**Assignment 6**

**1. Construct a triangle and its medians. Construct
a second triangle with the three sides having the lengths of the three medians
from your first triangle. Find some relationship between the two triangles.
(E.g., are they congruent? similar? have same area? same perimeter? ratio of
areas? ratio or perimeters?) Prove whatever you find.**

Given any triangle, we can place one vertex on the origin, and
give the other two vertices arbitrary coordinates, as shown in the diagram below.
Using the midpoint formula, we can also assign coordinates to the midpoints,
**D**, **E**, and **F**, of the triangle
edges. If we translate the point **F** by the vector going from
**A** to **D**, it's image **P** will
have the coordinates labeled below:

Similarly, if we translate the point **C** by the
vector going from **E** to **F**, it's image **Q**
will have the coordinates labeled below:

Notice that **P = Q**, which means that triangle
**FPC** is a triangle with sides congruent to triangle **ABC**'s
medians. (Because **FP** is congruent to **AD**, **PC**
is congruent to **BE**, and of course, **FC** is congruent
to itself.) We therefore have a simple construction of the required triangle.

I modeled this situation using this GSP file. After some experimentation, I made the following conjecture:

**The area of the median triangle is three-fourths the area
of the original triangle.**

My proof of this fact follows.

The original triangle **ABC**, and the second triangle
**FPC**, with its three sides having the the lengths of the three
medians of **ABC**, are labeled below:

We learned in Linear Algebra that the area of a triangle with
coordinates **(a,b)**, **(c,d)**, and **(e,f)**
is , so the area
of **ABC** is ,
and the area of **FPC** is ,
thus **area(FPC) = 3/4 * area(ABC)**.