[You might find it safer to use toothpicks instead, but I prefer matchsticks, and fire.]

This investigation is split into 2 parts. I will post the 2nd part tomorrow

In this investigation, we want to figure out the number of matchsticks that we need, to form several (unit) equilateral triangles. For simplicity, we will be restricting ourselves to a 2-D plane. (If you are brave, go ahead and try the 3-D version. It is quite challenging)

1) What is the minimum number of matchsticks that we need to form 1 equilateral triangle?

Clearly, we need at least 3, and 3 are sufficient.

2) What is the minimum number of matchsticks that we need to form 2 equilateral triangles?

Well, we could do it with $2 \times 3 = 6$. But if we allow them to share a common side, then we only need 5 matchsticks!

3) What is the minimum number of matchsticks that we need to form 3 equilateral triangles?

Continuing the above (almost like a snake), we see that we need $5 + 2$ matchsticks. Can we do better than that? I don't think so.

4) How can we (easily) form $N$ equilateral triangles using $2N+1$ matchsticks?

**Hint:** Do you spot a pattern above? Can you explain in detail how it works? Mathematicians use "Induction" as a way to formally express the pattern that they see.

5) What is the minimum number of matchsticks that we need to form 6 equilateral triangles?

From the previous question, we see that $13$ is enough. Can you do better than that?

**Hint:** Yes we can!

This investigation continues in Part 2.

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## Comments

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TopNewest5) By arranging the six triangles in a regular hexagon we only need 12 matchsticks.

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12

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9 matchsticks needed in a 3D pyramid form and 12 needed in 2D hexagon form...

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I've been wondering: What is Cosines Group and how can I join? It seems like a cool thing to do, posting simple concepts that people can learn from.

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A few weeks ago, Calvin sent emails out to ~20 people and were split into three groups: Cosines group for simple concepts, Torque group for olympiad math, and Goldbach's Conjurer's which investigates the most advanced math concepts out there (note that each group chose their own name)

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i like to join cosine group tnks

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#CosinesGroup. You can also interact with the current posts, add your comments to the notes or work on the problems.

Certainly. The easiest way to participate would be to make a post with a tag ofLog in to reply

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Remember that you can always "Edit this discussion" later on, if it doesn't display to your liking.

The same would apply for adding images to problems (and once we build up the "Edit this problem" feature.

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Hi Daniel,

Anyone can go ahead and start posting relevant material in the different #Groups, whenever they have something on their mind. If you wish to contribute on a regular basis (1-2 posts per week), send me an email (Calvin@Brilliant.org) and I'd help to get you set up. This way, having a core group of contributors allows us to continually present material.

You can get more information about the different groups in the Math Circles note.

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I'm a little iffy on posting 1-2 times a week. I might do it, and I might not; can I just go ahead and do it anyways without emailing you?

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For number 3) I see a way to do it with 6 matchstick (this does go beyond what the poster is covering, but it can be done for all you ambitious students out there). For number 5, you can do it with 7.

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Are you referencing the matchstick puzzle from another note? Keep in mind that the solutions are assuming that matchsticks don't overlap. Most of the solutions in the other note use overlap, although they are alternate solutions.

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Ahh. I was assuming that the matchsticks could overlap. You can do problem 5 with 10 matchsticks even with the restrictions, though.

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form a trigonal bypyramid and 7 eq triangles can be formed with only 9 matchsticks

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by a triangle pyramid on two sides 7 triangles can be formed from 9 sticks theyyunni

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7 match sticks are enough to form six equilateral triangles

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by overlaping match sticks we can form 6 equilateral triangles in 7 match sticks

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Overlapping doesn't count as we are in 2-D.

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form another equi triangle PQR with 3 sticks such tat one vertex P touches the middle of BC and the remaining side QR has Vertex A at its center.

Now ter will be two new vertices formed in between M and N..keep another stick XY horizontally .

You will get six equilateral triangles having side size of half the stick. BPM,PMN,CPN,AMQ,AMN,ANR.

This method is by overlapping. USING 7 STICKS

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The equilateral triangles should have side length equal to the side of the matchstick.

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they din mention as such frnd

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12 sticks. A hexagon cut into 6 equilateral triangles.

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3+2+2+2+2+1=12

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If we allowed overlapping (which we wouldn't for 2d but will for 3d) the we could actually create 6 equilateral triangles with 6 matchsticks. Seeing as there was no restriction to the creation of non-equilateral triangles you can just make the star of david (a six pointed star with a hexagon in the middle). Also if we allowed for the counting of triangles that are split into sections as well as the sections then this star of david would create 8 equilateral triangles. If you were restricting the creation of non-equilateral triangles but not restricting the counting of triangles that are split into sections as well as the sections then you could you another method and only use 6 matchsticks. As Karthik Dayal said: "form an equi triangle ABC with 3 sticks like one vertex A at bottom and remaining two B,C at top. form another equi triangle PQR with 3 sticks such tat one vertex P touches the middle of BC and the remaining side QR has Vertex A at its center." However, at this point we do not need to put a matchstick through M and N as we can just count the two larger triangles ABC and PQR.

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Read the next post - https://brilliant.org/discussions/thread/triangle-matchstick-investigations-2/.

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5 could be done with 8 matchsticks if you think in 3-D. One triangular pyramid which consists of 6 matchsticks will have 4 equilateral triangles and then 2 more matchsticks for the fifth.

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This investigation wants to generalize finding the minimum for $n$ triangles. The version in 3-D is slightly harder. You should give that a try.

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1 MATCHSTICK LENGTH IS 4.5 cm AND IT CAN BE EQUALLY DIVIDE INTO 3 PARTS WHERE EACH SIDE OF THE TRIANGLE IS 1.5 cm SO WE JUST NEED 4/1/3 OF THE MATCHSTICKS TO GET THREE EQUILATERAL TRIANGLES

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