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Another folded paper (Posted on 2012-01-03) Difficulty: 3 of 5
Start with a square of paper ABCD.

Fold corner A down to lie somewhere along side CD. Call x the portion of the way from C to D.

Corner B is outside of the original square and is now one side of a small triangle whose opposite side is along the remainder of side BC.

(A) Where would A be folded to make this small triangle isosceles?

(B) Find a formula in terms of x for the perimeter of this small triangle.

(C) Where would A be folded to maximize the area of this small triangle?

No Solution Yet Submitted by Jer    
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Solution Solution Comment 1 of 1

Let E and H be the points that corners
A and B are folded into respectivly.
Let F and G be the intersections of the
crease with sides AD and BC respectively.
Let I be the intersection of line EH
with side BC. Let s be the side length
of the square and x = |EC|.
Let beta = angle DAE. Then
                |DE|     |DC|-|EC|
   tan(beta) = ------ = -----------
                |AD|        |AD|
             = (s-x)/s.
Since F lies on FG (the perpendicular
bisector of AE, |AF| = |FE|.
Let alpha = angle DFE. Looking at
triangles DFE and AFE we see that
   alpha = 2*beta.
Therefore,
   cos(alpha) = cos(2*beta)
                 1 - tan(beta)^2
              = -----------------
                 1 + tan(beta)^2
                    x*(2s - x)
              = ------------------      (1)
                 2s^2 - 2sx + x^2
   sin(alpha) = sin(2*beta)
                   2*tan(beta)
              = -----------------
                 1 + tan(beta)^2
                    2s*(s - x)
              = ------------------      (2)
                 2s^2 - 2sx + x^2
Note that right triangles FDE, ECI, and
GHI are similar. Thus,
   |EC| = x and |EI| = x/cos(alpha),
   |GH| = |GI|cos(alpha), and
   |HI| = |GI|sin(alpha).
To tie FDE and ECI together,
   |EI| + |IH| = |EH|
        or
   x/cos(alpha) + |GI|sin(alpha) = s
        or
             s*cos(alpha) - x
   |GI| = -----------------------       (3)
           cos(alpha)*sin(alpha)
-------------------------------------------
(A) If triangle GHI is isosceles, then
       cos(alpha) = sin(alpha)
                  or
       x*(2s - x) = 2s*(s - x)
                  or
       x^2 - 4sx + 2s^2 = 0
                  or
       x = s*[2 - sqrt(2)].
-------------------------------------------
(B) Perimeter(GHI) = |GH| + |HI| + |IG|
    Combining this with (1)-(3) gives
    Perimeter(GHI) = x.
-------------------------------------------
(C) Maximize Area(GHI):
    Area(GHI) = |GH|*|HI|/2
                 |GI|^2*cos(alpha)*sin(alpha)
              = ------------------------------
                               2
    Combining this with (1)-(3) gives
                 sx^3 - x^4
    Area(GHI) = ------------            (4)
                 8s^2 - 4sx
    Maximum area when
      (8s^2 - 4sx)*d/dx(sx^3 - x^4) -
      (sx^3 - x^4)*d/dx(8s^2 - 4sx) = 0
             or
      (8s^2 - 4sx)*(3sx^2 - 4x^3) -
      (sx^3 - x^4)*(-4s) = 0
             or
      4sx^2*(3x^2 - 10sx +6s^2) = 0 
             or
      x = s*[5 - sqrt(7)]/3             (5)
     
    Combining (4) and (5) gives
                      s^2*[316 - 119*sqrt(7)]
    Max. Area(GHI) = -------------------------
                               54
QED
-------------------------------------------
Note the above results checked with
Geometer's Sketchpad.
 

  Posted by Bractals on 2012-01-03 13:23:18
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