Area of a critical triangle ABC if PA,PB known and PC unknown

Question

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 $ABC ?$

Answer 1

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.

Answer 2

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.

Answer 3

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.