An Exploration.Take any triangle ABC. Construct equilateral triangles externally on each side and locate the center of each equilateral triangle. Label these centers A', B', and C' for the triangle centers opposite angles A, B, and C. Construct

linesAA', BB', and CC'. [Note: Use lines rather than segments.] Observe.

Are AA', BB', and CC' concurrent?

Click here to open a GSP file to examine this construction and related constructions.

Although there is a convincing

demonstrationof concurrency, aproofis needed. For the moment this will be held as some unfinished business. Is the point of concurrency any of the orthocenter, centroid, incenter, or circumcenter of triangle ABC? In the following figure, the Euler line, centroid, orthocenter, circumcenter, and incenter are shown.The triangle A'B'C' is the First Napoleon Triangle. Hypotheses? The point of concurrence is called the First Napoleon point. What can you find out about this point?

For example, as vertex A is moved along a line parallel to side BC, what is the locus of the Napoleon point? Click here to open a GSP file for constructing the First Napoleon Triangle. Drag vertices around or add traces and animation to the file to examine features of the First Napoleon Point.

Extending the Exploration.Repeat the investigation but let A', B', and C' be the external VERTICES of the externally constructed equilateral triangles with A' off the side opposite vertex A, etc. Click here to open a GSP file for this construction. This produces a concurrency called the First Fermat point. Again, there is a convincing demonstration, but a proof is needed. When each angle of ABC measures less than 120 degrees, it is the point that minimizes the sum of the distances from the point to the three vertices of the triangle.

Going Further.Repeat the investigation using a

squareconstructed externally on each side of triangle ABC and with A', B', and C' being the centers of the squares. Click here to open a GSP file to construct squares externally on each side of ABC.

The four points of concurrency examined so far are curved between the centroid and the orthocenter, as can can be shown by displaying them at the same time.

Click here to open a GSP script for placing these four new concurrencies and the centroid and orthocenter in a triangle.

Going Inside.Repeat the exploration with the Napoleon Point and the Fermat Point with the equilateral triangles constructed toward the

interiorof triangle ABC. That is, the three triangles would overlap with ABC and each other.

Click here to open a GSP file for constructing three equilateral triangles toward the interior of ABC.

Repeat the exploration with the points of concurrency found by constructing squares on each side, but now with the squares constructed toward the

interiorof triangle ABC. That is, the three squares would overlap with ABC and each other.

Click here to open a GSP file for constructing squaresover the interior of ABC.

Now we have eight points of concurrency to examine, and with a little imagination, we can perceive that they lie along some curve or curves.

Click here to open a GSP file to construct these 8 points on the same triangle.

New start.What if similar isosceles triangles were constructed on each side of ABC? That is, each of the cases we have examined so far is a special case of a set of similar isosceles triangles constructed on the side of ABC. So setting such a construction and animating on the base angle from - 90 degrees to 90 degrees would give a locus for the point of concurrence and all of the eight points examined so far would be included. Further, a proof of this more general theorem -- a point of concurrency of

anyset of similar isosceles triangles so constructed -- will replace the need for proofs for the individual cases (although proofs and explorations for some special points might lead to other insights).

Click herefor a GSP file to show similar isosceles triangles construced on the three sides. The altitudes are is the same proportion as the sides of the triangle.

Click here to open a GSP sketch showing the hyperbola as a locus of the point of concurrency when ABC is an obtuse triangle. The vertices of ABC can be moved to produce any configuration.

Summary.When similar isosceles triangles are produced on each side of a triangle and the vertices of the triangle connected by a line to the vertex of the isosceles triangle on the opposite side, the lines are concurrent. As the base angle of the isosceles triangles are varied from 0 to 90 degrees, the locus is the arc of a hyperbola from the centroid when the base angle is 0 to the orthocenter when the base angle is 90 degrees. As the base angle is varied from 0 degrees to - 90 degrees, the concurrency lies along the balance of a hyperbola, with one leg passing through two vertices and the other leg passing through one vertex of the original triangle. Further exploration can lead to the discovery that the hyperbola is rectangular and its center lies on the nine-point circle.

Click here to open a GSP file to see this summary. The proofs are most readily developed with the use of trilinear coordinates, barycentric coordinates, or projective geometry.

References

Coxeter, H. S. M. (1961)

Introduction to Geometry. New York: Wiley.Eddy, R. H., and Fritsch, R. (1994) The conics of Ludwig Kiepert: A comprehensive lesson in the Geometry of the Triangle.

Mathematics Magazine, 67, 188-205.

**Return** to the EMAT 4600/6600 page

**Return** to the EMAT 6690 page