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 $\ce{HF}$."



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> Can someone elaborate the cause and mention other such reagents like $\ce{KF}$ which can be used for the same purpose?
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$\ce{AlF3}$ 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, $\ce{K3[AlF6]}$.
$$\ce{AlF3 + 3KF -> K3[AlF6]}$$


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Since it is pure $\ce{HF}$ we are talking about, there wouldn't be any hydrogen bonding in the first place. Therefore, there wouldn't be any dissociation to form $\ce{H+}$ and $\ce{F-}$ ions. 


Whereas for the $\ce{KF}$ solution, it would easy dissociate to form $\ce{K+}$ and $\ce{F-}$ ions since it is dissolved in water. I believe the crux to this question is the water; since you need it to produce $\ce{F-}$ ions which can form the $\ce{[AlF6]^3-}$ complex. This is compounded by the fact that $\ce{AlF3}$ is ionic in character since fluorine is very electronegative, the $\ce{Al-F}$ is relatively weaker and will easily break in the presence of water to form the complex.


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