After Medians Comes Nedians

On the day I wrote my last blog about some interesting properties of the medians of a triangle, I received a package of old Mathematics Teacher articles from Dave Renfro. One of the first I looked at was a January 1951 "Mathematical Miscellanea" edited by Phillip S. Jones. The article contained a contribution by John Satterly of the University of Toronto on a type of cevian that he called "Nedians". [My personal choice, since the "med" root is for the middle, would have just been to call them n-dians, but I'm sure that would have had a cultural backlash.]

For those who may not be familiar with the term "cevian", it refers to a segment in a triangle from a vertex to the opposite side (extended if necessary). Angle bisectors, medians, and altitudes are all cevians then, but a perpendicular bisector of a side would not be because it doesn't necessarily pass through a vertex. The name is in honor of Giovanni Ceva and was originated in France in 1888 and has spread from there.

Professor Satterly seems to have created the term "nedians" as a comparison for medians to describe a cevian that cuts the opposite side 1/n th of the way from one vertex to the next. [I shall use the notation 4-nedian for a nedian that cuts 1/4th of the way along the opposite side,] A median would be the 2-nedian.
In the image, triangle ABC has 3-nedians AD, BE, and CF where D is 1/3 of the way from B to C; E is 1/3 of the way from C to A, etc.

The intersections of the three Nedians of a triangle will form another triangle at their three points of intersection, called the nedian triangle. [JKL in the image].

Professor Satterly seems to have discovered several properties of the nedians, and their nedian triangles which I will give here; and then I have come up with several interesting properties of my own about them that I will add to this blog.

It is not too difficult using affine properties of a triangle to verify many of these.
Professor Satterly showed that the sum of the squares of the nedians would be times the sum of the squares of the sides of the original triangle. Notice that when n=2, this reduces to 3/4, the ratio given and proved for the medians.In any triangle, the sum of the squares of the medians is equal to 3/4 the sum of the squares of the three sides.
Students may wish to explore these properties by creating Sketchpad or Geogebra interactive models to confirm them, and then challenge themselves to prove them. Proving them is easier with a property of affine geometry. Every triangle is affine equivalent to every other triangle in the plane. Under affine transformations areas and lengths may change, but ratios of them are preserved... which means that you can choose any triangle... a right triangle or an equilateral triangle, to find a property about ratios of lengths or areas, and it will apply to any other triangle... and that is a BIG idea to lock away (I am not well schooled in the particulars of affine geometry, so if I have mis-stated that in some way, please advise).

Professor Satterly also stated that the Nedian triangle will have an area of \(\frac{(n-2)^2}{n^2-n+1}\) times the area of ABC. Note that for the median, or 2-nedian, the area diminishes to zero since the three medians intersect in a single point.

Professor Satterly suggested the term "backward nedian triangle" for a case in which the 1/n ratio went in the opposite order (Let D be 1/3 of the way from C to B insted). This can be eliminated if we simply allow any real number for the coefficient. Then the backward 3-nedian is just a 3/2-nedian in the regular order, and it seems that all his properties are still preserved. Notice that the areas of the 3 and 3/2-nedians are equal, but they are not congruent.

Exploring these constructions a little more, I came up with a few more properties that were not in the article. For example, the perpendicular distance from the three vertices of the nedian triangle to any side of ABC will equal the altitude of ABC to that same side. [I call these the sub-altitudes.]

Here the altitude is shown in bold red, and the three corresponding sub-altitudes are shown in dotted red. In the 3-nedian shown the distance from J to side B plus the distance from K to side B plus the distance from L to side B will equal the altitude from B to side B. A similar result exists for each altitude of the triangle. In addition, the three sub-altitudes will always partition the altitude in the same way. The shortest sub-altitude will be , while the next longer one will be and the longest will be .

It is also clear from the last statement that each of the three small triangles at the vertices of A, B, and C will be congruent. Their bases will each be 1/n of a base of the original triangle and their heights are so each of them is an equal fraction, of the original area of ABC. By similar reasoning we see that all three of the quadrilaterals will also have the same area.

I also observed that that each nedian is partitioned into three parts whose lengths, in order from the vertex to the opposite side, are .
For the 3-nedians in the image, for example, the nedian CF is partitioned so that CJ is 3/7of CF; JL is 3/7 of CF; and LF is 1/7 of LF. A similar partition holds for the other two 3-nedians AD and BE. In a 4-nedian, the partitions would be 4/14, 8/13, and 1/14.

I'm gonna call that a good days work...please advise of typo's or just plain bad math... and thanks Dave, for another stimulating journal article.

A Nice Property of the Medians of a Triangle

We begin with a simple triangle, ABC, with opposite sides a, b, and c; and median AD which we will call m.




The object of the exercise; to find the length of the median, m, given the length of the three sides, a, b, and c.
We begin by asserting a known, and easy to prove, theorem of parallelograms. The sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. The proof follows easily from the use of the law of cosines and the fact that the cosines of adjacent angles of the parallelogram are opposites of each other, one positive and one negative.

But how, you may ask, do parallelograms figure in the argument. If we make a congruent copy of ABC and rotate it 180 degrees around D, the midpoint of segment c we get a parallelogram and its two diagonals.



We observe that the diagonal AA’ has length 2m, and the other diagonal has length a. From the previously stated theorem it is then clear that :



Which leads simply to the result



The fact that this result was known before the birth of Christ does not diminish the simple elegance of the derivation.
To find the other two medians lengths, of course, it is only necessary to rotate the variables a, b, and c, .

.

Now collecting like terms we get .

And we see that in any triangle, the sum of the squares of the medians is three-fourths the sum of the squares of the sides. Pretty!

Another Go-Round with Archimedes

I just flew back to the UK from the US, and had time to read some translations from Archimedes' "On the Sphere and Cylinder."

The sad truth is that reading Archimedes always makes me feel a little stupid. It is not just that he was a genius, but that each time I read him, I realize some simple fact that I ought to have somehow noticed years before.

First some background notes: I wrote a while back about the fact that Archimedes discovered that the surface area of a sphere, between any two parallel cutting planes is exactly the same as the surface area of the cylinder with a diameter equal to the diameter of the sphere and a height equal to the distance between the two planes.


On this pass I noticed that he had given a very simple answer to a question that seems harder: What is the surface area of the segment of a sphere which has a height of h and a radius of the base is a. (see image)


The image gives a hint to how I would solve it. Use the given information to find the radius of the sphere, then use the fact that the surface of the segment is the same as the lateral surface of a cylinder with a diameter of 2r and a height of h, that is

If we followed up on this, we see that r² = a² + (r-h)²
If we expand that we get and subtracting r^2 from both sides gives and solving for r we get .. Now if we plug that in for the r in we get an area of . As Archimedes so cleverly determined, the area is independent of the sphere it was situated on. The area is the same as a circle whose radius is the distance from the vertex of the segment to a point on the circular base.

Wait, there is a little more. When I wrote about isoperimetric problems when bound against an edge (find the largest rectangular area that can be enclosed by n ft of fence built with a barn as one edge of the pen) I never thought about a simple 3d example of this same idea.... but Archimedes did. He observed that of all the spherical segments with a given surface area, the largest volume was for the one that enclosed a hemisphere. Very Clever.... another note about an "aha" moment in a future blog.

Happy New "Prime" Year

Well, it's the first prime day (the 2nd day of the year 2011) of a prime year that is the sum of a prime number of consecutive primes. In fact, it can be written as the sum of a prime number of consecutive prime numbers in at least two ways. The easy one is 2011 = 661+673+677, the other one that I know takes eleven consecutive primes...find it..I have been told there is not a year that can be expressed as a sum of successive primes in more than two ways for over a thousand more years That, my mathematical readers, is a lot of primes.

2003 was the last prime year, and it had the special property that the sum of its digits was also prime... not true for 2011.

Both 2017 and 2027 will be prime, but only 2027 is expressible as the sum of consecutive primes, but not a prime number of them.2081 is the next year that will be, like 2011, a prime that is the sum of a prime number of consecutive primes. If you forget, I'll remind you on Jan 2nd of that year... guess I better start working on a healthier diet... let's see, that will make me .... WOW, that IS a big number....
So what is the next year that can be expressed as a sum of consecutive primes starting with two... 2 + 3 + 5 + 7 ...... ???? If it is any help, I show the last year would have been 1988...

and the really hard problem... what is the next prime year that is the sum of consecutive prime digits starting with 2 + 3 + 5 + 7 ......?