Find the value of p, q, r if $\frac{p}{q + r - p} = \frac{q}{p + r - q} = \frac{r}{p + q - r} $

Question

I equated these 3 to k:
$\frac{p}{q + r - p} = \frac{q}{p + r - q} = \frac{r}{p + q - r} $ = k

and got k=1. After this, $\frac{p}{q + r - p} = k = 1 $, and hence i got:
$2p = q + r$ , $2q = p + r$ and $2r = p + q$.
Solving these three equations, above, give $p = q = r$.

I am unable to proceed further for the values of p, q, r.
Please guide me how to solve it, Thanks

Answer 1

The family $p = q = r$ is not the only set of solutions. Indeed from

$$0 = \dfrac{q}{p+r-q} - \dfrac{r}{p+q-r} = \dfrac{(p+q+r)(q-r)}{(p+r-q)(p+q-r)}$$

one deduces that either $p+q+r = 0$ or $q = r$. If $p+q+r \neq 0$ then indeed $p = q = r$. But the set $\{p,q,r\mid p+q+r = 0,\,p,q,r\neq 0\}$ satisfies the equations:

$$\dfrac{p}{q+r-p} = \dfrac{p}{p+q+r-2p} = \dfrac{p}{-2p} = -\dfrac{1}{2}$$

interestingly enough, this means that $k$ is either $1$ or $-1/2$.

The solutions can be expressed as $(p,p,p)$ or $(p,q,-(p+q))$ where $p$ and $q$ are any nonzero value.

Answer 2

As we see they are symmetrical we can say that any integral value would satisfy all the three. the reason is symmetry. so if values $(x,y,z)$ satify these equations then $s.(values)$ where s is any real constant will also satisfy all the three.