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| = 1-x
|DG| = x(1-x)
1-x
|FA| = -----
1+x
Therefore,
1-x x x(1-x)^2
[FBEG] = 1 - -------- - --- - ----------
2(1+x) 2 2
1+x+x^3-x^4
= -------------
2(1+x)
Finding the x that maximizes [FBEG]:
[2(1+x)][1+x+x^3-x^4]' = [1+x+x^3-x^4][2(1+x)]'
or
[2(1+x)][1+3x^2-4x^3] = [1+x+x^3-x^4][2]
or
x^2(3x^2+2x-3) = 0
Thus,
sqrt(10)-1
x = ------------
3
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Posted by Bractals
on 2008-10-16 15:36:53 |
Solution
