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
I can do it in
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.
Here is a solution in
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.
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 $4/11$.
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.
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.
A Solution in 2 folds
Pictorial Representation of the folds
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.