I just came across a curious identity about the angles of the “median triangle” of a given triangle, while I was reviewing a paper from a team participating in the Hang Lung Mathematics Award. Of course, I am not going to reveal the identity for the obvious reason.
Let me first describe the setting. Let be a triangle. (By abuse of notations, we regard (for example) both as a vertex, a vector (in or ), and the angle of the triangle at the vertex .) We can then draw the three medians on the triangle which, as is well-known, intersect at the so called centroid of the triangle. Let , and be the angles at the centroid as shown:
Theorem 1 We have the identity
This result certainly looks very elegant (and is new to me). However, the proof in that paper consists of several pages of computations which to me is not very enlightening. So I set out to write a proof myself, which will be described below. Nevertheless, I have to resort to coordinates to prove the result. It would be desirable to know if there is a more classical proof without using coordinates. (Of course, all the computations using coordinates can theoretically be translated to classical statements, e.g. the cosine law is just the expansion of the inner product . However, I am not quite willing to do such kind of line-by-line translation. )
Proof: First of all, we can without loss of generality assume that the centroid of is , i.e.
It turns out that it is better to regard the angles , and as angles at the vertices of the “median triangle” of , which we now explain. It is not hard to see that we can construct a triangle whose sides have the same lengths as the three medians of , which is called a median triangle.
In fact, with the assumption above, it is easy to see that the vertices can be chosen to be
As has centroid and the centroid divides the median in the ratio , this shows that is actually formed by parallel translating the medians of .
We can express the angles in terms of its area:
Note that by polarization and using , we have
By inspection, the coefficient of in the above expression vanishes and that of is 9. So by symmetry,
Therefore (2) becomes
Comparing this with (3), we conclude that
1. Another proof by Tsz Chiu Kwok
Proof: We will use the classical notations. In particular we do not use coordinates in this proof.
By Heron’s formula, we have
So (5) becomes
On the other hand, if we let to be the centroid of , then by parallelogram law,
So (6) becomes
Now we consider the triangle where the segment is obtained by parallel translating the segment .
It is easy to see that has angles at it vertices. ( is of course similar to the median triangle as described in the first proof). It is easy to see that
as each median divides a triangle into two triangles with the same area.
As OAG has sides , , , by applying (6) to , we get
Comparing to (7), we can get the result.
Remark 1 As a bonus of the proof, for a triangle with sides , we have
Of course a similar formula holds for the sum of squares of the cotangents.