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How will you create a 3 inch (within say plus or minus 0.05 inch) side on a standard piece of paper ( 8.5 x 11 inches) merely by Folding? No marking of any kind allowed. Only one paper available. Please explain why. Please try to do with minimum number of foldings. Of course no other measurement tool available

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  • $\begingroup$ Is the purpose to find any three inch length (between referenced points) anywhere on the paper? Or is the purpose to create a shape that has a "three inch side" between two of its adjacent corners? $\endgroup$ – Jasper Feb 24 '17 at 5:53
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    $\begingroup$ Why has a late answer with the same content as an existing answer been accepted? $\endgroup$ – Jonathan Allan Mar 8 '17 at 13:14
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I can do it in

2

folds, by

folding one corner down to the opposite side, then folding the top of the paper down to the bottom:
enter image description here
The diagonal line splits the right side into one line of 8.5" and another of 2.5", and the middle line splits the right side into two lines of 5.5". Thus the red line is 5.5" - 2.5" = 3" long.

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  • $\begingroup$ Either order should work. ​ ​ $\endgroup$ – user1579 Feb 19 '17 at 5:55
  • $\begingroup$ Amazing! Who would've thought it? Perhaps a mathematical explanation for why this works could be added. $\endgroup$ – DepressedDaniel Feb 19 '17 at 6:02
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    $\begingroup$ @DepressedDaniel I've included that below the picture. $\endgroup$ – DooplissForce Feb 19 '17 at 14:57
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    $\begingroup$ Agreed. Either way found this folding interesting $\endgroup$ – DEEM Feb 19 '17 at 15:14
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    $\begingroup$ Please do not forget to reward successful solvers with their $\color{green}{green \checkmark checkmark}$ to mark their answer Accepted! $\endgroup$ – Rubio Feb 22 '17 at 0:30
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Here's a solution with 2 folds.

Fold the 11" side in half to get a 5.5" X 8.5" rectangle. Fold the 4.5" side on the 8.5" side as shown.

[see image for better understanding]

https://i.stack.imgur.com/6BFu8.png

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Here's a solution in four folds.

Fold on a diagonal to create a 8.5 x 8.5 triangle with 2.5 inches sticking out.
Fold that 2.5 inch strip over and unfold the diagonal to create a single 2.5 inch fold at one end of the rectangle. The rectangle is now 8.5 x 9.5 inches.
Fold that 2.5 inch strip over itself again to create a 6 inch long side.

Fold the 6 inch side in half to create a 3 inch side.

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    $\begingroup$ It seems like it should be possible to do what you describe in three folds. $\endgroup$ – Peregrine Rook Feb 19 '17 at 20:28
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Here is a solution in

3

folds:

1. Fold the 8.5 inch dimension in half, to get a 4.25" x 11.00" rectangle.
2. Fold one of the two halves in half, to get a 2.125" wide strip.
3. Fold a raw corner (of the original sheet of paper) of the 2.125" wide strip to the crease, along a 45° angle (that starts at one end of the crease).
2.125" * sqrt(2) ~ 3.005", which is within one paper thickness of 3".
(There will also be error in the placement of the creases, but this algorithm will mitigate the cumulative error.)
Unfold the two strips.
The result is an 8.50 x 11.00" rectangle, with a 3.0" long dog-ear.

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  • $\begingroup$ Alternately, one can create this same length (again using only three folds) by folding a raw 8.5" edge to lie along a raw 11" edge, with the fold bisecting one of the corners of the page; and then folding the resulting diagonal edge in half twice. $\endgroup$ – Michael Seifert Feb 21 '17 at 15:22
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The following method can be used to turn the 11 inch long edge into a ruler with marks every inch. The three inch mark is found after 8 folds. A total of 15 folds makes a ruler with a mark every inch:

The labels are optional; they serve only to uniquely identify the points and creases. Label the ends of a short side NW and NE.
Label the corresponding ends of the other short side SW and SE.
Clockwise from the top right, the points are NE, SE, SW, and NW.
1. Crease between NW and SE. Unfold.
2. Pinch a mark halfway between NE and SE. This mark is point $1/2$.
3. Crease between SW and $1/2$. Unfold. Where this crease crosses NW-SE is point $1/3$.
4. Fold the raw edge SW-SE to lie on point $1/3$, such that raw corner SW lies along the NW-SW raw edge, and raw corner SE lies along the NE-SE raw edge. Pinch a mark at the NE-SE raw edge end of the fold. Unfold. This mark is point $1/6$.
5. Fold the raw edge SW-SE so that point SE lies on point $1/6$, and point SW lies along the NW-SW raw edge. Pinch a mark where the NW-SE crease crosses the new fold. Unfold. This mark is point $1/12$.
6. Crease from point SW through point $1/12$ to raw edge NE-SE. Unfold. The NE-SE end of this crease is point $1/11$.
7. Pinch a mark halfway between NE and $1/11$. This mark is point $6/11$.
8. Pinch a mark halfway between $6/11$ and SE. This mark is point $3/11$.
If folded perfectly, point $3/11$ is exactly 3 inches from the SE corner.

Finishing the ruler requires 7 more pinches:

a. $7/11$ is halfway between NE and $3/11$.
b. $9/11$ is halfway between NE and $7/11$.
c. $10/11$ is halfway between NE and $9/11$.
d. $5/11$ is halfway between SE and $10/11$.
e. $8/11$ is halfway between NE and $5/11$.
f. $4/11$ is halfway between SE and $8/11$.
g. $2/11$ is halfway between SE and $4/11$.

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Since people are posting alternative solutions, I might as well post what I had just as @DooplissForce was posting their answer. It's less elegant since it's not an exact answer.

Fold the paper in half to create midpoints on the shorter edges. Then fold one of the midpoints onto one of the opposite corners.

This creates a point each on the longer edges, one of which provides a ~3.04 inch measurement.

enter image description here

In the diagram above, we fold line CD on line BA to create G and E, then point E onto point A.

If we denote D as (0, 0), then E is (4.25, 11). FH is a perpendicular bisector of EA by the second fold, so their intersection is (6.375, 5.5) and the slope of FH is 4.25/11.

To find the height of F above D we take 5.5 - 6.375*(4.25/11), giving ~3.0369.

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  • $\begingroup$ Wow. Doopliss probably got the intended answer, but this is awesome. I kept on trying to work out a solution despite his answer since I believe the more answers the better. I feel it inculcates a sense of 'not settling' and discovering new techniques and solutions, even if there exists a perfect solution already. $\endgroup$ – Karan Atree Feb 22 '17 at 14:34
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A Solution in 2 folds

Pictorial Representation of the folds

enter image description here

Proof

By measurement (using a ruler and folding hence not 100% accurate)

AB = ?
BC = 5.9 cms
AC = 9.7 cms
AC² -BC² = 9.7² - 5.9² = 59.28
sqrt(59.28) = 7.699 cms = 3.03 inches

Could someone help me with the maths behind this? I'm not entirely sure how to present a general mathematical solution for this.

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  • $\begingroup$ wow. that is a new and interesting one $\endgroup$ – DEEM Feb 21 '17 at 15:16
  • $\begingroup$ This drawing does not have 17 x 22 proportions. The angles cannot be directly copied onto a sheet of paper. $\endgroup$ – Jasper Feb 21 '17 at 17:54
  • $\begingroup$ Is the procedure? Fold the northwest corner down to the southeast corner. (AC will be along this crease.) Fold the raw edge (that used to be the top edge, and now is to the right of the original rectangle) back toward the left, along the original right edge of the rectangle. (Point A will be where the original top edge crosses the AC crease.) Fold crease AB by making a crease perpendicular to the right edge of the original rectangle that goes through point A. Unfold to taste. $\endgroup$ – Jasper Feb 21 '17 at 18:08
  • $\begingroup$ This procedure requires three folds, not two. (The third fold is needed to mark point B.) $\endgroup$ – Jasper Feb 21 '17 at 18:10
  • $\begingroup$ After using six folds to make the 11.00" x 7.333" shape of the drawing, and following the three folds of the procedure I listed in the comments, I get distances AB ~ 2.6 inches, AC ~ 3.1 inches, and BC ~ 1.7 inches. $\endgroup$ – Jasper Feb 21 '17 at 19:51

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