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I’m kind of stumped. Logically if a hexagon has 6 sides and you split it and add one (the line) it will always result in two quadrilaterals. I may be off.

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    $\begingroup$ Welcome to puzzling :) What hexagons are allowed(convex, concave?) and what about non-Euclidean plane? $\endgroup$ – ABcDexter Mar 24 '18 at 12:07
  • $\begingroup$ Does the straight line have to perfectly divide the hexagon? $\endgroup$ – Arnav Borborah Mar 24 '18 at 12:09
  • $\begingroup$ Anything. The only requirement is that you do not get two quadrilaterals. $\endgroup$ – Mike Mar 24 '18 at 12:09
  • $\begingroup$ Although I think Euclidean geometry is implied. But I’d be satisfied with a non-Euclidean answer too. $\endgroup$ – Mike Mar 24 '18 at 12:12
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    $\begingroup$ Is a self-intersecting six edged figure still a hexagon? If so then I believe this would be a solution (where the third edge crosses back over the first to make a figure eight) $\endgroup$ – IanF1 Mar 24 '18 at 14:35
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This is way trickier than it seems.

That's why I'm not claiming with 100% certainty that this works, but I couldn't figure out a way to split this one:

enter image description here
(all the vertices lie on one (or both) of the two intersecting straight lines.)

Here's another, more symmetric candidate using the same idea (hope I didn't introduce any new problems by changing the layout):

enter image description here

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  • $\begingroup$ Awesome! I think you cracked it. $\endgroup$ – tom Mar 24 '18 at 15:21
  • $\begingroup$ love the second shape you found ! $\endgroup$ – tom Mar 24 '18 at 21:08
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    $\begingroup$ @tom yeah, me too! It kind of works like a magic trick; when you try splitting through one of the upper corners, and are trying to find a proper spot for the other end of the split on the horizontal bit: the part that isn't a quadrangle is pentagon, pentagon, still pentagon, soon going to lose a vertex but it's still a pentagon.. POOF it's a frigging triangle :-) A similar thing happens when trying to split through one of the concave corners, but in that case it's the quadrangle side that magically loses a corner at exactly the right moment. $\endgroup$ – Bass Mar 24 '18 at 21:25
  • $\begingroup$ The bottom "more symmetric" solution relies on the requirement to "split" the hexagon. Joining the bottom vertices results in an enclosing rectangle and a trapezium but doesn't split the original shape. $\endgroup$ – Lawrence Mar 24 '18 at 23:16
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    $\begingroup$ @Willtech splitting like that results in two pentagons (5 sides) and not quadrilaterals (4 sides) $\endgroup$ – Ivo Beckers Mar 25 '18 at 1:52
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Maybe I've misunderstood the question or I'm missing something obvious, but I think the following works:

enter image description here

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  • $\begingroup$ And with only one concave corner too! Very nice! $\endgroup$ – Bass Mar 25 '18 at 5:58
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Edit - I thought that there is no solution after @Rubio pointed out issues with version 3 - as he says if you can extend a line and cut along it then the version 3 figures fail --- I expected that the intented answer is Version 3 and that the cuts shown by Rubio were not allowed - or were not considered...

Now @Bass found a neat solution and @IanF1 has an interesting solution that might be allowed as well!! - all the answers here don't work.

Rubios images to show how things don't work

enter image description here enter image description here

Version 3

working finally, I think,

enter image description here

Note

that to get two quadrilaterals it is necessary to draw lines between 'opposite' corners... This fails for all three pairs in the image above as shown in the diagram below enter image description here

A more symmetrical(ish) hexagonal that works is

enter image description here

my first attempt at this is below, it is not so obvious

enter image description here

finally after 2 faied attempts I think this works....

thanks to Rubio for pointing out the problem with version two below

Rubio cut it like this....

enter image description here

Version 2

enter image description here

I think this is an answer {no it wasn't :_( }.... my first attempt below does not work - there is still one straight line that will split it into two quadrilaterals

Version 1

enter image description here

This can be split into two quadrilaterals - thanks to @Rubio for the image of the split

enter image description here

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    $\begingroup$ This? $\endgroup$ – Rubio Mar 24 '18 at 13:15
  • $\begingroup$ @Rubio - you beat me - I was just creating exacty the same image when you posted your comment $\endgroup$ – tom Mar 24 '18 at 13:18
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    $\begingroup$ ... This? $\endgroup$ – Rubio Mar 24 '18 at 13:19
  • $\begingroup$ @Rubio - ok - think this is ok now - see last edit and 3 attempts to cut the hexagon. -- bu a big bigt THANK YOU. i would not have got there without your helpful comments $\endgroup$ – tom Mar 24 '18 at 13:40
  • $\begingroup$ I knew that the answer had to entail all possible lines going outside the area of the hexagon in order for the solution to be valid but I just couldn’t visualize it. $\endgroup$ – Mike Mar 24 '18 at 13:44
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Here's mine:

hexagon[1]
The opposite vertices, AD, BE and DF, are made invisible to each other by the edges BC, CD and DE. Then all that remains is to ensure F can see both E and A.

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  • $\begingroup$ this is very similar to my version 3, which @rubio has proved is not a solution... extend your line CD till it cuts the hexagon in 2 then you get two quadrilaterals.... $\endgroup$ – tom Mar 24 '18 at 13:55
  • $\begingroup$ You're also in the same camp as the folks commenting on the main post: is extending one of the original hexagon's sides allowed? If so, this one is not a solution either, because this. $\endgroup$ – Rubio Mar 24 '18 at 13:56
  • $\begingroup$ It’s not specified but for the purpose of a challenge let’s assume extending the sides is not allowed. $\endgroup$ – Mike Mar 24 '18 at 14:04
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    $\begingroup$ i just saw it as a soft version of en.wikipedia.org/wiki/Art_gallery_problem $\endgroup$ – JonMark Perry Mar 24 '18 at 14:08
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    $\begingroup$ @Mike If extending a side is not allowed then there have been several figures already posted that cannot be divided into two quadrilaterals by a single straight line. Your "question" doesn't really ask a question - if the challenge is literally the title, then it's too broad because there's going to be a lot of ways to do it (assuming side extension is forbidden). You should either make the question "Can it be done?", or come up with tighter criteria so that there aren't an infinite number of solutions. $\endgroup$ – Rubio Mar 24 '18 at 14:12

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