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 1We 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

This triangle also has centroid and its sides are represented by the vectors

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

Therefore

By inspection, the coefficient of in the above expression vanishes and that of is 9. So by symmetry,

Therefore (2) becomes

Now we consider . By regarding the triangles to be lying on the -plane in and using the cross product, we compute the area of the median triangle from (1):

noting that

On the other hand, by (2), (1) and (4),

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.

Let , and be the sides of opposite to , and respectively. As , we have

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 1As 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.

I have another observation. Not really simpler though. Heron’s formula gives that for a triangle with sides , and , we have

Now (your equation (2)). So by the above formula,

Now consider triangle ABC with centroid O, we have

since

and

by your argument.

So are the same.

Thanks! That’s a nice application of Heron’s formula, and is closer to what I expect as a classical proof. (However, it seems that the burden of expanding the expression involving etc. cannot be avoided, though in your case, it’s hidden inside the Heron’s formula )

By the way I have slightly modified your comment to make the math easier to read.

Thanks for correcting my mistake of leaving the 1 in the formula.

I think considering the triangle with side OA, OB and OC is a little bit simpler (of course you will need to observe that the corresponding angles matches). For example $\mathrm{area}(OA, OB, OC) = \mathrm{area}(OAB) = 1/3 \mathrm{area}(ABC)$ can be derived with little calculations. The first equality is by considering the parallelogram consisting of sides OA and OB (both sides are half the area of the parallelogram), and the second is by $\mathrm{area}(OAB) = \mathrm{area}(OBC) = \mathrm{area}(OCA)$.

I see your point, and have modified the post accordingly. (I was a bit reluctant to introduce the new traingle which has side OA, OB and OC as we already got a median triangle in the first proof, but I agree that this makes the computation cleaner.)

By the way, the way to use latex here is to type

$latex your-latex-code-here$