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How to Prove $a^3 \cos (B-C) + b^3 \cos (C-A)+ c^3 \cos (A-B)=3abc$?

If $A,$ $B,$ and $C$ are the vertices of the Triangle ABC, and $a,$ $b$ and $c$ is the side opposite to the sides of the triangle.

You may use -

Projection Formula:

$$a=b \cos C + c \cos B$$

and Sine and Cosine Relation too.

My try

$$a^3 \cos (B - C)+ b^3 \cos (C-A) + c^3 \cos (A-B)=3abc$$ LHS = $$ 3(b \cos C + c \cos B)(a \cos C + c \cos A)(a \cos B + b \cos A)$$ $$ 3[ ab \cos B \cos ^2 C + abc \cos A \cos B \cos C +a^2c \cos ^2 B \cos C +ac^2 \cos A \cos ^2B ]+[b^2 \cos A \cos^2 C + b^2c \cos^2 A \cos A+abc \cos A \cos B \cos C + ac^2 \cos A \cos^2 B] $$

Now I can't solve it further because there seems no way to reduce that further.

Thanks :)

saulspatz
  • 53,131

1 Answers1

1

It can be proved with sine-rule and cosine rule but it is really ugly.

$${\rm LHS} - {\rm RHS} = \sum_{cyc} ( a^3\cos(B-C) - abc ) = \sum_{cyc}( a^3(\cos B\cos C + \sin B\sin C) - abc)$$ By sine rule,

$$a : b : c = \sin A : \sin B : \sin C \quad\implies\quad \begin{cases} a^3\sin B\sin C = abc\sin A^2\\ b^3\sin C\sin A = abc\sin B^2\\ c^3\sin A\sin B = abc\sin C^2 \end{cases}$$ This leads to

$$\begin{align}{\rm LHS} - {\rm RHS} &= \sum_{cyc}( a^3 \cos B \cos C - abc\cos^2 A)\\ &= \sum_{cyc}\left[a^3\left(\frac{a^2+c^2-b^2}{2ac}\right)\left(\frac{a^2+b^2-c^2}{2ab}\right) - abc\left(\frac{b^2+c^2-a^2}{2bc}\right)^2\right]\\ &= \frac{1}{4abc}\sum_{cyc}a^2(\underbrace{a^4 - (b^2-c^2)^2 - (b^2+c^2-a^2)^2}_{I}) \end{align} $$ Notice what's in the square bracket equals to $$\require{cancel}I =\cancel{a^4} - (b^2-c^2)^2 - (b^2+c^2)^2 + 2a^2(b^2+c^2) - \cancel{a^4} =2(a^2(b^2+c^2) - b^4-c^4)$$ We obtain

$$\begin{align}{\rm LHS} - {\rm RHS} &= \frac{1}{2abc}\sum_{cyc} a^4b^2 + \color{red}{a^4}\color{green}{c^2} - a^2 b^4 - \color{blue}{a^2}\color{magenta}{c^4}\\ &= \frac{1}{2abc}\sum_{cyc} a^4b^2 + \color{red}{b^4}\color{green}{a^2} - a^2 b^4 - \color{blue}{b^2}\color{magenta}{a^4}\\ &= 0 \end{align}$$

achille hui
  • 122,701
  • A beautiful equation hiding under an Ugly Proof! –  Jul 24 '18 at 16:43
  • @AbhasKumarSinha I agree the equation itself is beautiful, I suspect there is a geometric proof behind it. – achille hui Jul 24 '18 at 16:44
  • I think that this is theoretical proof only because we are taught derivation and the formula only today, because next month we will be high school –  Jul 24 '18 at 16:50
  • But how you Broke that all cosine ratios to convert difference of angles to single angle? –  Jul 24 '18 at 16:54
  • Oh okay, But I think that answer can be presented in more simple manner, but heck yeah! Complicated ones look better –  Jul 24 '18 at 16:56
  • Does using $\sigma$ instead of long chains of expression while proving '0' always works? –  Jul 24 '18 at 16:58
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    $\sum_{cyc}$ refers to cyclic sum over $a,b,c$. Because the sum is cyclic, for any expression $f(a,b,c)$, you have $\sum_{cyc}f(a,b,c) = \sum_{cyc}f(b,c,a) = \sum_{cyc} f(c,a,b)$. That's why you can replace the colored terms in the expression by corresponding terms (and cancelled them in this case). – achille hui Jul 24 '18 at 17:00
  • Can you try this question on Quora? - https://www.quora.com/unanswered/If-the-sides-of-the-Triangle-are-a-b-sqrt-a-2-ab-b-2-then-find-greatest-angle-find-the-measured-value-of-the-angle-not-which-one-is-greatest-See-Comment-for-more-details-about-the-Question –  Jul 24 '18 at 17:09
  • @AbhasKumarSinha I don't use quora. – achille hui Jul 24 '18 at 17:19