Finding a triangle ratio. 
In the triangle ABC, the point P is found on the side AB. AC = 6 cm, AP = 4 cm, AB = 9 cm. Calculate BC:CP.

For some reason, I cannot get this even though I tried for half an hour.
I got that, $BC/CP < 17/10 = 1.7$ by the triangle inequality. $AP/PB = 4/5$
But that does not help one but, I'm very stuck!
 A: Let $\angle{BAC}=\theta$. Then, by the law of cosines, 
$$\begin{align}BC:CP&=\sqrt{9^2+6^2-2\cdot 9\cdot 6\cos\theta}:\sqrt{4^2+6^2-2\cdot 4\cdot 6\cos\theta}\\&=\sqrt{9(13-12\cos\theta)}:\sqrt{4(13-12\cos\theta)}\\&=\color{red}{3:2}\end{align}$$
A: The given problem cannot be solved.
If points $P,C,B$ are connected by an elastic rubber band varying angle at $A$ by hinging about $A$ indefinitely many solutions can exist.
EDIT 1/2
It turns out the problem has no $unique$ solution as I originally expected, but is a locus.
EDIT3

Sketched here what I had in mind when I said what I said about locus. It is not relevant to the problem or its solution. However allow me only to say in modest any understanding or context of what we are saying I believe is essential in mathematical considerations.
A: Notice, let $BC=x$ & $CP=y$ now, applying cosine rule in $\triangle ABC$ $$\cos\angle BAC=\frac{AB^2+AC^2-BC^2}{2(AB)(AC)}=\frac{9^2+6^2-y^2}{2(9)(6)}=\frac{117-y^2}{108}\tag 1$$
similarly, applying cosine rule in $\triangle APC$
$$\cos\angle BAC=\frac{AP^2+AC^2-PC^2}{2(AP)(AC)}=\frac{4^2+6^2-x^2}{2(4)(6)}=\frac{52-x^2}{48}\tag 2$$
Equating (1) & (2), we get 
$$\frac{117-y^2}{108}=\frac{52-x^2}{48}$$ 
$$468-4y^2=468-9x^2$$
$$\frac{y^2}{x^2}=\frac{9}{4}\implies \frac{y}{x}=\frac{3}{2}$$
$$\color{red}{BC:CP=3:2}$$
A: Interestingly enough, this problem can also be solved by treating the triangle as three points and knowing that $C$ must be $6$ units away from $A$. If you put $A$ at $(0,0)$ $P$ at $(4,0)$, and $B$ at $(9,0)$, $C$ must be somewhere on the circle $x^2 + y^2 = 36$.
If $C$ is on that same line at $(-6,0)$, $BC=15$ and $CP=10$.
If $C$ is on the line at $(6,0)$ instead, $BC=3$ and $CP=2$.
I will say that the Law of Cosines solution from MathLove is probably the way to go about this though.
