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### Dividing a Square in Thirds

```
Date: 03/27/2001 at 14:23:24
From: Frank Prager
Subject: Equal areas

I want to take a square and divide it into three equal pieces using
three lines radiating from the center of the square.

I determined that each line has to have an angle of 120 degrees to
complete a circle.

The difficulty is in getting the areas of the square to be equal.
```

```
Date: 03/27/2001 at 16:26:36
From: Doctor Rob
Subject: Re: Equal areas

Thanks for writing to Ask Dr. Math, Frank.

There is more than one way to do this. Here are two:

A(-1,1)          G             B(1,1)
o-----------+-----------o
| -_        .           |
|   -_      .           |
|     -_    .         _,o E
|       -_  .     _,-'  |
|         -_. _,-'      |
H + - - - - - o - - - - - + I
|          /.O          |
|         / .           |
|        /  .           |
|       /   .           |
|      /    .           |
o-----o-----+-----------o
D(-1,-1)   F     J             C(1,-1)

The first way is to make one line from A to O. Then the area of AGO
and AHO are both 1/2 square unit. That means the area of GBEO and HDFO
both have to be 5/6 square unit to make a total of 4/3 square unit for
ABEO and ADFO. That leaves OEI and OFJ with an area of 1/6 square
unit, so E must have coordinates (1,1/3) and F(-1/3,-1). Then:

<AOF = <EOA = arccos(-1/sqrt[5]) = 116.565 degrees
<FOE =        arccos(-3/5)       = 126.870 degrees

These angles can be found by using the Law of Cosines and the lengths
of the sides of the triangles AOE and EOF.

<hr>

A(-1,1)          G             B(1,1)
o-----------+-----------o
|           .           |
|           .           |
F o._         .         _,o E
|  `-._     .     _,-'  |
|      `-._ . _,-'      |
H + - - - - - o - - - - - + I
|           |O          |
|           |           |
|           |           |
|           |           |
|           |           |
o-----------+-----------o
D(-1,-1)         J             C(1,-1)

The second way is to put one line from J to O. Then the areas of OIE
and OHF must be 1/3 unit, so the coordinates of E and F are E(1,2/3)
and F(-1,2/3). Then the angles are:

<JOE = <FOJ = arccos(-2/sqrt[13]) = 123.690 degrees
<EOF =        arccos(-5/13)       = 112.620 degrees

These angles can be found by using the Law of Cosines and the lengths
of the sides of the triangles JOE and EOF.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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