A curious identity on the median triangle

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 {\Delta ABC} be a triangle. (By abuse of notations, we regard (for example) {A} both as a vertex, a vector (in {\mathbb R^2} or {\mathbb R^3}), and the angle of the triangle {\Delta ABC} at the vertex {A}.) 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 {D, E}, and {F} be the angles at the centroid as shown:triangle1.png

Theorem 1 We have the identity

\displaystyle \frac{1}{\sin^2 D}+ \frac{1}{\sin^2 E}+ \frac{1}{\sin^2 F} =\frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C}.

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 { |A-B|^2}. 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 {\Delta ABC} is {0}, i.e.

\displaystyle A+B+C=0.

It turns out that it is better to regard the angles {D}, {E} and {F} as angles at the vertices of the “median triangle” of {\Delta ABC}, 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 {\Delta ABC}, which is called a median triangle.


The triangle ΔABC and its median triangle.

In fact, with the assumption above, it is easy to see that the vertices can be chosen to be

\displaystyle D=\frac{1}{2}(A-B),\quad E=\frac{1}{2}(B-C),\quad F=\frac{1}{2}(C-A).

This triangle also has centroid {0} and its sides are represented by the vectors

\displaystyle \frac{3}{2}A, \quad \frac{3}{2}B,\quad \frac{3}{2}C. \ \ \ \ \ (1)

As {\Delta ABC} has centroid {0} and the centroid divides the median in the ratio {2:1}, this shows that {\Delta DEF} is actually formed by parallel translating the medians of {\Delta ABC}.

We can express the angles in terms of its area:

\displaystyle \begin{array}{rl} \sin A= \frac{2\mathrm{Area }(\Delta ABC)}{|B-A||C-A|}. \end{array}


\displaystyle \begin{array}{rcl} \frac{1}{\sin^2 A} =\frac{ |B-A|^2 |C-A|^2 }{4 \mathrm{Area}^2(\Delta ABC)}. \end{array}

We then have

\displaystyle \begin{array}{rcl} &&\frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C}\\ &=&\frac{1}{4 \mathrm{Area}^2(\Delta ABC)} (|B-A|^2 |C-A| ^2+ |A-B| ^2 |C-B|^2+ |A-C|^2 |B- C| ^2). \end{array} \ \ \ \ \ (2)

Note that by polarization and using {A+B+C=0}, we have

\displaystyle \begin{array}{rcl} |B-A|^2 = 2|B|^2+2|A|^2-|B+A|^2 = 2|A|^2+2|B|^2-|C|^2 . \end{array}


\displaystyle \begin{array}{rl} &|B-A|^2 |C-A|^2+ |A-B|^2 |C-B|^2+ |A-C|^2 |B-C|^2\\ =& (2|A|^2+2|B|^2-|C|^2)(2|A|^2+2|C|^2-|B|^2)+\\ &(2|A|^2+2|B|^2-|C|^2)(2|B|^2+2|C|^2-|A|^2)+\\&(2|A|^2+2|C|^2-|B|^2)(2|B|^2+2|C|^2-|A|^2). \end{array}

By inspection, the coefficient of {|A|^4} in the above expression vanishes and that of {|A|^2|B|^2} is 9. So by symmetry,

\displaystyle \begin{array}{rl} &|B-A|^2 |C-A|^2+ |A-B|^2 |C-B|^2+ |A-C|^2 |B-C|^2\\ =&9(|A|^2|B|^2+|B|^2|C|^2+|C|^2|A|^2). \end{array}

Therefore (2) becomes

\displaystyle \frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C} =\frac{9(|A|^2|B|^2+|B|^2|C|^2+|C|^2|A|^2)}{4 \mathrm{Area}^2(\Delta ABC)}. \ \ \ \ \ (3)

Now we consider {\Delta DEF}. By regarding the triangles to be lying on the {xy}-plane in {\mathbb R^3} and using the cross product, we compute the area of the median triangle {\Delta DEF} from (1):

\displaystyle \mathrm{Area}(\Delta DEF) =\frac{1}{2}\left|\frac{3}{2}A\times \frac{3}{2}B\right| =\frac{3}{4}\left(\frac{3}{2}\left|A\times B\right|\right) =\frac{3}{4}\mathrm{Area}(\Delta ABC), \ \ \ \ \ (4)

noting that

\displaystyle 3|A\times B|= 3|B\times C|=3|C\times A|=2\mathrm{Area }(\Delta ABC).

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

\displaystyle \begin{array}{rcl} &&\frac{1}{\sin^2 D}+ \frac{1}{\sin^2 E}+ \frac{1}{\sin^2 F}\\ &=&\frac{1}{4 \mathrm{Area}^2(\Delta DEF)} \left(\left|\frac{3}{2}A\right|^2 \left|\frac{3}{2}B\right| ^2+ \left|\frac{3}{2}B\right|^2 \left|\frac{3}{2}C\right| ^2+ \left|\frac{3}{2}C\right|^2 \left|\frac{3}{2}A\right| ^2\right)\\ &=&\frac{9(|A|^2|B|^2+|B|^2|C|^2+|C|^2|A|^2)}{4 \mathrm{Area}^2(\Delta ABC)}. \end{array}

Comparing this with (3), we conclude that

\displaystyle \frac{1}{\sin^2 D}+ \frac{1}{\sin^2 E}+ \frac{1}{\sin^2 F} =\frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C}.


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 {a}, {b} and {c} be the sides of {\Delta ABC} opposite to {A}, {B} and {C} respectively. As {\frac{1}{2}bc\sin A= \mathrm{Area }(\Delta ABC)}, we have

\displaystyle \frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C} =\frac{ b^2c^2+c^2a^2+a^2b^2 }{4\mathrm{Area}^2(\Delta ABC)}. \ \ \ \ \ (5)


By Heron’s formula, we have

\displaystyle \begin{array}{rl} 16 \mathrm{Area}^2 (\Delta ABC) =&-(a^4+b^4+c^4)+2(a^2b^2 + b^2c^2+c^2a^2)\\ =&-(a^2+b^2+c^2)^2+4(a^2b^2 + b^2c^2+c^2a^2). \end{array}

So (5) becomes

\displaystyle \frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C} =1+\frac{(a^2+b^2+c^2)^2}{16\mathrm{Area}^2(\Delta ABC)}. \ \ \ \ \ (6)


On the other hand, if we let {O} to be the centroid of {\Delta ABC}, then by parallelogram law,

\displaystyle a^2 = 2OB^2 + 2OC^2 - OA^2.

So (6) becomes

\displaystyle \frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C} =1+\frac{9(OA^2+OB^2+OC^2)^2}{16\mathrm{Area}^2(\Delta ABC)}. \ \ \ \ \ (7)


Now we consider the triangle {\Delta OAG} where the segment {AG} is obtained by parallel translating the segment {OB}.


It is easy to see that {\Delta OAG} has angles {D, E, F} at it vertices. ({\Delta OAG} is of course similar to the median triangle as described in the first proof). It is easy to see that

\displaystyle \mathrm{Area}(\Delta OAG)=\frac{1}{3}\mathrm{Area}(\Delta ABC)

as each median divides a triangle into two triangles with the same area.

As {\Delta }OAG has sides { OA}, {OB}, {OC}, by applying (6) to {\Delta OAG}, we get

\displaystyle \begin{array}{rl} \frac{1}{\sin^2 D}+ \frac{1}{\sin^2 E}+ \frac{1}{\sin^2 F} =&1+\frac{( OA ^2+ OB ^2+ OC ^2)^2}{16\mathrm{Area}^2(\Delta OAG)}\\ =&1+\frac{9(OA^2+OB^2+OC^2)^2}{16\mathrm{Area}^2(\Delta ABC)}. \end{array}

Comparing to (7), we can get the result.


Remark 1 As a bonus of the proof, for a triangle {\Delta ABC} with sides {a, b, c}, we have

\displaystyle \frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C} =1+\frac{(a^2+b^2+c^2)^2}{16\mathrm{Area}^2(\Delta ABC)}.

Of course a similar formula holds for the sum of squares of the cotangents.

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4 Responses to A curious identity on the median triangle

  1. Tsz Chiu Kwok says:

    I have another observation. Not really simpler though. Heron’s formula gives that for a triangle with sides a, b and c, we have
    4(a^2b^2 + b^2c^2 + c^2a^2) - (a^2 + b^2 + c^2)^2 = 16 \,\mathrm{ area}^2.
    Now \sum 1/\sin^2 (\mathrm{angles}) = (a^2b^2 + b^2c^2 + c^2a^2) / (4 \,\mathrm{area}^2) (your equation (2)). So by the above formula,
    \sum 1/\sin^2 (\mathrm{angles}) = (a^2 + b^2 + c^2)^2 / (16 \,\mathrm{ area}^2).

    Now consider triangle ABC with centroid O, we have
    (BC^2 + CA^2 + AB^2) = 3 (OA^2 + OB^2 + OC^2) since BC^2 = 2 OB^2 + 2 OC^2 - OA^2
    \mathrm{area}(BC, CA, AB) = 3 \,\mathrm{area}(OA, OB, OC) by your argument.
    So \sum 1/\sin^2 (\mathrm{angles}) are the same.

  2. KKK says:

    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 a^2b^2 etc. cannot be avoided, though in your case, it’s hidden inside the Heron’s formula 4(a^2b^2 + b^2c^2 + c^2a^2) - (a^2 + b^2 + c^2)^2 = 16 \mathrm{Area}(\Delta ABC)^2)

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

  3. Tsz Chiu Kwok says:

    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 $\mathrm{area}(OAB) = \mathrm{area}(OBC) = \mathrm{area}(OCA)$.

  4. KKK says:

    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 $latex your-latex-code-here$

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