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
If angle EBC is designated alpha, then x = tan(alpha), so in the description below, remember that references to x are also references to tan(alpha).
Angle C'BE is also equal to alpha, so ABA' is 90  alpha degrees, and each of ABF and FBA' is 45  alpha.
Since C'EB and BEC are each 90  alpha, they add to 180  2*alpha, and it follows that each of C'ED and GED is alpha.
The area of ABCD is 1, and from this we need to subtract the areas of triangles ABF, EBC and DGE.
From the angles mentioned above and the unit lengths of the sides AB and BC, AF = tan(45  alpha) and EC = x. But AF = tan(45  alpha) can be written as (1x)/(1+x). The areas of the corresponding triangles are just half these lengths as they are the heights of right triangles whose bases are 1.
The base of right triangle DGE is 1  x, so DG is x*(1  x) = x  x^2, and the area of the triangle is thus (1x)(xx^2)/2.
So the area to be maximized is 1  (x/2 + (1x)/(2*(1+x)) + (1x)(xx^2)/2).
By doubling, taking the derivative and doing some algebra, this can be maximized by solving 2  2/(1+x^2)  4x + 3x^2 = 0.
I chose instead to write a program to evaluate the area at regular intervals and home in to a maximum value.
The program initially was:
DEFDBL AZ
PRINT
start = .1#: fin = .9#
FOR x = start TO fin STEP (fin  start) / 40#
a = 1  (x / 2 + (1  x) / (2 * (1 + x)) + (1  x) * (x  x * x) / 2)
PRINT USING "#.######## #.##############"; x; a
NEXT x
giving
x Area
0.10000000 0.50040909090909
0.12000000 0.50067885714286
0.14000000 0.50103501754386
0.16000000 0.50148303448276
0.18000000 0.50202637288136
0.20000000 0.50266666666667
0.22000000 0.50340386885246
0.24000000 0.50423638709677
0.26000000 0.50516120634921
0.28000000 0.50617400000000
0.30000000 0.50726923076923
0.32000000 0.50844024242424
0.34000000 0.50967934328358
0.36000000 0.51097788235294
0.38000000 0.51232631884058
0.40000000 0.51371428571429
0.42000000 0.51513064788732
0.44000000 0.51656355555556
0.46000000 0.51800049315069
0.48000000 0.51942832432432
0.50000000 0.52083333333333
0.52000000 0.52220126315789
0.54000000 0.52351735064935
0.56000000 0.52476635897436
0.58000000 0.52593260759494
0.60000000 0.52700000000000
0.62000000 0.52795204938272
0.64000000 0.52877190243902
0.66000000 0.52944236144578
0.68000000 0.52994590476190
0.70000000 0.53026470588235
0.72000000 0.53038065116279
0.74000000 0.53027535632184
0.76000000 0.52993018181818
0.78000000 0.52932624719101
0.80000000 0.52844444444444
0.82000000 0.52726545054945
0.84000000 0.52576973913043
0.86000000 0.52393759139785
0.88000000 0.52174910638298
Variables start and fin were modified to home in on the answer.
In its final state it looked like:
DEFDBL AZ
PRINT
start = .720755#: fin = .72076#
FOR x = start TO fin STEP (fin  start) / 40#
a = 1  (x / 2 + (1  x) / (2 * (1 + x)) + (1  x) * (x  x * x) / 2)
PRINT USING "#.######## #.##############"; x; a
NEXT x
and found (among other values not shown):
0.72075900 0.53038081089283
0.72075913 0.53038081089284
0.72075925 0.53038081089284
0.72075938 0.53038081089284
0.72075950 0.53038081089282
indicating an x ~= .7207593 to give an area of .53038081089284.

Posted by Charlie
on 20081016 15:44:40 