One eighth. 1/8th
The easy way to find 1/8th of the length of a line segment is to divide 3 times. On a square, fold two adjacent corners together to get half, fold the corner to the center that you just found and get a quarter, and finally one more time, fold paper so the corner lands on the one-quarter mark. That's three steps.
An interesting shortcut requires only two steps if you have a square. Make a pinch or mark halfway between two adjacent corners, let's say on the right side: upper right corner and lower right corner are lined up and the spot halfway between them is marked with a pinch. Take a corner on the opposite side of the square, say the upper left corner, and put it on top of the midpoint of the right side that you just marked. You don't have to make a full fold, but just find the place (this will be on the bottom of the square if we are using the above landmarks) where the fold intersects the bottom edge. Pinch this spot and you will have marked 1/8th of the distance along the bottom of the square. That's two steps.
Challenge: prove this using the equation of a line going through two points where the slope of the line is expressed by the equation:
m = ( y0 - y )/( x0 - x )
where (x0, y0) is a specific point and x and y are variables on the x-y coordinate system. For a straight line, m a constant.
Hints: Using the above references, and starting with a unit square (length 1 on each side), the midpoint between the top right and bottom right is (1, .5)(let's call that point B) and the upper left corner of the square is (0,1)(let's call this point A). The midpoint between those is -- (.5, .75)(let's call this C). The line that forms the fold is the perpendicular bisector of the line between the upper left and midpoint of the right side.
The slope of the line through A and B is m = -1/2. Its perpendicular is -1/m which would be m1 = 2. The perpendicular bisector of segment AB would have a slope of 2, going through point C (.5, .75). The equation for that line would be m = (y - .75)/(x - .5) = 2.
Sooo, 2x -1 = y - 3/4
y = 2x -1/4 = 0 where the square is on the x axis.
2x = 1/4
x = 1/8
Voila, the fold is exactly 1/8th of the length of the side of the square, in two folds, if you like, or two pinches, since there is no need to actually fold the crease, just pinch and mark the spot.
More: if you mark two of these on the same square, and put one of the 1/8th marks on top of the other (that's five pinch marks), the part where the last fold intersects a side, is 3/7th of the length of the side of the square. This five-pinch procedure yields a particularly interesting artifact, that the side of a square can directly and exactly arrive at an odd numbered subdivision of a square. Other exact odd-numbered divisions from folding are demonstrated in http://www.langorigami.com/wp-content/uploads/2017/09/origami_constructions.pdf.
Extra credit: prove with algebraic equivalents that the last fold is actually 3/7 of the distance along one of the edges of the square.
