If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$

Question

If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
show that $$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$

where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c$ are sides of $\triangle ABC$.

Actually, I don't have any idea. Please someone help. It would be great if someone who answers include how he/she found where to begin.

Answer 1

\begin{align} \text{If }\quad \frac{b+c}{2k-1}&=\frac{c+a}{2k}=\frac{a+b}{2k+1} \tag{1}\label{1} ,\\ \text{show that }\quad \frac{\sin A}{k+1}&=\frac{\sin B}{k}=\frac{\sin C}{k-1} \tag{2}\label{2} . \end{align}

From \eqref{1} we have by the rules based on componendo and dividendo,

\begin{align} \frac{b+c}{2k-1}&=\frac{c+a}{2k}=\frac{a+b}{2k+1} \\ &= \frac{-(b+c)+(c+a)+(a+b)}{-(2k-1)+(2k)+(2k+1)} = \frac{2a}{2k+2} = \frac{a}{k+1} \\ &= \frac{(b+c)-(c+a)+(a+b)}{(2k-1)-(2k)+(2k+1)} = \frac{2b}{2k} = \frac{b}{k} \\ &= \frac{(b+c)+(c+a)-(a+b)}{(2k-1)+(2k)-(2k+1)} = \frac{2c}{2k-2} = \frac{c}{k-1} . \end{align}

Now we have \begin{align} \frac{a}{k+1}&= \frac{b}{k}= \frac{c}{k-1} \tag{3}\label{3} . \end{align}

Recall that by the sine rule for $\triangle ABC$ with the radius of circumscribed circle $R$

\begin{align} a&=2R\sin A,\quad b=2R\sin B,\quad c=2R\sin C, \end{align}

so \eqref{3} becomes \begin{align} \frac{2R\sin A}{k+1}&= \frac{2R\sin B}{k}= \frac{2R\sin C}{k-1} \end{align}

and \eqref{2} follows.

Answer 2

Hint:

$$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}=\frac{(b+c)+(c+a)+(a+b)}{(2k-1)+(2k)+(2k+1)}=\frac{a+b+c}{3k}$$

$$\frac{b+c}{2k-1}=\frac{a+b+c}{3k}=\frac{(a+b+c)-(b+c)}{(3k)-(2k-1)}$$

You can easily find $a:b:c$ and apply the sine formula.

Answer 3

Let $m$ be the common value of the initial fractions, so that $$\begin{align} a + b &= ( 2 k + 1 ) m \tag{1} \\ b + c &= ( 2 k - 1 ) m \tag{2}\\ c + a &= ( 2 k \phantom{+1\;\,}) m \tag{3} \end{align}$$

We can combine these equations to get $$\begin{align} \phantom{-}(1)-(2)+(3):&\quad 2 a = 2(k+1)m \\ (1)+(2)-(3):&\quad 2b= 2(k\phantom{+1\;\,})m \\ -(1)+(2)+(3):&\quad 2c= 2(k-1)m \end{align} \tag{4}$$

By the Law of Sines, $$\sin A : \sin B : \sin C = a : b : c = k + 1 : k : k - 1 \tag{5}$$ and the result follows. $\square$