# Make 4 congruent equilateral triangles with 6 matches in 2 dimensions [closed]

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.

• @ 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. Apr 4 '16 at 19:42
• Having fun trying
– humn
Apr 4 '16 at 19:46

My solution:

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

• Is it me or is this puzzle too simple? Apr 4 '16 at 16:12
• Updated for a prettier image. Apr 4 '16 at 16:22
• 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 Apr 4 '16 at 16:34
• 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. Apr 4 '16 at 16:40
• There wouldn't have to be, but now I'm not sure. It would essentially be a version of your answer I guess. Whoops. Apr 4 '16 at 16:48

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

*matches and angles aren't exact because I lack drawing skills.

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

• I also did not see this solution. 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.

Now edited: modified to fit correct constraints.

• @ charfellow - Those violate that they have to be four ** congruent ** equilateral triangles. (edit) Apr 4 '16 at 20:12
• Whoops. Missing a word makes you miss quite a bit. Apr 4 '16 at 21:19
• @OliveStemforn Removed those answers. This one should work. Apr 4 '16 at 21:31
• @ charfellow -----> +1 Apr 4 '16 at 21:59
• 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. Apr 4 '16 at 22:54

Sorry for ugly drawing :P

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

• Same size as PrisonMonkeys': both have side length 1/3 of matchstick length.
– Deusovi
Apr 5 '16 at 3:39
• This was my solution, excepting if it was rotated and/or flipped over from this. Apr 5 '16 at 6:09
• Also pretty much my #2 solution ;) Apr 5 '16 at 13:36
• @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. Apr 5 '16 at 17:10