Solution to Assignment #4


1. The bottom right-hand corner of a legal-size (8.5" by 14") sheet of paper is folded over so that it just touches the left edge of the sheet. What is the least possible length of the crease c?

Skeleton Solution. The first thing to notice is that the problem breaks down into three cases, depending on whether the crease runs from (i) the top edge to the bottom edge of the paper, (ii) the right edge to the bottom edge, or (iii) the right edge to the left edge of the sheet of paper.

[folded sheets]

The general idea is to find the minimum length of the crease in each case separately and compare the results to find the least possible length.

N.B.: Some people got confused because the question on the assignment included the phrase "as in the diagram below", and the diagram depicted an instance of case (ii). It was only intended to illustrate the bottom right-hand corner touching the left edge of the sheet, and not to to exclude other ways in which the paper might be folded over to get the job done. In the event, since the other two cases are surprisingly similar to each other, and quite easy compared to case (ii), they counted for only 2 points out of 10 between them.

First, let's tackle case (ii), the hard one. Consider the following diagram:

[diagram for case (ii)]

We'll actually minimize c2 (this makes the algebra a little easier), and then get the minimum value of c by taking the square root. Note that by the Pythagorean Theorem, c2 = x2 + y2.

To put c2 in terms of just one variable -- so we can throw calculus at it -- we'll solve for y in terms of x via h using various of the right triangles in the diagram. This will also require that we work out the possible values of x for this case. We'll leave that as an exercise! (Hint: Consider the special cases where the crease passes through one of the upper-right or lower-left hand corners.) Obviously, x is less than or equal to 8.5, and it turns out that x must be bigger than about 5.05.

Note that x2 = h2 + (8.5 - x)2 from the small triangle in the bottom left-hand corner, so that

h = [x2 - (8.5 - x)2]1/2 = [17x - 8.52]1/2.

From the larger triangle above it we have y2 = (y-h)2 + 8.52, so we get

y2 = (y - [17x - 8.52]1/2)2 + 8.52

after we substitute in for h. Simplifying and solving for y now gives

y = 8.5x / [17x - 8.52]1/2.

This lets us put c2 in terms of x alone:

c2 = x2 + y2 = x2 + (8.52 x2 / [17x - 8.52]) = 2x3 / [2x - 8.5] ,

after simplifying as much as possible.

It follows that there is critical point when

0 = (c2)' = [6x2 (2x - 8.5) - 4x3] / [(2x - 8.5)2] = [x2(8x - 6·8.5)] / [(2x - 8.5)2] ,

that is, when x= 0 or x = 3·8.5/4 = 6.375. x = 0 falls outside the range for x (about 5.05 to 8.5), but 6.375 is in it.

We need to evaluate c2 at the critical point and the endpoints to find the minimum value:

Thus the minimum value of c in case (ii) is achieved at the critical point x = 6.375, this minimum value being approximately about 11.04 (the square root of 121.9). (Inches...)

Next, what about cases (i) and (iii)? As was said before, these are surprisingly similar: in both cases you are looking to minimize the length of a line segment joining two parallel lines.

[parallel lines]

After a little geometry (or even calculus!), it shouldn't be hard to convince yourself that the closer the angle theta in the diagram is to pi/2 radians or 90 degrees (i.e. to a right angle), the smaller will the length of the line segment be.

How close can it get in case (i)? It can actually be a right angle if we fold the sheet vertically down the middle, so that the bottom right-hand corner ends up on he bottom left-hand corner. In this case, the parallel lines are 14 inches apart, and so the minimum length of the crease is 14 inches for this case. Case (ii) still gives the overall minimum...

How close can it get in case (iii)? A little reflection will show that the angle is closest to a right angle when the bottom right-hand corner is touching the upper left-hand corner. Working out the length of the crease in this instance, using the sort of geometric tricks we used in case (ii), gives a little over 9.9 inches. This is the minimum length of the crease for case (iii), and also the overall minimum.

Assignment #4
Math logoDepartment of Mathematics  Trent crestTrent University
Maintained by Stefan Bilaniuk. Last updated 1998.08.22.