Solve for Rationals $p,q,r$ Satisfying $\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1$. Find all rational solutions $(p,q,r)$ to the Diophantine equation
$$\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1\,.$$
At least, determine an infinite family of $(p,q,r)\in\mathbb{Q}^3$ satisfying this equation.  If you can also find such an infinite family of $(p,q,r)$ with the additional condition that $0<p,q,r<1+\sqrt{2}$, then it would be of great interest.
This question is related to Three pythagorean triples.

If you combine the system into a single equation, you find that $r$ satisfies the quadratic equation
\begin{equation*}
r^2-\frac{4(p^2q+p(q^2-4q-1)-q)}{p^2(q^2-1)-4pq-q^2+1}r-1=0
\end{equation*}
For $r \in \mathbb{Q}$, the discriminant must be a rational square, so there must exist $d \in \mathbb{Q}$ such that
\begin{equation*}
d^2=(p^2+1)^2q^4-32p^2q^3+(2p^4-32p^3+68p^2+32p+2)q^2+32p^2q+(p^2+1)^2
\end{equation*}
The quartic has a rational solution when $q=0$, and so is birationally equivalent to an elliptic curve.
Using Mordell's method with a symbolic algebra package, we find this elliptic curve to be
\begin{equation*}
V^2=U^3+2(p^4-4p^3+10p^2+4p+1)U^2+(p^2-2p-1)^4U
\end{equation*}
with the reverse transformation
\begin{equation*}
q=\frac{(p^2+1)V+8p^2U}{(p^2+1)^2U+(p^2-2p-1)^4}
\end{equation*}
Numerical experiments suggest that the torsion subgroup of the curve is isomorphic to $\mathbb{Z4}$, and we find $(0,0)$ is a point of order $2$, and
$(-(p^2-2p-1)^2, \pm 4p(p^2-2p-1)^2)$ are of order $4$.
These experiments also suggest that the rank of the curves is always at least $1$, and it is not hard to find that $(-(p^2+1)^2, \pm 8p^2(p^2+1)\,)$
are of infinite order.
Taking the negative $V$ value gives
\begin{equation*}
q=\frac{2p(p^2+1)^2}{(p+1)(p-1)(p^4-2p^3+2p^2+2p+1)}
\end{equation*}
and, if we slot this into the quadratic for $r$ we get
\begin{equation*}
r=\frac{(p-1)(p^6-2p^5+7p^4-p^2+2p+1)}{(p+1)(p^6-2p^5-p^4+7p^2+2p+1)}
\end{equation*}
Because there is a point of infinite order, we can generate an infinite number of parametric solutions, though they become increasingly complicated.
Hope this helps, and a Merry Christmas to all. 
