Assignment 4

I chose to write up activity #6:

Take any triangle. Construct a triangle connecting the three midpoints of the sides. This is called the MEDIAL triangle. It is similar to the original triangle and one-fourth of its area. Construct G (the centroid), H (the orthocenter), C (the circumcenter), and I (the incenter) for this new triangle. Compare to G, H, C, and I in the original triangle.

I first created a GSP file that contained a triangle, its medial triangle, and the points G, H, C, and I of each of the two triangles. This file can be downloaded here. After playing with this sketch, I noticed several interesting things about these points, which I prove below.

1. The original triangle's circumcenter is equal to the medial triangle's orthocenter.

First, notice that FE is parallel to CA. We see why this is true by noticing that angle FEB is congruent to angle DAE, since all four of the smaller triangles are congruent. By the same argument, we can show that DE is parallel to CB, and that DF is parallel to AB. This means that that the perpendicular bisector of BC is the same line as the altitude of the medial triangle through F. Similarly, the perpendicular bisectors of AB and AC are the same as the altitudes of the medial triangle through E and D, respectively. This proves our hypothesis.

2. The original triangle's centroid is equal to the medial triangle's centroid.

I claim that the median BD of the original triangle intersects FE at the midpoint H of FE, therefore HD is a median of the medial triangle. This is true, because since FE is parallel to CA, triangle HEB is similar to triangle DAB. Since EB is half of AB, HE must be half of DA, so H must be the midpoint of FE. By the same logic, FG and IE are also medians of the medial triangle, and so the centroids of the two triangles are equal.

3. The Euler Segment of the medial triangle is half as long as the Euler Segment of the original triangle.

In the following diagram, I have labeled the original triangle's centroid, orthocenter, and circumcenter as G, H, and C. Similarly, the medial triangle's centroid, orthocenter, and circumcenter are labeled G', H', and C'. G lies on CH, which is the original triangle's Euler Segment, such that length(GH) = 2*length(GC). Similarly G' lies on C'H' such that length(G'H') = 2*length(G'C').

This means that length(C'H') = 3*length(C'G') and length(HC) = 3*length(G'H'). But since length(G'H') = 2*length(C'G'), then length(HC) = 6*length(C'G'). Therefore the length of the medial triangle's Euler Segment (C'H') divided by the length of the original triangle's Euler Segment (CH) is [3*length(C'G')] / [6*length(C'G')] = 1/2.