Classic Iterated Function Systems

Larry Riddle, Agnes Scott College

Barycentric Coordinates Details

Barycentric
Coordinates

In a triangle $ABC$, any point $P$ can be defined by its barycentric coordinates $(w_A, w_B, w_C)$. These coordinates represent the "weights" assigned to each vertex such that \begin{equation} P = w_A A + w_B B + w_C C \end{equation} subject to the constraint $w_A + w_B + w_C = 1$.

The barycentric coordinate $w_A$ of a point $P$ is defined by the ratio of the area of the sub-triangle formed with the base $BC$ to the total area of triangle $ABC$. \begin{equation} w_A = \frac{\text{Area}(\triangle PBC)}{\text{Area}(\triangle ABC)} \end{equation}

Suppose $P$ is on the line that connects the midpoints of the sides $AB$ and $AC$.

Triangle PBC inside triangle ABC

Because triangle $PBC$ and triangle $ABC$ share the same base $BC$, the ratio of their areas is equal to the ratio of their heights. Let $H$ be the altitude of vertex $A$ from base $BC$, and let $h$ be the distance of point $P$ from base $BC$. A line parallel to the base $BC$ that passes through the midpoints of $AB$ and $AC$ is exactly at height $h = H/2$. Therefore \begin{equation} w_A = \frac{\frac{1}{2} \cdot \text{base} \cdot h}{\frac{1}{2} \cdot \text{base} \cdot H} = \frac{h}{H} = \frac{1}{2} \end{equation}

For points along the base $BC$, $h = 0$ implies $w_A = 0$, and at vertex $A$, $h = H$ implies $w_A = 1$. Then above the midline, $w_A \gt \frac{1}{2}$, and below the midline $w_A \lt \frac{1}{2}$.

Midline of triangle dividing it into sections where weight A is less than and greater than 1/2.

Medial Triangle

Suppose a point $P$ is in the medial triangle of a triangle $ABC$, i.e. the triangle with vertices at the midpoints of the triangle's sides $AB$, $AC$, and $BC$.

Point P inside the medial triangle of triangle ABC.

Let the barycentric coordinates of $P$ be $(w_A,w_B,w_C)$. Then the medial triangle is the region where the point $P$ is "below" the midline for every vertex. Thus as shown above, this triangle is the intersection of the regions where $w_A \lt \frac{1}{2}$, $w_B \lt \frac{1}{2}$, and $w_C \lt \frac{1}{2}$. So the barycentric coordinates of every point in the medial triangle is less than 1/2.