The "Nedial" Triangle

I wrote recently about the concept of a Nedian; an extension of the idea of medians created by Professor John Satterly of the University of Toronto. I have played around a little more with the concept and come up with a calculation for the area of what I call a nedial triangle, extending the idea of a medial triangle.



Most High School students are introduced to the Medians of a triangle, and it is quite easy to show using basic high school geometry that the triangle whose vertices are the three feet of the medians has an area equal to 1/4 the area of the original triangle. In fact, it is pretty easy to show that the original triangle can be dissected into four congruent copies of the medial triangle.



If instead we use the feet of the three nedians, we get a triangle I have called the nedial triangle. I have worked out that the area of such a triangle for an n-nedian (the cevian that cuts the opposite side 1/n th of the way along the edge) will have an area of
times the area of the original triangle ABC. Since the median is the 2-Nedian, this gives the correct ratio for the medial triangles.

The image below shows the 4-nedial triangle, which has an area of 7/16 the area of ABC.



The area of the three triangles at the vertices of the original triangle will each be
times the area of ABC.

The area of the nedial triangle will grow from a minimum of 1/4 the area of ABC increasing toward a limit of one as n approaches infinity. This can be confirmed by L'Hopital's rule or simple division of the terms. The fact that the limit is also one as n approaches one points out that when the base of the nedian is more than half way along the opposite side and the "n" in the nedian must be less than 2 but more than one we are measuring the area of what Satterly called the "backward nedians" which will create a nedial triangle of the same area. n=3/2 gives the exact area of n=3 (although the triangles are NOT congruent). In general for every k-nedial the area given by the nedial with index equal to ⁿ/_n-1 has the same area.

Centroids of Inscribable Polygons

The term centroid, or center of gravity, of a triangle usually refers to the center of gravity obtained by treating the vertices as point masses with a mass of one unit. This leads to the centroid noramally found in geometry by intersecting the three medians. If we have points (xA,yA), (xB,yB), and (xC,yC) it is well known, and easily derived, that the centroid would be .

There are other types of centroids, and one of them has an interesting relation to the class of polygons mentioned in the title. Another center of gravity can be found if we use the boundary of the polygon and assume it is made up of uniform rods. For a triangle, this produces a center of gravity that is at the intersection of the angle bisectors of the medial triangle (the triangle with vertices at the midpoints of the original triangle). This point is also called the Spieker Center of the triangle. To distinguish these two types of centroids, I will call the center of gravity of the vertices G_p (for point) and the center of gravity of the boundary G_b.

Some interesting sidebars about the angle bisectors of the medial triangle before I continue about centers of gravity.... The angle bisectors of the medial triangle are always parallel to the angle bisector of the opposite vertex. Also, these lines are each a "cleaver" of the triangle; a term created by Dov Avishalom in a Mathematics Magazine article for a line that passes through the midpoint of one side and divides the perimeter of the triangle into two equal halves.

I recently came across a paper by Tom Apostle and Mamikon A. Mnatsakanian that pointed out that in any triangle the two centroids and the incenter were always collinear. They also show that the ratio of the distances of the two centroids from the incenter is always in a ratio of 3/2. The ratio G_b/ G_p= 3/2.


Then in a really nice generalization, they showed that the same property applies to any inscribable polygon, one in which each edge is tangent to a common circle, such as the quadrilateral below. They use the clever term "circumgons" in the paper for such shapes. This is really a nice term since "inscribable" and "circumscribable" are used both to describe the circle inscribed in a polygon, or the polygon circumscribed about the circle... even I get confused. . Since it is often much easier to find G_p and the incenter of inscribable polygons, this provides a clever way to find G_b.

Students may be interested to know that Archimedes used the formula Area= 1/2 P r for the area of a circle, where P is the perimeter (circumference). This is, of course, simply and alternate form of the formula they learned in grade school. They may be even more surprised to find that the Archmedian formula gives the area of any inscribable polygon if P is the perimeter of the polygon, and r is the radius of the inscribed circle. The proof is very easy, and is often provided in high school textbooks for the limited case of regular polygons which would always be inscribable. The image below provides non-subtle clues to the proof.