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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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