Given Ratio, Find Sum

$\sum_{cyc} a\cos A = \sum_{cyc} 2R\sin A\cos A = R\sum_{cyc} \sin 2A =4R \prod_{cyc} \sin A \dots (1) \\ \dfrac{[ABC]}{R}=4 \Rightarrow \dfrac{abc}{4R^2}=4 \Rightarrow 8R^3\prod_{cyc} \sin A = 16R^2 \Rightarrow \prod_{cyc} \sin A = \dfrac{2}{R} \dots (2)$

Substitute $(2) \ in \ (1)$ ,

$\sum_{cyc} a\cos A = 4R\times \dfrac{2}{R} =\Large \boxed{8}$

First I found out that acosA+bcosB+ccosC=8 Δ² / ( abc )

Proof:

Let Δ = area of ΔABC.

bc sin A = sin B = ab sin C = 2Δ ................ (1) sin A = 2Δ / bc. & sin B = 2Δ / ca & sin C = 2Δ / ab. ........... (2)

LHS= a cos A + b cos B + c cos C, where a/sin A =m

= (m/2) [ 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ]

= (m/2) [ ( sin 2A + sin 2B ) + sin 2C ]

= (m/2) [ 2 sin (A+B)· cos(A-B) + sin 2C ]

= (m/2) [ 2 sin C. cos(A-B) + 2 sin C cos C ]

= m sin C [ cos(A-B) + cos C ] here cos C = cos [ π - (A+B) ] = - cos (A+B)

= m sin C [ cos(A-B) - cos(A+B) ]

= m sin C [ 2 sin A sin B ]

= 2 ( m sin A ) sin B sin C

= 2 a sin B sin C

= 2a [ 2Δ / ca ] [ 2Δ / ab } from (2)

= 8 Δ² / ( abc ) ................................................... (M)

= RHS

Now given Δ/R=4 so Δ=4R=abc/Δ implies Δ²=abc .................(N). [because (R=abc/4Δ)] substitute the value of (N) in (M) and we will get 8