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### 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.

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

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 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/
```
Associated Topics:
High School Geometry
High School Polyhedra
High School Triangles and Other Polygons

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