In a triangle, prove that $a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$

Question

I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$ where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS: $$ \begin {align} a\cos A&+b\cos B+c\cos C\\ &= a\left[2\cos^2\left(\frac A2\right)-1\right]+b\left[2\cos^2\left(\frac B2\right)-1\right]+c\left[2\cos^2\left(\frac C2\right)-1\right]\\ &=a\left[2 \frac{s(s-a)}{bc} -1\right]+b\left[2 \frac{s(s-b)}{ac} -1\right]+c\left[2 \frac{s(s-c)}{ab} -1\right]\\ &=2a\left (\frac{s(s-a)}{bc}\right)+2b\left (\frac{s(s-b)}{ac}\right)+2c\left( \frac{s(s-c)}{ab}\right)-2s\\ &=\frac{2s}{abc}[a^2(s-a)+b^2(s-b)+c^2(s-c)-abc] \end{align}$$ where $$s=\frac{a+b+c}{2}$$

I want to convert that last equation into Heron's formula: $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

But I'm stuck there. Any help please?

Answer 1

From the cosine rule:

$$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$

So

$$\begin{align}a\cos(A)+b\cos(B)+c\cos(C)&=a\frac{b^2+c^2-a^2}{2bc}+b\frac{a^2+c^2-b^2}{2ac}+c\frac{a^2+b^2-c^2}{2ab}\\ &=a^2\frac{b^2+c^2-a^2}{2abc}+b^2\frac{a^2+c^2-b^2}{2abc}+c^2\frac{a^2+b^2-c^2}{2abc}\\ &=\frac{a^2b^2+a^2c^2-a^4+a^2b^2+b^2c^2-b^4+a^2c^2+b^2c^2-c^4}{2abc}\\ &=\frac{-(a-b-c)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\ &=\frac{(b+c-a)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\ &=\frac{(2s-2a)(2s-2c)(2s-2b)(2s)}{2abc}\\ &=\frac{8(s-a)(s-c)(s-b)(s)}{abc}\\ &=\frac{8\Delta^2}{abc}\\ \end{align}$$

Answer 2

The polynomial $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}$$ vanishes for $c:=a+b$ and because it is cyclic it vanishes also at $a:=b+c$ and at $b:=a+c$. Notice that the same is true for $$(s-a)(s-b)(s-c).$$

In fact, for the second polynomial it is clear and for the first putting $c:=a+b$ in the first polynomial we get

$$\frac{2ba^2+2ab^2-2ab(a+b)}{8}\equiv 0.$$

Finally the two polynomials have the same leading coefficient $\frac{1}{8}$.

Therefore $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}=(s-a)(s-b)(s-c).$$

Answer 3

$t=a^2(s-a)+b^2(s-b)+c^2(s-c)-abc=a^2(s-a)+b^2(s-a)+c^2(s-a)+a(b^2+c^2-bc)-(b^3+c^3)\\b^3+c^3=(b+c)(b^2+c^2-bc)\\t=(s-a)(a^2+b^2+c^2)-(b^2+c^2-bc)(b+c-a),b+c-a=2(s-a)\\t=(s-a)(a^2+b^2+c^2-2(b^2+c^2-bc))=(s-a)(a^2-(b^2+c^2-2bc))=(s-a)(a^2-(b-c)^2)=(s-a)(a+c-b)(a+b-c)=4(s-a)(s-b)(s-c)$

Answer 4

HINT:

Use Law of sines $a=2R\sin A$ etc.

Now Double-Angle Formula $2\sin A\cos A=\sin2A$

Then Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle

Finally use $\triangle=\dfrac{abc}{4R}$ (Proof)