How would you make exactly four congruent equilateral triangles with just six matches of equal length in two dimensions?

No other triangles may be created when you are done.

Matches may not be bent, torn, or separated into other matches.

Match ends do not necessarily have to join other match ends. Specifically speaking, certain match ends might be free-standing.

Matches may rest across/intersect other matches.

The figure must possess exactly two lines of symmetry.

  • $\begingroup$ @ humn - Because the matches may rest across/intersect other matches, then that can include overlapping. If you have a potential solution with overlapping, it can be looked at. $\endgroup$ – Olive Stemforn Apr 4 '16 at 19:42
  • $\begingroup$ Having fun trying $\endgroup$ – humn Apr 4 '16 at 19:46

My solution:

enter image description here

(please excuse the non-exactness of the diagram, but it should still convey the answer)

  • $\begingroup$ Is it me or is this puzzle too simple? $\endgroup$ – Shuri2060 Apr 4 '16 at 16:12
  • $\begingroup$ Updated for a prettier image. $\endgroup$ – Shuri2060 Apr 4 '16 at 16:22
  • $\begingroup$ After seeing your answer, yes it seems pretty easy. But my mind didn't go to that solution, so it would be difficult to me $\endgroup$ – Raystafarian Apr 4 '16 at 16:34
  • $\begingroup$ Perhaps you are right then. I just felt that there surely would be many solutions given that there were few restrictions to what the matches could do. I mean, you even are allowed to put matches to one side if they aren't needed. $\endgroup$ – Shuri2060 Apr 4 '16 at 16:40
  • $\begingroup$ There wouldn't have to be, but now I'm not sure. It would essentially be a version of your answer I guess. Whoops. $\endgroup$ – Raystafarian Apr 4 '16 at 16:48

A solution that works for any x triangles with x+2 matches

enter image description here
*matches and angles aren't exact because I lack drawing skills.

  • $\begingroup$ I also did not see this solution. $\endgroup$ – Olive Stemforn Apr 4 '16 at 18:15
  • 5
    $\begingroup$ I quite like the extendability of this one. $\endgroup$ – Shuri2060 Apr 4 '16 at 18:35
  • $\begingroup$ Interesting; this is the first answer that doesn't involve three pairs of parallel lines. $\endgroup$ – Peregrine Rook Apr 5 '16 at 8:13

In addition to Question Asker's answer, here is another solution that satisfies the constraints:

enter image description here

  • $\begingroup$ I also did not see this solution. $\endgroup$ – Olive Stemforn Apr 4 '16 at 17:31

Three more solutions; Number one, like Rod's, extends to X triangles with X+2 matches, when X is even. Based loosely on Question Asker's answer.

Number two is also extendable, but less beautifully so.

And number three is extendable (X triangles with X+2 matches) as well.

enter image description here

Now edited: modified to fit correct constraints.

  • $\begingroup$ @ charfellow - Those violate that they have to be four ** congruent ** equilateral triangles. (edit) $\endgroup$ – Olive Stemforn Apr 4 '16 at 20:12
  • $\begingroup$ Whoops. Missing a word makes you miss quite a bit. $\endgroup$ – charfellow Apr 4 '16 at 21:19
  • $\begingroup$ @OliveStemforn Removed those answers. This one should work. $\endgroup$ – charfellow Apr 4 '16 at 21:31
  • $\begingroup$ @ charfellow -----> +1 $\endgroup$ – Olive Stemforn Apr 4 '16 at 21:59
  • $\begingroup$ 1 is practically the same as what Question Asker already provided, and 3 is a rotated version of what Rod has done. 2 is correct, although I think you might run into problems with the "no other triangles may be created" when you try to extend it. $\endgroup$ – PrisonMonkeys Apr 4 '16 at 22:54

Here is other answer

enter image description here

Sorry for ugly drawing :P

  • $\begingroup$ It's not an "ugly"drawing to me. I would guess it to be relatively difficult to approximate equilateral triangles with intersecting line segments. $\endgroup$ – Olive Stemforn Apr 5 '16 at 6:14
  • $\begingroup$ Seems a few of us ended up thinking along the same lines (ha, lines!) for this one. $\endgroup$ – charfellow Apr 5 '16 at 13:37
  • $\begingroup$ Yes, "great minds think alike" — this is almost identical to my answer, and, by extension (pun intended), similar to charfellow's solution #2. $\endgroup$ – Peregrine Rook Apr 5 '16 at 17:16

I know it isn't a contest, but my solution seems to have larger triangles than most of the others:

  • 1
    $\begingroup$ Same size as PrisonMonkeys': both have side length 1/3 of matchstick length. $\endgroup$ – Deusovi Apr 5 '16 at 3:39
  • $\begingroup$ This was my solution, excepting if it was rotated and/or flipped over from this. $\endgroup$ – Olive Stemforn Apr 5 '16 at 6:09
  • $\begingroup$ Also pretty much my #2 solution ;) $\endgroup$ – charfellow Apr 5 '16 at 13:36
  • $\begingroup$ @charfellow: Yes, similar, but all six lines are different (two are extended/compressed and four are displaced/slid).  Mine is clearly derived from a regular hexagram; your inspiration seems to have been something different. And, as PrisonMonkeys pointed out, your #1 solution is pretty similar to Question Asker's answer. $\endgroup$ – Peregrine Rook Apr 5 '16 at 17:10

How about this?

enter image description here

Please don't mind the blurry lines as I am not that good at PS.

  • $\begingroup$ Sorry I missed this line in question: No other triangles may be created when you are done. $\endgroup$ – Developer107 Apr 5 '16 at 4:54
  • $\begingroup$ Maybe you will reattempt it. $\endgroup$ – Olive Stemforn Apr 5 '16 at 6:18
  • $\begingroup$ * * * * * I would like to have an honorable mention for the following: mymathforum.com/math/… . . . . . . figure in post #9 * * * * * $\endgroup$ – Olive Stemforn Apr 6 '16 at 14:55

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