Why is AlF3 soluble in anhydrous HF only in the presence of KF? I am pondering over that what must be the reason that $\ce{AlF3}$ is insoluble or less soluble in anhydrous $\ce{HF}$? And pageants it is soluble to a good extent only in the presence of $\ce{KF}$? Is there any other reagent in presence of which $\ce{AlF3}$ is soluble in $\ce{HF}$? 

I guess a probable answer for the cause of insolubility could be:
"the nonavailability of $\ce{F-}$ ions because of the presence of intermolecular hydrogen bonding in $\ce{HF}$, which makes $\ce{AlF3}$ insoluble in $\ce{HF}$."


  Can someone elaborate the cause and mention other such reagents like $\ce{KF}$ which can be used for the same purpose?
$\ce{AlF3}$ is insoluble in the anhydrous $\ce{HF}$ because the $\ce{F-}$ ions are not available in intermolecular hydrogen bonded $\ce{HF}$, but it becomes soluble in the presence of $\ce{KF}$ due to the formation of soluble complex, $\ce{K3[AlF6]}$.
$$\ce{AlF3 + 3KF -&gt; K3[AlF6]}$$