Matchstick Puzzle
How many triangles at most can you make using six matchsticks?
Details and Assumptions:
- A matchstick is assumed to be a unit segment. You cannot break the matchsticks.
- The triangles don't necessarily have the same size, but they may not be degenerate.
- Vertices of triangles must be intersections of matchsticks.
———
Yes, very obviously inspired from this .
The answer is 20.
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4 solutions
Moderator note:
Simple standard approach.
You did not say it must be the maximum triangles.
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they asked how many triangles AT MOST YOU CAN MAKE? got it? :D
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The original version of the problem didn't state "at most"; I added it after this comment to clarify that intention.
It was implied to be so, but edited to make it clearer.
Where are 20 triangles...............?
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Any three matchsticks together form a triangle.
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I know this but how six matchsticks can make 20 triangles
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@Qasim Qureshi – Take any three matchsticks; they form a triangle. There are 2 0 sets of three matchsticks in total.
I remember doing it in our geometry class. The only difference was instead of matchsticks it was straight lines ;)
You're right. I counted more than 12 in some seconds.
Did it the same way.But I don't get why its a level 4 problem.Anyways I've up-voted your solution :)
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I don't understand why it is Level 4 either.
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I like that picture btw explaining how you can achieve 20 triangles from just 3 matchsticks.
Shouldn't this be in Combinatorics?
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It probably should be, but I like to put it in Logic and see how people that are so used to those matchstick problems immediately answer it without a more careful reading of the problem.
In that picture there is only 6 triangles
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I think you are only counting the small triangles.
It's hard to count all 20 triangles. But you can try it with less matchsticks like 3, 4 & 5. There will be 1, 4 & 10 triangles.
I know the problem was edited to make it more clear, but it is still unclear. A prima facie reading of the problem suggests that the matchsticks should be formed once into a structure which yields a certain number of triangles. In that case, the maximum number is 4.
After reading the solutions, it's become clear to me that you are asking "how many triangles can be formed by subsequent non-identical combinations of the matchsticks?" Not fair.
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Yes, the matchsticks are formed once into a single structure. The conditions that differentiate this from the usual problems of this type are that you may potentially have parts of matchsticks that aren't used, and that the vertices of the triangles don't have to be the endpoints of the matchsticks. 20 triangles are present in the above example.
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My mistake. I misunderstood one piece of the puzzle...I thought it was saying the vertices needed to be the endpoints. Re-reading it, I see they merely need to be an intersection. Good point. I should pay more attention :)
Additionally, even with that assumption, the correct answer is not 20, but 30 (20 combinations of 3, 6 combinations of 4, 3 combinations of 5, and 1 combination of 6). All of these combinations can produce a triangle according to the rules given. My vote: poorly constructed question, and incorrect answer.
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If two matchsticks together form a single side of a triangle, it's impossible to have triangles where the two matchsticks form different sides. Thus making extra triangles by using more matchsticks comes with the cost of removing some of the original triangles.
since 3 matchsticks forms a triangle then num of triangles formed 6C3 =20
I put the same solution but when I saw yours one I delete my solution,
:D
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You do realise that @ghanshyam Sharma 's solution is a copy of the same solution that @Ivan Koswara sir posted. Generally ( r n ) is the same as n C r
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i didn't copy anyone's solution ,,,i just posted my own ,,,it may be similar to someone's
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@Ghanshyam Sharma – Your solution is similar to the Ivan koswera sirs solution. You probably didn't notice it because you might have not known that ( r n ) is the same as n C r .
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@Athiyaman Nallathambi – i knew that these notations are same but i didn't see anyone's solution , i just posted mine
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@Ghanshyam Sharma – Well, we should look at the existing solutions before posting a new one. Many of us know what the solution is, but if each of us posts the similar one it won't be anything more than a garbage.
You must have seen some problems (specially famous ones) that has numerous similar solutions in the top of the solution list. The problem there is if anyone posts even an extraordinary solution which differs from the others, most likely it won't get any views. So, we should all keep that in mind to keep the solution section clean and to add more perfection in Brilliant.
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@Md Omur Faruque – ok ,,i know ...that was my first day on brilliant and now i know all these things that you are saying .so end this discussion here because it doesn't make any sense now
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@Ghanshyam Sharma – you are active since 2014 and you are saying it is my first day in brilliant !! this make no sense
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@Syed Baqir – yes i have created brilliant account in 2014 but i was not using it from that time ,,i just started using it at the day when i posted this solution...and it makes sense =D
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@Ghanshyam Sharma – Lets just end this discussion.Next time once you solve a question , check the solutions before posting your own solution.
to make up a triangle we need Three Edges... that means Three Matchsticks in this case...... So the number of ways of doing that are ... Selecting Three Matchsticks out of the total Six given.. that is ... 6C3 ..ways ..
we can achieve 20 Triangles of Different Sizes, If we assume Intersection of Matchsticks as Vertices of Triangle And Make it as some what like Five pointed Star with some more triangles in it
Clearly any three matchsticks can only form one triangle, so the maximum number is ( 3 6 ) = 2 0 triangles. On the other hand, achieving this is very easy:
A few examples of the triangles: