In Problem 1, RMO 2004 there is a particular choice of length which leads to the solution, the length being that of the tangent from the foot of the perpendicular to the circle.
Just a rundown of the problem: Find $Q$ on $OR$ such that $x = y$ for every $P$ on line $l$.
The most important part of the solution is considering tangent $XR$. From there we construct the equations
$$y^2 + r^2 = z^2$$ $$(r^2 + u^2) + t^2 = z^2$$ $$u^2 + t^2 = x^2$$ Note $r$ is radius. From there we substitute to get $$r^2 + x^2 = z^2$$ $$y^2 + r^2 = z^2$$ Combining gives $y^2 = x^2$ which is what is desired. Of course this solution is nowhere near possible without the introduction of $XR = u \:$. By considering this tangent of fixed length $u$ and fixing $Q$ on $OR$ such that $QR = u$ the solution unfolds. However this particular choice of $Q$ and $u$ doesn't seem very intuitive to me. What could be the logic and thought, or rather "steps" that go into choosing such a $Q$?
Simple: $X$ is a special case of $Y$ and $u$ is a corresponding special case of $y$ in the case of $P=R$.
Then $Q$ is found so that $PQ = RQ$ becomes equal $PY = RX$.
EDITED – fixed typos.