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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1$\begingroup$ Welcome to puzzling :) What hexagons are allowed(convex, concave?) and what about non-Euclidean plane? $\endgroup$– ABcDexterMar 24, 2018 at 12:07
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$\begingroup$ Does the straight line have to perfectly divide the hexagon? $\endgroup$– Arnav BorborahMar 24, 2018 at 12:09
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$\begingroup$ Anything. The only requirement is that you do not get two quadrilaterals. $\endgroup$– MikeMar 24, 2018 at 12:09
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$\begingroup$ Although I think Euclidean geometry is implied. But I’d be satisfied with a non-Euclidean answer too. $\endgroup$– MikeMar 24, 2018 at 12:12
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1$\begingroup$ 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) $\endgroup$– IanF1Mar 24, 2018 at 14:35
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4 Answers
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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:
(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):
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1$\begingroup$ @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. $\endgroup$– BassMar 24, 2018 at 21:25
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$\begingroup$ 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. $\endgroup$– LawrenceMar 24, 2018 at 23:16
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4$\begingroup$ @Willtech splitting like that results in two pentagons (5 sides) and not quadrilaterals (4 sides) $\endgroup$– IvoMar 25, 2018 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:
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$\begingroup$ And with only one concave corner too! Very nice! $\endgroup$– BassMar 25, 2018 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
Version 3
working finally, I think,
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
A more symmetrical(ish) hexagonal that works is
my first attempt at this is below, it is not so obvious
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....
Version 2
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
This can be split into two quadrilaterals - thanks to @Rubio for the image of the split
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$\begingroup$ @Rubio - you beat me - I was just creating exacty the same image when you posted your comment $\endgroup$– tomMar 24, 2018 at 13:18
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$\begingroup$ @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 $\endgroup$– tomMar 24, 2018 at 13:40
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$\begingroup$ 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. $\endgroup$– MikeMar 24, 2018 at 13:44
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Here's mine:
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.
![hexagon[1]](https://i.stack.imgur.com/BUEQw.png)
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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.... $\endgroup$– tomMar 24, 2018 at 13:55
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$\begingroup$ 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, 2018 at 13:56
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$\begingroup$ It’s not specified but for the purpose of a challenge let’s assume extending the sides is not allowed. $\endgroup$– MikeMar 24, 2018 at 14:04
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2$\begingroup$ i just saw it as a soft version of en.wikipedia.org/wiki/Art_gallery_problem $\endgroup$– JMPMar 24, 2018 at 14:08
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2$\begingroup$ @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, 2018 at 14:12
