Essay Number Two
Trisecting the Area of a Triangle
Parallel Lines

Collaborative Work with Chris Romano

Picture 1

The Approach:

We will start by trying to construct the segment FG. Once we do that, we can bisect the area of triangle AFG to complete our trisection. In order to find the length of segment FG, we will need to set up some equations....

Finding the Length of Segment FG:

Let A1 = Area (ABC).
Let A2 = Area (AFG).
Let h1 = height (ABC).
Let h2 = height (AFG).

Then A1 = (1/2)(BC)(h1).
Also A2 = (1/2)(FG)(h2).

Since we are trying to trisect the area of this triangle, (2/3)A1 = A2.

We also know that triangle ABC is similar to triangle AFG (Corresponding Angles are congruent and then AA Similarity).

Then we can set up some proportions based on this similarity. BC/FG = h1/h2.

Let's do some algebra:
(2/3)A1 = A2
(2/3)(1/2)(BC)(h1) = (1/2)(FG)(h2)
(2/3)(BC)(h1/h2) = FG
(2/3)(BC)(BC/FG) = FG
(2/3)BC2 = FG2
sqrt(2/3)(BC) = FG

This means that if we can construct the square root of (2/3) times the length of BC, then we can find the length of FG. Since we can construct the length of any real number, this is not a problem.

Constructing the Length of Segment FG:

Since sqrt(2/3) = sqrt(6)/3, we will first construct the square root of 6 times BC, and then trisect that segment.

Constructing a spiral will give us the square root of 6 times BC. The red segment is sqrt(6)(BC).
Picture 2

Now we must take this length and trisect it to get (sqrt(6)/3)(BC). The blue segment is this length.
Picture

Placing Segment FG in the Triangle:

So now that we have the correct length for segment FG, we need to place it in the correct spot in the triangle. That is, it needs to be parallel to the base, BC. The correct placement of FG is in green.
Picture 4

Constructing the Length of Segment DE:

All we need to do is now bisect triangle AFG in order to find the last part of our trisection. To find the length of segment DE, see this page: http://jwilson.coe.uga.edu/EMT725/Half/analysis.html (provided courtesy of Jim Wilson). Therefore, we know that DE = sqrt(2)/2*(FG).

So all we need to do is construct the square root of 2 times FG, then bisect that segment and we will have the correct length of DE. The process is similar (and simpler) than the construction of FG. The red segment is the correct length of DE.
Picture 5

Placing Segment DE in the Triangle:

So now that we have the correct length for segment DE, we need to place it in the correct spot in the triangle. That is, it needs to be parallel to the base, BC. Since FG is also parallel to the base, we can simply place DE parallel to FG. The correct placement of DE is in blue.
Picture 6

Seeing It All With GSP:

To check this construction out and see the areas measured, here is a GSP Sketch of this construction.

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