PQR is an isosceles triangle where PQ = PR. X is point on the circumcircle of ΔPQR, such that it being in the opposite region of P with respect to QR. The normal drawn from the point P on XR intersects XR at point Y. (see figure)
If XY=12, then find the value of QX+RX.
Copy the diagram left of the line $PX$, and rotate the triangle $\triangle PQX$ about $P$ so that the image of $PQ$ coincide with $PR$. Let the rotated image of $X$ be $X'$.
Consider the cyclic quadrilateral $PQXR$, $$\begin{align*} \angle PQX + \angle PRX &= 180^\circ\\ \angle PRX' + \angle PRX &= 180^\circ\\ \end{align*}$$
So $XRX'$ is a straight line.
Also, $PX = PX'$, so $\triangle PXX'$ is isosceles.
$Y$ is the base of the altitude drawn from $P$ onto the (possibly extended) straight line $XR$, and is always between the base vertices $XX'$. So $Y$ is the midpoint of $XX'$, and
$$\begin{align*} QX + XR &= RX' + XR\\ &= XX'\\ &= 2 XY\\ &= \underline{\underline{24}} \end{align*}$$