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Given a strip of 10 equilateral triangles, how many folds are necessary to reduce it to a single equilateral triangle? You may only fold along the grid lines. Multiple folds along collinear segments are not counted as one fold. Here's an illustration of a 10 triangle strip, for reference:

image

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    $\begingroup$ If you fold the paper along two colinear segments, does that count as one or two folds? $\endgroup$
    – 2012rcampion
    Commented Apr 26, 2023 at 19:46
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    $\begingroup$ @2012rcampion That's a great question. I've just started thinking about the problem today, so I'm on the fence. Which do you think would be more interesting? $\endgroup$
    – PuzzleAndy
    Commented Apr 26, 2023 at 20:13
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    $\begingroup$ Nice puzzle and good answers. If you do further puzzles of this form, consider counting origami folds. If the fold can be accomplished by moving a single chunk, it would count as one fold, even if the other side of the fold has multiple flaps. Hopefully this makes the count unambiguous. $\endgroup$
    – Lawrence
    Commented Apr 28, 2023 at 3:20
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    $\begingroup$ @Lawrence Ah, thank you! Much appreciated. $\endgroup$
    – PuzzleAndy
    Commented Apr 28, 2023 at 4:41
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    $\begingroup$ Intuitively, an estimate or upper-bound is ceiling(log_2(N)). Because each folding at best halves the number of triangles. $\endgroup$
    – smci
    Commented Apr 29, 2023 at 19:54

4 Answers 4

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Here is a solution that does it in 6 folds using colinear folds. Without, it would be 8.

First, fold along the halfway line

10 triangles, with a red line along the centermost fold

This takes us down to 9 visible triangles. Next we fold along the end trapezoidal shape.

Some folded triangles, with a red line indicating fold

We are down to 7 visible triangles. Folding along the long central edge...

An alien ship shape

Brings us to 5. Folding either of the inner lines...

5 triangles with a red line over on of the center folds

Gives us 4 triangles, and easy symmetry makes the last two steps clear.

4 and 2 triangles, each with a red line over their center folds.

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I know that the answer has already been accepted, but the question doesn't appear to require that the "single equilateral triangle" is the same size as one of the "10 equilateral triangles".

So there is a solution that only requires four folds as follows:

enter image description here

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  • $\begingroup$ This is genius. $\endgroup$
    – Ted Whiting
    Commented Apr 27, 2023 at 12:26
  • $\begingroup$ Agreed! Genius! $\endgroup$
    – PuzzleAndy
    Commented Apr 27, 2023 at 15:04
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With some "creative reading of the rules", I managed to create a single equilateral triangle folding only

3 times. (Although the co-linear fold counting rule may give this a different score)

Like so:

enter image description here

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    $\begingroup$ Woah dude! And the coloring? Smart! $\endgroup$
    – PuzzleAndy
    Commented Apr 27, 2023 at 15:05
  • $\begingroup$ Clever, but the OP presumably meant "a single equilateral triangle of sidelength 1" $\endgroup$
    – smci
    Commented Apr 29, 2023 at 19:53
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I believe this is optimal, but I have no proof.

Six folds marked in red, with shadow of previous step in grey.
Six folds

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    $\begingroup$ I'm going to give the answer to the other guy, just cause he answered first and may have been constructing his answer before I made the edit. But thank you for editing my post so quickly and thank you for your contributed answer. I really like the way you showed the fold lines and where the pieces were previously. $\endgroup$
    – PuzzleAndy
    Commented Apr 26, 2023 at 22:07
  • $\begingroup$ This answer is a real 6-folder even with the collinearity rules. Nice one! $\endgroup$
    – justhalf
    Commented Apr 29, 2023 at 10:27

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