Task E - modified version

Take any triangle ABC. Construct equilateral triangles internally on each side. Let A', B' and C' be the external vertices of the internally constructed equilateral triangles.

Construct lines AA', BB' and CC'. Observe. Are they concurrent? Is the point of concurrency any of the orthocenter, centroid, incenter or circumcenter of triangle ABC? If not what can you find out about this point.
Note this first version is in fact the construction for the second fermat point. This point has been dealt with in the discussion of Task D.

As we discussed in Task D in many ways these problems are similar and as such the original phrasing of the Task E is simply another case of the isosceles triangles already dealt with in task D and we expect the point of concurrency of the lines to lie on the hyperbola discussed in task D.

Original phrasing - Task E:

Take any triangle ABC. Construct equilateral triangles internally on each side. Let A', B' and C' be the centroids of the internally constructed equilateral triangles.

Now E is the point of concurrency of the of the lines joining the centroids of the internally constructed equilateral triangles and the vertices of the original triangle.

E, as is to be expected, is a point on the locus of D. However, what is more interesting is that E is a member of the same branch of the hyperbola as the other points that have positive base angles (see construction description Task D). This raises the interesting question for what values of the base angle will the point D be a member of each of the two branches of the hyperbola?

From the work done we know that the one branch contains the following points:

H-- orthocenter: base angle = 90 deg
F -- Fermat point: base angle = 60 deg
S -- concurrency point for the centers of squares problem: base angle = 45 deg
P -- concurrency point for the centroids of equilateral triangle problem: base angle = 30 deg
G-- centroid: base angle = 0 deg
E -- concurrency point for the centroids of internally constructed equilateral triangle problem: base angle = -30 deg

While the other branch contains the point F2 -- second Fermat point: base angle = -60 deg.

This question is left as a problem for the reader.

To investigate the problem suggested above and to watch a GSP drawing of the locus click here.


Back