Assignment 6

By: Olamide Alli

Exploration:

Consider any triangle ABC.  Find a construction for a point P such that the sum of the distances from P to each of the three vertices is a minimum.

Initial Assumptions

1.  The point P is going to be inside the triangle. Any point outside the triangle will be equidistant to the nearest vertex inside and will also have a bigger distance from the vertex that is farthest away.  The extra distance will produce a greater sum than what is required or needed.

2. The point P is a center of a triangle and is equidistant from each vertex.

From the picture above we are going to examine the points C, H, G, and I which are the circumcenter, orthocenter, centroid, and incenter respectively. Considering that the triangle ABC is an acute triangle, we can state the point C, H, G, and I are all centers of the triangle ABC. We can also state that all of the angles at the vertex are acute angles as well which further affirm that the point C, H, G, and I are centers of the triangle ABC.
Next, let’s tackle constructing the perpendicular bisectors for each side of the triangle ABC. The common point for each perpendicular bisector is going to be the point C, which is subsequently the circumcenter of the triangle ABC.

The red dashed lines in the figure above represent the perpendicular bisectors for each side of the triangle ABC. Please notice that there are three newly developed segments inside of the triangle ABC connecting to a new point, C’. The segments AC’, CC’, and BC’, which are represented by green dashed lines, are congruent to each other. Since the original circumcenter C is equidistant to each vertex of the triangle ABC, then the point C’ is the point “P” in the initial conjecture. This is the point where the sums of the segments from that point to the vertices are a minimum.