Pumping triangles

Question

On the inside of the triangle (0,0),(1,0),(0,1), define the "pretty useless map" or pump by the following prescription:

Given a point A find its projections x0,y0 to the x and y axes and also x1,y1 to the same axes shifted by one unit. Connect x1 to y0 and y1 to x0. The intersection of these two straight lines shall be pump(A)

Question:

Given a triangle ABC show that the area of triangle pump(A)pump(B)pump(C) can be written as f(X) for some fixed scalar function f where X is the area of ABC.

Most elegant solution wins.

Hint 1:

A purely geometric, zero algebra solution exists.

Hint 2 (Warning: this really is a spoiler):

Source: Me.

Answer 1

The alternative solution:

First apply a dual transformation. That is, we take the dual of the configuration by replacing each point with a line and each line with a point, as shown.
(If you've never heard of this trick and it sounds unbelievably wacky, see https://en.wikipedia.org/wiki/Duality_(projective_geometry) and scroll down to the Poles and Polars section for a construction of such a transformation. I got this particular transformation by swapping poles with polars around a small circle whose center lies in the upper-right "quadrant" of the rectangle.)
Under this transformation, it suffices to show that the three dark blue (violet?) lines are concurrent.

Then:

We apply Pappus's Theorem on the two thin black lines (these happen to be the duals of two infinity points that were present in the original diagram). Pappus states that if $A,B,C$ lie on a line $L_1$ and $X,Y,Z$ lie on a line $L_2$, then the points $AY \cap BX,BZ \cap CY,AZ \cap CX$ are collinear! They don't have to lie in any particular order, so we choose this order: Applying Pappus to the thin black lines and labelling the six points in question in this specific order, a direction application of Pappus gets that the three light blue points are collinear. This is sufficient to show that the three dark blue (violet?) lines are concurrent.

Answer 2

Here's a short but rather unenlightening algebraic proof. (I thought I had a nice simple more-geometric proof but it has a hole, so while I think about filling the hole I'll write this one out so there's a solution.)

The function f is the constant function with value zero, because the area is always 0, because all points pump(P) lie on a single fixed line, namely y=x. Proof: if P is at (x,y) the lines we're interested in are -yX+(1-x)Y=-xy and (1-y)X-xY=-xy. The point pump(P) is on their intersection and therefore also satisfies -X+Y=[-yX+(1-x)Y]-[(1-y)X-xY]=(-xy)-(-xy)=0 or X=Y.