Area of a critical triangle ABC if PA,PB known and PC unknown help me to solve this this problem please:
In a triangle $ABC$, $\angle BAC$ = $60\,^{\circ}$,$AB=2AC$.Point P is inside the triangle such that $PA=\sqrt{3}$,$PB=5$. What is the area of triangle $ABC ?$
 A: You can apply the cosine rule:


*

*By applying it on $\bigtriangleup APB$, $c^2=25+3-2\cdot5\cdot\sqrt{3}\cdot\cos\angle APB$

*Now, $\bigtriangleup ABC=\frac{1}{2}\cdot cb\cdot \sin60=c^2\frac{\sqrt{3}}{8}$ [as $c=2b$]

*So we must need $\angle APB$. If it is given we are done.

A: The shape of $ABC$ is determined up to similarity, without the information on $P$, so the question is effectively asking for a removal of the scaling ambiguity, which would mean a complete determination of all distances between the points (or the reduction to a very small finite number of possibilities, such as having a "$P$ inside $ABC$" and a "$P$ outside" solution, each with one or two possibilities for the shape and size of the figure $ABCP$).  This is impossible:
If the information is given in the form of the data on $ABC$ and the ratio $PA/PB$, the whole diagram is determined only up to similarity, and the locus of $P$ (for fixed positions of $A,B,C$) with that ratio of distances is a circle.  Thus, one can continuously change the scale of $ABC$ (and the circle) in some range, while at the same time adjusting the position of $P$ on the circle to keep $PA$ constant (therefore equal to $3$, in some adapted but constant system of units) and this will also maintain the correct value for $PB$, satisfying all conditions of the problem (both $PA$ and $PB$, and not only their ratio, will have the correct value) while varying the shape of the figure $ABCP$.
Because there is a continuous deformation of the shape of the solutions, they are not unique.
A: 
It's trivially found that $\triangle ABC$
is the right triangle with 
$\angle BCA=90^\circ$ and 
$\angle ABC=30^\circ$.
Let's fix the point $A$ at the origin
with two rays $AB$ and $AC:$
$\angle CAB=60^\circ$.
Since $|PA|=\sqrt3$ is fixed, 
the point $P$ must be located 
on the circular arc $P_{\min}P_{\max}$
centered at $A$.
In order to fit the other condition
$|PB|=5$, 
the point $B$ must be located 
anywhere
on the segment $B_{\min}B_{\max}$,
\begin{align} 
|AB_{\max}|&=|PA|+|PB|=5+\sqrt3
,\\
|AB_{\min}|&=|AD|+|DB_{\min}|
=\tfrac{\sqrt3}2+\sqrt{5^2-\tfrac94}
=\tfrac12\,(\sqrt3+\sqrt{91})
.
\end{align} 
Since $\angle BCA=90^\circ$,
the point $C$ is found at the intersection 
between the ray $AC$ and the circle centered at the middle of $AB$
with radius $\tfrac12\,|AB|$.
The area $S$ of $\triangle ABC$
must be between $S_{\min}$ and $S_{\max}$,
\begin{align} 
S_{\min}&=\tfrac{\sqrt3}8\,|AB_{\min}|^2
=\tfrac1{16}\,(47\,\sqrt3+3\,\sqrt{91})
\approx 6.876535251
,\\
S_{\max}&=\tfrac{\sqrt3}8\,|AB_{\max}|^2
=\tfrac{\sqrt3}8\,(5+\sqrt3)^2
=\tfrac14\,(15+14\,\sqrt3)
\approx 9.812177826
,
\end{align}
so at best we can say that
\begin{align} 
S&=
\tfrac1{32}\,(60+103\,\sqrt3+3\,\sqrt{91})
\pm
\tfrac3{32}\,(20+3\,\sqrt3-\sqrt{91})
,\\
&\approx
8.344356539\pm 1.467821289
.
\end{align}
A: Let $a:=AC$ ; therefore $BC=a\sqrt{3}$, $AB=2a$.
Let $b=PC$ and $\alpha:=a^2$, $\beta:=b^2$.
Let us express that the Cayley-Menger determinant of the four points $A,B,C,P$ is zero (see here):
$$\begin{vmatrix}0 & 4\alpha & \alpha & 3 & 1\\
   4\alpha & 0 &3\alpha &25& 1\\
   \alpha & 3\alpha &0& \beta& 1\\
     3& 25 &\beta &0 &1\\
     1&1 &1 &1& 0\end{vmatrix}=0\tag{1}$$
Expanding, one obtains :
$$-8\alpha(3\alpha^2 - 42\alpha + \beta^2 - 17\beta + 163)=0\tag{2}$$
Dropping the case $-8\alpha=0$ (which is the impossible case of a triangle reduced to a point), (2) is equivalent to :
$$3(\alpha-7)^2+(\beta-\dfrac{17}{2})^2=\left(\dfrac{15}{2}\right)^2\tag{3}$$
which is the equation of an ellipse represented here :

Fig. 1 : The ellipse with equation (3) with respect to axes $\alpha=a^2$ and $\beta=b^2$.
As we are looking for a maximal value for the triangle's area :
$$S=\dfrac{1}{2} a \times \sqrt{3}a = \dfrac{\sqrt{3}}{2}a^2=\dfrac{\sqrt{3}}{2}\alpha\tag{4}$$
It suffices to consider the maximal value $\alpha_{max}$ of $\alpha$ which is reached for the rightmost point of the ellipse (indicated by a little star on Fig. 1) giving 
$$\alpha_{max}=7+\sqrt{3}\dfrac{15}{2}\tag{5}$$
Using (4) and (5), we  can assert that the maximal area is

$$\alpha_{max}\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}+\dfrac{45}{4}\approx 9.8121778...$$

which coincides with the value obtained by G.Kov, and corresponds to the extreme case where point $P$ is on side $AB$.
Remarks : 
1) In fact, we have worked using "necessary arguments". We have checked that the solution obtained is actually a "feasible solution". But, there are constraints that we haven't handled, which in particular forbid some parts of the ellipse, but without impact on this solution. 
2) Equation (2), when we take $\beta=2^2$ gives back equation $a^4-14a^2+37=0$ in a solution of the fully constrained problem indicated by Calvin Lin.
A: There is not enough information to solve the problem.  By drawing the line from $C$ to the midpoint of $AB$, we know $ABC$ is $30-60-90$.  We don't have enough to get $AB$.  We know $5-\sqrt 3 \lt AB \lt 5+\sqrt 3$ and some of the low end may drive $P$ outside the triangle, but there are triangles of various areas that satisfy the requirement.
