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This is a problem which baffled me for years. Many possibly know the solution. See the figure below. $ABCD $ is some arbitrary quadrilateral. Each side is divided evenly into three parts. It is asserted that the quadrilateral shaded has an area 1/9 of that of the $ABCD$.

My first idea was using affine or projective geometry. But the affine transform cannot transform $ABCD$ into a square. The projective transform can, but it does not preserve the ratio (no longer evenly divided).

So, could you provide the solution for me?

enter image description here

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marked as duplicate by Chris Culter, Hans Lundmark, achille hui, user354271, Lord Shark the Unknown Aug 31 '17 at 9:57

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  • $\begingroup$ I think we should consider only convex quadrilaterals. Or do you think the statement holds also for concave quadrilaterals? $\endgroup$ – Crostul Aug 31 '17 at 7:33
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    $\begingroup$ @Crostul With help of an CAS, it also seems to work for concave quadrilaterals. $\endgroup$ – achille hui Aug 31 '17 at 7:46
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    $\begingroup$ There's a claimed positive proof here: books.google.com/books?id=vlxi-RwZ8lcC&pg=PA112 Liong-shin Hahn (2005) New Mexico Mathematics Contest Problem Book problem 141, solution on page 112 $\endgroup$ – Chris Culter Aug 31 '17 at 8:17
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    $\begingroup$ And here: geometer.org/mathcircles/Area.pdf Tatiana Shubin and Tom Davis (2014) "Geometry (Mostly Area) Problems" problem 6, solution on page 29. $\endgroup$ – Chris Culter Aug 31 '17 at 8:21
  • $\begingroup$ I see. It has noting to do with affine or projective geometry! $\endgroup$ – Beamer Aug 31 '17 at 8:26