, 2012

Since $\angle MON=90^{\circ}$, the product of the slopes of $NO$ and $OM$ is $-1$. The slope of $NO$ is $\dfrac{n^2-0}{n-0}=n$, since $n\neq0$ (points $N$ and $O$ are distinct). The slope of $OM$ is $\dfrac{\tfrac{1}{4}-0}{\tfrac{1}{2}-0}=\dfrac{1}{2}$. Thus, $n\times\dfrac{1}{2}=-1$ or $n=-2$. Since $\angle ABO=90^{\circ}$, the product of the slopes of $BA$ and $BO$ is $-1$. The slope of $BA$ is $\dfrac{b^2-4}{b-2}=\dfrac{(b-2)(b+2)}{b-2}=b+2$, since $b\neq2$ ($A$ and $B$ are distinct). The slope of $BO$ is $\dfrac{b^2-0}{b-0}=b$, since $b\neq0$ ($B$ and $O$ are distinct). Thus, $(b+2)\times b=-1$ or $b^2+2b+1=0$. Factoring, $(b+1)(b+1)=0$ and so $b=-1$. Since $\angle PQR=90^{\circ}$, the product of the slopes of $PQ$ and $RQ$ is $-1$. The slope of $PQ$ is $\dfrac{p^2-q^2}{p-q}=\dfrac{(p-q)(p+q)}{p-q}=p+q$, since $p\neq q$ ($P$ and $Q$ are distinct). The slope of $RQ$ is $\dfrac{r^2-q^2}{r-q}=\dfrac{(r-q)(r+q)}{r-q}=r+q$, since $r\neq q$ ($R$ and $Q$ are distinct). Thus, $(p+q)\times (r+q)=-1$. Since $p,q$ and $r$ are integers, then $p+q$ and $r+q$ are integers. In order that $(p+q)\times (r+q)=-1$, either $p+q=1$ and $r+q=-1$ or $p+q=-1$ and $r+q=1$ (these are the only possibilities for integers $p,q,r$ for which $(p+q)\times (r+q)=-1$). In the first case, we add the two equations to get $p+q+r+q=1+(-1)$ or $2q+p+r=0$. In the second case, adding the two equations gives $p+q+r+q=-1+1$ or $2q+p+r=0$. In either case, $2q+p+r=0$, as required.