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maximum square folding (Posted on 2008-10-16) Difficulty: 2 of 5
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

  Submitted by Daniel    
Rating: 3.5000 (2 votes)
Solution: (Hide)

Bractals offers an excellent analytical solution
and Charlie offers a great computer solution

answer to 1,

the easiest way I found to compute the area is to find
the area for the 3 triangles BCE,EGD, and ABF and subtract
them from the area of the original square.

area of BCE:
well BC=1 CE=x and angle BCE=90 thus its area is simply x*1/2=x/2

area of EGD:
from this point on let < signify angle
thus thus triangles BEC and EGD are similar
CE/DG=BC/ED
x/DG=1/(1-x)
DG=x*(1-x)
ED=(1-x)
area is DG*ED/2=x*(1-x)^2/2

area of ABF:
Tan( Tan(45- (Tan(45)-Tan( Tan(45)=1
Tan( (1-x)/(1+x)=AF
AF=(1-x)/(1+x)
AB=1
thus the area is AB*AF/2=(1-x)/(2x+2)

adding up these areas we get

(1/2)*(x+x*(1-x)^2+(1-x)/(1+x))
and to get the area of FBEG we subtract this from one
thus if A(X) is the area of FBEG then
A(x)=1-(1/2)*(x+x*(1-x)^2+(1-x)/(x+1))


now to answer part 2,

the border values are A(0)=1-(1/2)=1/2
and A(1)=1/2

now to try and find an extreme value betwen 0 and 1
we solve A'(x)=0
A'(x)=1/2*(-1-(1-x)^2+2*(1-x)*x+(1-x)/(1+x)^2+1/(1+x))=0
1+(1-x)^2-2*(1-x)*x-(1-x)/(1+x)^2-1/(1+x)=0
(1+x)^2+(1-x^2)^2-2*(1-x^2)(1+x)x-(1-x)-(1+x)=0
3x^4+2x^3-3x^2=0
x^2(3x^2+2x-3)=0
x=0 already considered this value above
3x^2+2x-3=0
x=(-1+-sqrt(10))/3
x can't be negative so we are left with
x=(sqrt(10)-1)/3
this gives A(x)=(125-35*sqrt(10))/27=0.530381
which is greater than either border value

thus A(x) is maximized when x=(sqrt(10)-1)/3=0.720759
giving a maximum area of 0.530381
this corresponds to

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Puzzle AnswerK Sengupta2022-06-27 02:09:45
Solutionnumerical solutionCharlie2008-10-16 15:44:40
SolutionSolutionBractals2008-10-16 15:36:53
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