In $\triangle ABC$, $(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = $?

Question

In $\triangle ABC$, $$(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = \text{?}$$

I tried solving this by cosine rule but it is becoming too long. Any short step solution for this?

Answer 1

Solving this problem by Law of Cosines isn't short but not that unmanageable.

The key is the observation of pattern within the expression:

$$(b^2-c^2)\cot A = \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R} = \frac{R}{abc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]$$ This has the form $\phi(a,b,c) - \phi(b,c,a)$ where $$\phi(a,b,c) = \frac{R}{abc}(b^4 - a^2b^2)$$ Other two terms have a similar form. They can be obtained from above by replacing $(a,b,c)$ with $(b,c,a)$ and $(c,a,b)$ respectively.

When you sum over all 3 terms, you are performing a cyclic sum over $a,b,c$. Terms of the form $\phi(a,b,c) - \phi(b,c,a)$ will simply cancel each other. The end result is $0$.

$$\sum_{cyc} (b^2-c^2)\cot A = \sum_{cyc} \left( \phi(a,b,c) - \phi(b,c,a)\right) = \sum_{cyc}\phi(a,b,c) - \sum_{cyc}\phi(a,b,c) = 0$$

In cases when you need to present a complete derivation but you don't want to become too verbose, you can do something like this:

Let $R$ be the circumradius, we have $$\begin{align} \sum_{cyc} (b^2-c^2)\cot A &= \sum_{cyc} \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R}\\ &= \frac{R}{abc} \sum_{cyc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]\\ &= \frac{R}{abc} \left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(c^4 - b^2c^2)\right]\\ &= \frac{R}{abc}\left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(b^4 - a^2b^2)\right]\\ &= 0 \end{align} $$

Answer 2

Using Proof of the sine rule

and Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$(b^2-c^2)\cot A$$

$$=4R^2(\sin^2B-\sin^2C)\cot A$$

$$=4R^2\sin(B+C)\sin(B-C)\cdot\dfrac{-\cos(B+C)}{\sin(B+C)}$$

$$=-4R^2\sin(B-C)\cos(B+C)=2R^2(\sin2C-\sin2B)$$