In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled

Question

In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.

My Attempt:

Given: $$\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$$ $$\dfrac {\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+b^2-c^2}{2ab}}{\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+c^2-b^2}{2ac}}=\dfrac {b}{c}$$ On simplification, $$\dfrac {ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3}{ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3}=\dfrac {b}{c}$$

Answer 1

Alternate approach:
Cross multiply
$\cos A \sin C + \sin 2C = \sin B \cos A + \sin 2B$
$\cos A (\sin C - \sin B) - (\sin 2B - \sin 2C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} - 2* \sin (B-C) \cos(B+C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2* \sin (B-C) \cos A=0$
$2\cos A [\sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2 \sin \frac{C-B}{2} \cos \frac{B-C}{2}] = 0$
$2\cos A *\sin \frac{B-C}{2}[\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2}]=0$

IF: $\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2} =0 \implies \tan \frac B2 \tan \frac C2 = -\frac 13$ and $\frac B2<90 $ and $\frac C2 < 90$

Answer 2

You are very close to answer.

• cross multiply.

• subtract 1 from both side.

• take b-c common from Numerator.

• It's done!! :)

Answer 3

In your work try to factor out a term like $(b-c)$ from the expression:

$$c(ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3) = b(ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3)\\ ab^2c+ac^3-a^3c + 2a^2c^2 +2b^2c^2 - 2 c^4 = ab^3 + abc^2 -a^3b + 2a^2 b^2 + 2b^2 c^2 -2b^4\\ abc(b-c) + a(c^3 - b^3 ) +a^3 (b-c)+2a^2 (c^2 - b^2) +2(b^4-c^4)$$

From here either $b=c$, or

$$abc - a(c^2+cb+b^2)+a^3 - 2a^2(c+b) + 2(b^3 + b^2c+bc^2 +c^3)=0\\ - ac^2 -ab^2 +a^3 -2a^2c - 2a^2 b + 2b^3 + 2b^2c+2bc^2 + 2c^3 = 0 \\ -2c(a^2-b^2 -c^2)+a(a^2-b^2-c^2)-2b(a^2-b^2-c^2) = 0\\ (a-2b-2c)(a^2-b^2-c^2)=0$$

This is only possible if $a^2 = b^2 +c^2 $ as $a-2b-2c$ is always negative due to triangle inequality.