# How many matchsticks need to be removed so there are no equilateral triangles?

The puzzle is as follows [with minor copy edits for grammar]:

By only using matchsticks of equal length a set of equilateral triangles has been built as indicated in the figure from below. With that information, what is the minimum number of matchsticks that must be removed so that there aren't any of those triangles?

The alternatives given are:

1. 4
2. 5
3. 6
4. 7

This seems to be a variation of the typical puzzle involving the construction of a certain figure, as the task here is to delete the triangles.

I tried all sorts of combinations in removing the matchsticks, and my answers keep being off by one.

Since I'm not very good at this, could someone show me with a way to solve this sort of puzzle without much fuss?

My first guess was to remove systematically all the sides, but this would require moving many matchsticks. Then I thought to remove only the critical ones.

This is a matter of several trials. Needless to say that I ended up removing more than what the alternatives say.

Perhaps someone with more experience than me at solving this sort of puzzle could provide a strategy on where to look for removals first?

As I mentioned removing the edges of the hexagon inside don't help much and removing the exterior matchsticks also don't help as it increases the number.

What's the trick here? It would help me if answers included a drawing so I could understand the necessary removals.

This was obtained from my Puzzle challenges book which from its looks seems to be an adaptation from an IQ test from APA exams of the 1980s.

• Looks like there are many optimal solutions. We don't need another answer at this point, but the first that occurred to me was to rot13(erzbir gur erthyne urkntba). – aschepler Feb 25 at 0:08
• @aschepler I was already wondering why no-one's posted the rot13(Zvgfhovfuv ybtb) solution, but here it was, hidden in the comments :-) – Bass Feb 25 at 6:52
• Is removing just one so that it can be struck and used to ignite the others cheating? – Andrew Morton Feb 25 at 10:55
• Now that we have a number of valid answers, it would be interesting to see if we can calculate how many different ways (excluding rotations and reflections) there are to eliminate all triangles by removing the minimum number of matches... – Darrel Hoffman Feb 25 at 14:16

@hexomino's answer is correct and well-reasoned, as always. Here's another approach, which to me feels much more.. "axe-to-the-head" is what I'd call it in my native language, so I thought it might be interesting enough to warrant posting.

Lower bound: (This is what @trolley813's encrypted comment is saying.)

At least six. Every highlighted triangle needs to lose a match.

Upper bound:

Six is enough. A triangle has three sides, no two of which are parallel with each other. If there are only matches in two orientations, then it's just plain impossible to create a triangle even by moving or adding matches.

So if we remove, say, all the six horizontal matches, we won't even have to look at the picture to know that there aren't going to be any surprise triangles lurking in the shadows anywhere.

• This is the most elegant and the coolest solution to me :) – Bubbler Feb 24 at 23:07
• @Bass Thanks for that nice explanation. I was about to ask what was troller's encrypted message mean but your wordy answer made it all clear. Needless to say this answer is the most logical thus I'm accepting it. – Chris Steinbeck Bell Feb 25 at 2:06
• I think the most geometrical solution is to remove the regular hexagon. – obscurans Feb 25 at 7:01
• @obscurans sure, the Mitsubishi solution is very pretty and symmetrical. It still feels like looking for vampires though, and then finding the most efficient way to drive a stake through all their hearts (the triangles). The answer here, on the other hand, constructs a universe where vampire hearts cannot possibly exist, which is extremely effective but almost certainly overkill, both of which are characteristic properties of an "axe-to-the-head" solution. (I have no idea what the corresponding English term might be, if there is one.) – Bass Feb 25 at 7:14

I think the answer is

6

Reasoning

Note firstly that if we wish to destroy the triangles on the outside of the shape, we'll have to remove one match from each of these shapes. By removing the innermost match in each case we also destroy three more inner triangles thereby maximising the affect of this removal as shown in the image below.

After that, we see there are still three completed triangles in the middle. Since no two of these share a match, we are forced to remove three more matches to complete the task. This can be done in the following way,

As trolley813 mentions in the comments

6 is the lower limit here as there are 6 triangles with side length 1 which have no match in common (the triangles whose bases are horizontal) so we must, at least, remove one match from each of these.

• One small addition: rot13(Gur ybjre yvzvg vf 6, fvapr gurer ner 6 gevnatyrf (jvgu fvqr 1) juvpu qb abg unir nal pbzzba zngpurf, fb ng yrnfg bar zngpu cre gevnatyr unf gb or erzbirq.) – trolley813 Feb 24 at 11:14
• @trolley813 That's a good point, thanks. – hexomino Feb 24 at 11:15

Six again.

Explanation/Approach:

In this alternative set of solutions, you remove 3 alternate matches from the central wheel, and then one each from the corner (small) triangles while making sure that at least one of them is from the outer boundary (else, the big triangle formed by the outer boundary remains). An example solution from this set is:

• You only provided half of a solution. You showed a way to do it by removing X matches, but you didn't prove that there's no way to do it by removing X-1 matches. – Joseph Sible-Reinstate Monica Feb 25 at 16:59

My answer is 6 matches. If we remove three radii of the regular hexagon then it easy to remove three sides of triangles. By so doing, no equilateral triangle remains.

• Please add spoiler tags to your answers – samm82 Feb 24 at 23:01
• You only provided half of a solution. You showed a way to do it by removing X matches, but you didn't prove that there's no way to do it by removing X-1 matches. – Joseph Sible-Reinstate Monica Feb 25 at 16:59
• @ Joseph Sible-Reinstate Monica. When we remove one diagonal and one radius of the hexagon, 5 triangles have been eliminated. It is self evident. If we remove 3 sides of the hexagon no triangles are left. These facts prove that the minimum number of matches to be removed is 6. – Vassilis Parassidis Feb 27 at 1:58
• Those facts prove that IF you start by removing a diagonal and a radius of the hexagon, THEN it is possible to make all triangles disappear by only removing 3 more matches. Proving that removing 6 is enough isn't the same as proving that removing 6 is necessary. – Bass Mar 5 at 22:35
• @Bass. I am saying exactly what you are saying, that to remove 3 sides of the hexagon means moving 3 matches.. The most simple thing to say is to remove the six sides of the hexagon, but I wanted to make my answer more sophisticated. – Vassilis Parassidis Mar 6 at 1:05