Question

A rectangular piece of paper is 12 inches high and six inches wide. The lower right-hand corner is folded over so as to reach the leftmost edge of the paper (See diagram.).

Find the minimum length of the resulting crease.

Answer 1

@wio

Answer 2

It's the analysis that is the harder part.

Answer 3

You want to find the length of the crease in terms of... I suppose the height of the corner.

Answer 4

You have to use trig.

Answer 5

no trig, just a visual short cut

Answer 6

"Find the minimum length of the resulting crease."

Answer 7

oops i made a mistake
lol

Answer 8

Are you sure the crease only gets smaller as you move up?

Answer 9

the minimum would be when it is folded at a 45 degree angle making a 45-45-90 triangle at which time you would use a^2 + b^2 = c^2
a and b are 6"

Answer 10

wio he is trying to troll u O_O don't go into iht o-o

Answer 11

What he said makes intuitive sense.

Answer 12

"Are you sure the crease only gets smaller as you move up?"
yes, move it all the way down and it is folded in helf, the crease is 12" long, move it up and it decreases.

Answer 13

wio, just be carefull o-o your dealing with L here o-o

Answer 14

if you move it up past 45 degrees the fold will start to overlap and remain the same length

Answer 15

wild idea, grab a piece of paper and see what it does when you mimic the test.

Answer 16

btw square root of 72 = 8.49

Answer 17

not enough data there to solve, plus the fold is not at the location of the shortest crease/

Answer 18

There is enough data.

Answer 19

you have 2 lengths and no angles......

Answer 20

We have:\[
L^2=x^2+y^2
\]We also know \(3\leq x \leq 6\)

Answer 21

Standing by, maybe I will learn something here.

Answer 22

I still stand by my answer, assuming that the corner must remain touching the left side of the paper, the shortest crease would be the square root of 72".

Answer 23

hmm I gather the bottom side "x" which is 6 inches long, can only go UP that much
which will make the side upper-y edge pretty much horizontal, and thus a bisector to the 12-inch length

Answer 24

at that point, you'd end up with a 45-45-90 triangle in the lower-right bottom of the paper as @ehuman said

Answer 25

Here is a more thorough labeling of it all.
We may be able to get some nice equations from this.