How to find a,b and c in terms of the semi perimeter (s) from this equation?

Question

I have an equation: $(s-a)/4=(s-b)/3=(s-c)/2$ , from this I'm supposed to get a,b and c (the sides of a triangle) in terms of s.. If I substitute $s=(a+b+c)/2$ in the given system of equations I get three other equations : $a+3b-2c=0$ $2a+b-c$ and $4a-3b+c=0$ . But still I don't see any quick way to find a,b,c in terms of s. Is the only way to get the them in terms of s, to actually solve for a,b and c and then get them in terms of s?
*by getting them in terms of s, I mean being able to write them as $a = (constant)*s$ etc..

Answer 1

If $\frac a b=\frac c d$ then $\frac a b=\frac c d=\frac {a+c} {b+d}$ and similarly for equality for three fractions. Hence $\frac {s-a} 4=\frac {s-b} 3=\frac {s-c} 2=\frac {s-a+s-b+s-c} {4+3+2}=\frac {3s-2s} 9$. Now you can get your expressions for $a,b,c$ easily.

Answer 2

Let \begin{align} \tfrac14(s-a)=\tfrac13(s-b)=\tfrac12(s-c)&=k \quad\text{for some }k\in\mathbb{R} \end{align}

Then we have a linear system of four equations in four variables

\begin{align} \left[ \begin{matrix} 1&0&0&4 \\ 0&1&0&3\\ 0&0&1&2\\ 1&1&1&0 \end{matrix} \right] \cdot \left[ \begin{matrix} a\\b\\c\\k \end{matrix} \right] &= \left[ \begin{matrix} s\\s\\s\\2s \end{matrix} \right] , \end{align}

which can be easily solved with any standard method and has a solution in terms of $s$:

\begin{align} a &= \tfrac59\,s ,\quad b = \tfrac23\,s ,\quad c = \tfrac79\,s ,\quad k = \tfrac19\,s . \end{align}

Answer 3

A variation of Kavi Rama Murthy's answer. Denote the fractions by $k$: $$(s-a)/4=(s-b)/3=(s-c)/2=k \Rightarrow \\ s-a=4k;\\ s-b=3k;\\ s-c=2k.$$ Add them to find $k$: $$3s-2s=9k \Rightarrow k=\frac{s}{9}.$$ Hence: $$s-a=\frac{4s}{9} \Rightarrow a=\frac{5s}{9};\\ s-b=\frac{3s}{9} \Rightarrow b=\frac{2s}{3};\\ s-c=\frac{2s}{9} \Rightarrow c=\frac{7s}{9}.$$