ratio based question

Question

If p,q and r are three distinct real numbers such that $(pq+1):(qr+1):(rp+1)$ is $q:r:p$,then prove that $|pqr|=1$.

Answer 1

We have $\dfrac{pq+1}{q} = \dfrac{qr+1}{r} = \dfrac{rp+1}{p}$, or $p + \dfrac1q = q+\dfrac1r = r+\dfrac1p$. Considering each pair of expressions separately, we get the three equations $$\left\{\begin{aligned}p - q &= \dfrac{1}{r} - \dfrac{1}{q} = \dfrac{q-r}{qr} \\ q - r &= \dfrac1p - \dfrac1r = \dfrac{r-p}{pr} \\ r-p &= \dfrac1q - \frac1p = \dfrac{p-q}{pq} \end{aligned}\right.$$ Now multiply: $(p-q)(q-r)(r-p) = \dfrac{(q-r)(r-p)(p-q)}{p^2q^2r^2}.$ Since $p, q, r$ are distinct, we can cancel $(p-q)(q-r)(r-p)$ on both sides to get $p^2q^2r^2 = 1$. Then $|pqr| = 1$, as requested.