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

enter image description here

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)


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):

enter image description here

Source: Me.

  • $\begingroup$ @hexomino Nope, the picture is correct. What may be off is my verbal description. By shifting an axis by one unit I mean, for example, that the equation for the x axis (y=0) becomes y=1 after shifting. So if A=(x,y) then x0 = (x,0), x1 = (x,1). $\endgroup$ – loopy walt Mar 28 at 18:01
  • $\begingroup$ Yep, my apologies, I understand now. Thanks for clarifying. $\endgroup$ – hexomino Mar 28 at 18:06

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.
dual (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.


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: pappus 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.

  • $\begingroup$ Yes, that's what I had in mind. Well done! Btw., apparently it is called Pappus' hexagon theorem in English and the wikipedia article even has a section on its dual. And there is a picture labelled dual theorem: affine form. Note: I only found this after publishing the puzzle. I may be lazy wrt puzzle design (Take famous theorem and obfuscate.) but not that lazy. $\endgroup$ – loopy walt Apr 1 at 6:32
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    $\begingroup$ You being a bit of a projective geometry expert, do you agree that @Gareth's answer is essentially a proof of Pappus? $\endgroup$ – loopy walt Apr 1 at 6:52
  • $\begingroup$ I suppose so. The original problem was essentially a purely projective one in that it can be obtained from the general case by sending some points to infinity and using the other two degrees of freedom to make the square. Then solving this specific case will solve the general one via projective transformations, which then leads to Pappus via duality. $\endgroup$ – greenturtle3141 Apr 1 at 19:33

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.

  • $\begingroup$ Yes, they should. It occurs to me that there's a geometrical proof that's kinda equivalent to this one, which I'll edit into my answer. $\endgroup$ – Gareth McCaughan Mar 28 at 14:23
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    $\begingroup$ Yeah, there's a definite projective-geometry vibe to this. (I take it that's part of what your recent hint is gesturing towards.) In particular, of course the result is unchanged if you apply a projective transformation to the whole picture. And the exact picture in your second hint is one that I've had in my head, though mostly as a mistake because of course it doesn't quite directly match the pumping process. Unfortunately I'm kinda rubbish at projective geometry :-). $\endgroup$ – Gareth McCaughan Mar 30 at 0:16
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    $\begingroup$ Actually, I'm getting a bit nervous that I made some basic mistake, seeing as now two formidable maths buffs seem unable to make the intended simple last step. But assuming I didn't, using the principle @greenturtle3141 mentioned Gareth's algebraic argument could I think be made into a nice proof of the thing I seem to have hidden so darn well. $\endgroup$ – loopy walt Mar 30 at 10:33
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    $\begingroup$ Quite fun watching this discussion unfolds, haha. Geometry is one of my weakest subject too, so I don't expect myself to be able to contribute much to this already great answer. Haha. Hopefully someone will see through loopy's trick soon! $\endgroup$ – justhalf Mar 30 at 12:51
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    $\begingroup$ ... Oh, the bounty. I see. No, no objections to that :-). $\endgroup$ – Gareth McCaughan Apr 1 at 10:30

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