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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


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