Last night, I couldn’t sleep. And what does a geek do when he can’t sleep? He creates math puzzles in his head, that’s what. For some reason, when I closed my eyes, I visualized a square. Then I began splitting up the square using straight lines and taking note of the resultant shapes. (No, I’m not on drugs. And yes, it helped me sleep, thank you for asking.) When I woke up, I jotted down what I remembered, and a little tweaking created today’s Brain Game.
If you’re not a math geek, hang on. It could get bumpy. And you’ll probably want to look at the image to follow along. Click to go on. >>
I noted that by using two traversing lines, you could create three triangles (for 9 sides total), two triangles and a quadrilateral (for 10 sides), one triangle and two quadrilaterals (for 11 sides), and so on all the way to 16. But there was one exception: the one that would result in shapes with a total of 13 sides.
Is there an answer? I’m not certain. I might be overlooking something obvious. Of course, our site doesn’t have the ability to attach files to comments. But if you DO think you have an answer, sketch it out to a GIF or JPG file, then send it to trivia@mentalfloss.com and we’ll take a look at it. If you’re the first to successfully solve the problem, I’ll send you the mental_floss T-shirt of your choice.
And if no one can figure it out, then I henceforth decree that this puzzle shall now and forever be known as Sandy’s Paradox, and I officially throw my hat into the ring for a Fields Medal. Woohoo!
Comments (11)
Comment
- subscribe /
- give a gift /
- terms of service /
- about /
- contact us /
- write for us /
- ad info /
- subscriber services /
- press /
- privacy policy /
- rss /
© 2012 Mental Floss Inc. All Rights Reserved. Mental Floss is a registered trademark owned by Felix Dennis. mentalfloss.com is a trademark owned by Felix Dennis.
Disclaimer: All submissions made to mental_floss become the property of Mental Floss Inc and may be used without further permission by the submitter and without compensation to the submitter. If you do not agree to these terms, please do not submit anything to mental_floss.
you need three traingles and one quad to total thirteen sides. the first line is from on corner to its opposite corner. the second line is from one of the remaining corners to an opposing side.
posted by Bill on 7-15-2008 at 6:57 am
Like this:
h**p://xs229.xs.to/xs229/08292/13973.gif
posted by Marco on 7-15-2008 at 7:14 am
@Bill – Ha. You’re so right. This is cool.
But I’m sorry you don’t get a paradox named after you, Sandy. That would have been amazing.
Recaptcha: prizewinner maybe (Nope.)
posted by Eileen on 7-15-2008 at 7:15 am
Yep. Bill’s got it!
posted by Brendan on 7-15-2008 at 7:22 am
Bill’s absolutely correct. One line running from one corner to the opposite corner, then another line running from a third corner to one of the opposite sides, would give 13 sides (three triangles and a quadrilateral).
Congratulations, Bill. I’ll reach you via email and will arrange the delivery of the T-shirt of your choice. Great job!
(And thanks, Eileen. My wife still thinks I’m a paradox.)
posted by Sandy on 7-15-2008 at 7:24 am
I’m gong to send Marco a shirt as well, for taking the time to draw the diagram. It seems kinda obvious looking at it… I doodled a few dozen shapes, but none like that one.
Ah, well. No Fields Medal for me this year. I guess I’ll have to shoot for a county fair ribbon.
posted by Sandy on 7-15-2008 at 7:33 am
The most edges I can do with three lines is 28, can anyone best this? This problem seems to become more intractable with higher number of lines, yet I remember there being a discussion of this kind of problem before.
posted by Leadhyena on 7-15-2008 at 10:53 am
Followup: all numbers from 12 to 28 can be done with three lines, and I’m pretty sure that 28 is the max. Here’s a rough proof: It should be evident that the best you can do with zero lines is 4 sides. For each line, you are splitting 0, 1, or 2 edges on the boundary square, and you are creating two edges for each polygon that you cut across (that line segment counts for both polygons), and two edges for each line that you split (you create two edges that count twice from one edge that will count twice, and 4-2 is 2). Therefore the maximum number of sides that you can make is bounded by the sum of the number of exterior edges you create (0,1,2) and the number of interior edges you create (2*(polys cut+edges cut)). Since you can’t subtract sides with a cut, you can start with the max from the previous maximum number of edges for the given number of lines. So, if I start with zero edges and add one, the max number of internal regions I can cut is 1, so with cutting two external edges, the mx for one line is 4+2+(2*(1+0)) or 8. With two lines, I can now cut across two regions and one internal edge, so the max is 8+2+(2*(2+1)) or 16. With three edges, we take the max for two, and now we can cut across three regions and two internal edges, so the max for three would be 16+2+(2*(3+2)) or 28. If this is right, then the max for n edges would be the solution to the recurrence relation F_n+1=F_n+2+2*(n+n-1) or F_n=2n^2+2n+2.
Unsolved problems at this point are least number of edges for n lines and if all in the boundary between min and max can be accomplished. I’m not crazy enough to do four lines until I know I’ll get somewhere with it, and I have other work to do. This is a fascinating problem… Thanks Sandy!
posted by Leadhyena on 7-15-2008 at 11:51 am
Seems pretty likely that if n is the number of lines, the least number of edges would be 3(n+1). If all the lines intersect at the midpoint of one side of the square, two of the lines going to opposite corners of the square, and the remaining lines going anyplace, then the square would be divided into (n+1) triangles and 3(n+1) edges would be achieved. And it seems clear that you couldn’t get fewer edges than this.
posted by Mark on 7-15-2008 at 12:49 pm
Yay! I solved this too… It took me 4 doodles. :)
posted by Nerak on 7-15-2008 at 2:09 pm
@Mark: You are correct, and I’m surprised I didn’t see this simple representation. This would create the least number of edges, but it seems difficult to prove. With my proof on the max, I lean upon the result that the number of regions produced by n non-parallel lines intersecting in non-collinear points is n*(n+1)/2+1. There is no similar result for minimum number of regions, although it seems that if you can always produce n+1 regions all with 3 sides, that this would be the bare minimum, because producing polygons with less sides is against the rules.
Still up in the air is the question about whether there’s a number of lines and a number of edges between the max and min for that number of lines that isn’t achievable. This is an extremely difficult problem because I can only imagine a constructive proof for this, and it would be unbelievably messy.
posted by Leadhyena on 7-15-2008 at 10:58 pm