Cutting a Square into Five Equal Pieces

Date: 07/12/99 at 09:05:56
From: Shinichi
Subject: Square problem...

Hi Dr. Math,

I was wondering how it is possible to divide a square cake into five 
equal parts. There are some restrictions: you cannot divide the cake 
into ten parts and give two to each person, and it must be cut through 
the center point.

       ___________
       |         |
       |         |
       |    .    |
       |         |
       |         |
       -----------

What I have done is these but they were rejected.


       ___________               ___________
       |    |    |               |    |    |
       |   _|__  |               | \  |  / |
       |---|  |---       &       |   \|/   |
       |   |__|  |               |   / \   |
       |    |    |               |  /   \  |
        ---------                 ---------

Can you help me please? 
Thank you, Doctor.

Date: 07/12/99 at 11:27:39
From: Doctor Rob
Subject: Re: Square problem...

Thanks for writing to Ask Dr. Math.

Your last drawing above contains the right idea. Measure the length of 
the sides of the square, which is the base of the cake. Call that s. 
The perimeter of the square is then 4*s. Divide that by 5. Now from 
any starting point (such as the center of the top edge in your 
picture), measure around the edge a total distance of 4*s/5. Mark this 
point, and measure around the edge again a total distance of 4*s/5. 
Repeat this until you return to the starting point. Now from the 
center, make cuts to all five points marked.  That will divide the 
cake into five equal volume pieces, and furthermore, each piece will 
have the same amount of icing, too.

To prove that the volumes of the pieces are all the same, it is enough 
to prove that the areas of the four quadrilaterals and one triangle 
are all the same. Draw the diagonals of the square. That will cut the 
quadrilaterals into two triangles each. All the nine triangles will 
have height s/2, and their bases will be s/2, 3*s/10, 7*s/10, s/10, or 
4*s/5. Starting at the starting point and moving around the perimeter 
of the square, you'll see that

        s/2 + 3*s/10 = 4*s/5
     7*s/10 +   s/10 = 4*s/5
               4*s/5 = 4*s/5
       s/10 + 7*s/10 = 4*s/5
     3*s/10 +    s/2 = 4*s/5

Starting at one corner and moving around the perimeter of the square, 
you'll see that

               s/2 +    s/2 = s
            3*s/10 + 7*s/10 = s
     s/10 +  4*s/5 +   s/10 = s
            7*s/10 + 3*s/10 = s

Thus the triangle areas will be s^2/8, 3*s^2/40, 7*s^2/40, s^2/40, or 
s^2/5. When you add up the areas of the triangles to get the areas of 
the quadrilaterals, all quadrilaterals will have area s^2/5, or 1/5 of 
the total area of the square:

        s^2/8 + 3*s^2/40 = s^2/5
     7*s^2/40 +   s^2/40 = s^2/5
                   s^2/5 = s^2/5
       s^2/40 + 7*s^2/40 = s^2/5
     3*s^2/40 +    s^2/8 = s^2/5

Now the volume is the area of any of these polygons times the height 
of the cake, so all volumes are equal to h*s^2/5, and all icing areas 
are equal to (h+s)*s/5.

Notice that if you had started at, say, a corner point, the 
quadrilaterals would be different shapes, but the area of each would 
still be s^2/5, because the triangles forming them have common 
altitude s/2 and the sum of their bases is 4*s/5, so the total area is

     A1 + A2 = a*b1/2 + a*b2/2
             = a*(b1+b2)/2
             = (s/2)*(4*s/5)/2
             = s^2/5

For the same reason, any starting point could have been used.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/