Start with a square piece of paper. Label the vertices ABCD. Pick a point on CD and label it E. Fold along the line BE. Label the new location of C as C'. Find the point F on AD such that when folding along BF it makes the new location of A coincide with C'. Now lastly find a point G on AD such that when folding along EG it makes the new location of D lie on EF (either EC' or A'F). After all 3 of these folds are completed you should have a new irregularly shaped quadrilateral FBEG.
For simplicity's sake assume the original square
is of unit length. Now the 2 problems are:
1) If x is the length of CE, then give an equation
for the area of FBEG based on x.
2) Find the x that maximizes the area of FBEG
Let [PQ...W] denote the area of polygon PQ...W.
Then
[FBEG] = [ABCD]  [ABF]  [BCE]  [EDG]
Finding the segment lengths in terms of x:
ED = 1x
DG = x(1x)
1x
FA = 
1+x
Therefore,
1x x x(1x)^2
[FBEG] = 1      
2(1+x) 2 2
1+x+x^3x^4
= 
2(1+x)
Finding the x that maximizes [FBEG]:
[2(1+x)][1+x+x^3x^4]' = [1+x+x^3x^4][2(1+x)]'
or
[2(1+x)][1+3x^24x^3] = [1+x+x^3x^4][2]
or
x^2(3x^2+2x3) = 0
Thus,
sqrt(10)1
x = 
3

Posted by Bractals
on 20081016 15:36:53 