So, the triangles are the same size, they are equilateral triangles. Can you use an odd number of these, to create a rectangle? If so how many, if possible post an image of your answer!

Answer with the least amount of triangles gets Best Answer

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    $\begingroup$ I can't even see how you could make a rectangle with any number of equilateral triangles. $\endgroup$ – itdoesntwork Oct 7 '14 at 23:08
  • $\begingroup$ The answer is: no. $\endgroup$ – Florian F Oct 7 '14 at 23:20
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    $\begingroup$ @itdoesntwork draw a square, draw an X through it $\endgroup$ – Red Alert Oct 7 '14 at 23:33
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    $\begingroup$ @RedAlert Those wouldn't be equilateral. $\endgroup$ – jlars62 Oct 7 '14 at 23:45
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    $\begingroup$ Since a rectangle requires 90-degree corners, and every angle inside an equilateral triangle is 60-degrees, it's impossible. The only multiple of 90 that you can make is 180-degrees (or any multiple of it). You can make rhombs, but no rectangles $\endgroup$ – Joe Oct 7 '14 at 23:59

Honest answer: no, since a rectangle requires 90-degree corners, and every angle inside an equilateral triangle is 60-degrees. The only multiple of 90 that you can make is 180-degrees (or any multiple of it). You can make rhombs, but no rectangles.

Getting funky and a bit creative, yes... in a way.

Since I don't know of any proper tools for this, I've used Euclid: the game for it. Here's a screenshot of the rectangle CGJK made from equilateral triangles, circles and intersections.


It occurs to me that only 3 triangles are needed since CEJ* is a rectangle, since CE is the radius of circle C, but CJ is shorter than that. The answer below follows from before I noticed that, and thought you needed 5 triangles.

It uses a total of:

  • 5 equilateral triangles
    • ABC
    • BCE
    • BEF
    • EFG
    • FGH
  • 7 circles
  • 2 points (A, B)
  • 9 intersections (C, D, E, F, G, H, I, J, K)

To create it, you do the following (my mathematical notation is terrible, so this is just mostly English and not proper math-speak):

  • Create any 2 points (A, B)
  • Join them with a segment (AB)
  • Plot a circle outwards from A to a radius of AB, likewise for B
  • Plot points C and D where they meet
  • Join C and D with a segment (I used a ray, but meh)
    • this creates the 2 right angles on the left hand side
  • Create a circle outwards from C with a radius BC
  • Plot point E on the intersection of B's circle and C's circle
  • Create another equilateral triangle BCE
  • Create a circle outwards from E with a radius EB
  • Keep plotting intersections, triangles and circles until you get to H
  • Plot intersection point I at the bottom intersection between the circles F and H
  • Connect GI with a segment to create the right-hand edge of the rectangle
  • Plot points J and K just to be neat
  • $\begingroup$ Wow! I'd like to see a better response then that, I would've accepted just the first part but +1 for getting creative! Well done! $\endgroup$ – warspyking Oct 8 '14 at 0:57
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    $\begingroup$ I don't understand why this qualifies as creating rectangle from triangles. $\endgroup$ – justhalf Oct 8 '14 at 4:21
  • $\begingroup$ @justhalf because the rectangle is composed entirely of triangles, and it's been constructed purely from geometry. It's possible to say with certainty that all the triangles are equilateral, and that the rectangle angles are correct, and that the tops and sides are different lengths. It's not the best question, since the actual answer is "no", but it's the closest you can get to a "yes" $\endgroup$ – Joe Oct 8 '14 at 8:06
  • $\begingroup$ Oh, also to clear one technicality up @justhalf, the question says the triangles must form the rectangle, and they must be equilateral. It doesn't say you can't use circles, points and segments to help you plot the triangles (although the vertical segments/rays is pushing it) $\endgroup$ – Joe Oct 8 '14 at 10:32
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    $\begingroup$ @justhalf I agree, it's not possible with just the triangle - my answer is basically a get-out-of-jail-free card for a question that doesn't have an interesting answer :P It was fun trying to think of something, though $\endgroup$ – Joe Oct 8 '14 at 10:43

You didn't say they can't overlap. I put a crude example - 5 triangles. Recall that a square is a rectangle whose sides are of equal length.

enter image description here

Think outside the equilateral triangle warspy

  • $\begingroup$ Lol that last statement :D $\endgroup$ – warspyking Oct 8 '14 at 12:45
  • $\begingroup$ I knew you'd appreciate it lol $\endgroup$ – d'alar'cop Oct 8 '14 at 12:47
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    $\begingroup$ So, I could be overlooking something, but that should be 4 triangles. Right? $\endgroup$ – Xrylite Oct 8 '14 at 23:13
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    $\begingroup$ @Xrylite Yes :) I just slapped an extra copy of the triangle on top - because we needed an odd number :p $\endgroup$ – d'alar'cop Oct 8 '14 at 23:14
  • $\begingroup$ @d'alar'cop I spent too long reading and thinking that I eventually discarded the 'odd' requirement. I knew it wouldn't be that simple to go through the work of creating the art, but mess up the structural answer. ^^ $\endgroup$ – Xrylite Oct 9 '14 at 0:06

Another way of interpreting the question treats it as a classic Dudeney problem about dissections, with solution:

Dissection of square and equilateral triangle

Image from Wikimedia, released into the public domain by its creator.


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