Let $\Gamma$ be the graph of the parabola $y = x^2$ in the $xy$-plane. Let $\gamma: \mathbb{R} \to \mathbb{R}^2$ be the map
$$ x \mapsto (x, x^2). $$
It is actually a diffeomorphism from $\mathbb{R}$ onto $\Gamma$. Let $p_1$, $p_2$ and $p_3$ be $3$ distinct points on $\Gamma$. The formula for the $3\times 3$ Vandermonde determinant then tells us that the area $A(p_1, p_2, p_3)$ of the triangle with vertices $p_1$, $p_2$ and $p_3$ is given by
$$ A(p_1, p_2, p_3) = \frac{1}{2}|x_2 – x_1| |x_3 – x_1| |x_3 – x_2|,$$
where $p_i = \gamma(x_i)$ for $1 \leq i \leq 3$.
The problem I propose is to find a geometric proof of the above formula for $A(p_1, p_2, p_3)$. It is a problem in $2$-dimensional Euclidean geometry, so there may be a nice geometric proof of it. Thoughts and/or comments are welcome!
Edit 1: Thanks for @Jean Marie for pointing out the link with another question, namely Area of triangle inscribed in a parabola. The questions are definitely very closely related. I didn't think about using tangents actually, while the question there does, which is good, since it allows for more general parabolas such as for instance $y = a x^2$, where $a$ is not necessarily $1$. The reason why I think I should keep this post, is that the answers there at the time of writing do not provide a geometric proof. One proof uses calculus, and I think the other one uses a Vandermonde determinant.
I already know one proof using Vandermonde determinants, but I am still interested in a geometric proof. You see, I am interested in a more complex problem (the Atiyah problem on configurations), and having a geometric proof of the formula here may help, if it generalizes to rational normal curves instead of the $\gamma$ here. But I am trying to go step by step.
The next problem I would like to propose is actually an area inequality, closely related to this one, and which implies the $n = 3$ special case of the Atiyah-Sutcliffe problem on configurations of points. I have created a post for that problem for those who are interested: A lower bound on an area of a triangle defined using $6$ distinct points on a parabola.
Edit 2: Here is an approach I'd be interested in. In the other post Jean Marie linked to (Area of triangle inscribed in a parabola), they equate the area of a triangle inscribed in a parabola with twice the area of the triangle formed by the tangents to the parabola at the 3 vertices. I am convinced this is related to some kind of projective duality in the plane between points and lines, and this is the kind of algebro-geometric ideas I would like to see at play (as these would generalize to higher degree rational normal curves). So I propose the same problem as that other post (or the I guess equivalent formula I propose here, though I don't immediately see the equivalence) using ideas from classical algebraic geometry such as duality and rational normal curves.
Edit 3: I keep getting messages that the post Area of triangle inscribed in a parabola may answer my questions. The setup is pretty much the same, a triangle inscribed in a parabola, and in that post, they are also interested in the area of that triangle. The right-hand side in what they are trying to show in that post, namely that the area of such a triangle is twice the area of the triangle defined by the tangent lines to the parabola at the $3$ vertices of the triangle, is not very obviously equal to the right-hand side in the formula here. I mean, I am sure there is some nice argument for why the $2$ right-hand sides are equal, but it is not very obvious.
If you guys decide to close this post, it is fine with me, but I hope that at least @Intelligenti pauca will write a similar nice answer to that other post along similar geometric lines. That answer is nice and deserves to be somewhere on Math.SE!
Best Answer
If $h_k$ are the projections of $p_k$ onto $x$ axis, then we have: $$ A(p_1,p_2,p_3)=A(h_1,h_3,p_3,p_1)-A(h_1,h_2,p_2,p_1)-A(h_2,h_3,p_3,p_2), $$ where I assumed $x_1<x_2<x_3$ to avoid using absolute values. The areas of the three trapezoids on the r.h.s. can be computed and the result is: $$ A(p_1,p_2,p_3)= -{1\over2}\big[(x_1^2+x_3^2)(x_1-x_3)+(x_1^2+x_2^2)(x_2-x_1) +(x_2^2+x_3^2)(x_3-x_2)\big], $$ which is the same as the expression given in the question.