurred to me during writing the question is the following: let the line be $y=mx+e$, the cubic as in the question. Then we want to solve $$ y= mx+e = ax^3+bx^2+cx+d, $$ or $$ ax^3+bx^2+(c-m)x+(d-e)=0. $$ Being a cubic equation, this has three roots, $x_1,x_2,x_3$, which may be repeated, but we have the factorisation $$ a(x-x_1)(x-x_2)(x-x_3), $$ from which we discover the quadratic term is $ a(x_1+x_2+x_3) $, which is equal to $b$ by the form we originally had. So this does the line $\implies$ sum case, in more generality (since the roots can coincide), but not the converse, as far as I can see.


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