Crux problem #39 with vectors approach

Question

It's given a point $P$ inside an equilateral triangle $ABC$ such as segment lengths $PA$, $PB$, $PC$ are $3$, $4$ and $5$ units respectively. Calculate the area of $\triangle ABC$.

So, let $b:=\overrightarrow{AB},\,c:=\overrightarrow{AC},\, \overrightarrow{AP}=xb+yc$, $$\begin{cases} (\overrightarrow{AP})^2&=x^2b^2+y^2c^2+2xybc=9\\ (\overrightarrow{AP}-\overrightarrow{AB})^2&=(x-1)^2b^2+y^2c^2+2(x-1)ybc=16\\ (\overrightarrow{AP}-\overrightarrow{AC})^2&=x^2b^2+(y-1)^2c^2+2x(y-1)bc=25 \end{cases}$$ As we have equilateral triangle, $b^2=c^2=2bc=:\frac{1}{a^2}$, then $$\begin{cases} x^2+y^2+xy=9a^2\\ (x-1)^2+y^2+(x-1)y=16a^2\\ x^2+(y-1)^2+x(y-1)=25a^2 \end{cases}$$ $$\begin{cases} x^2+y^2+xy=9a^2\\ -2x+1-y=7a^2\\ -2y+1-x=16a^2 \end{cases}$$ $$\begin{cases} x^2+y^2+xy=9a^2\\ -2x+1-7a^2=y\\ -2(-2x+1-7a^2)+1-x=16a^2 \end{cases}$$ $$\begin{cases} 9x^2+9y^2+9xy=81a^2\\ -2x+1-7a^2=y\\ 3x-1=2a^2 \end{cases}$$ $$\begin{cases} (2a^2+1)^2+(1-25a^2)^2+(2a^2+1)(1-25a^2)=81a^2\\ y = \frac{1 - 25 a^2}{3}\\ x=\frac{2a^2+1}{3} \end{cases}$$ $$193 a^4 - 50 a^2 +1=0$$ $$\left[ \begin{array}{l} \begin{cases} a^2 = \frac{1}{193}\left(25 - 12 \sqrt{3}\right)\\ x = \frac{1}{193}\left(81 - 8 \sqrt{3}\right)\\ y = \frac{4}{193}\left(25 \sqrt{3} - 36\right) \end{cases}\\ \begin{cases} a^2 = \frac{1}{193}\left(25 + 12 \sqrt{3}\right)\\ x = \frac{1}{193}\left(81 + 8 \sqrt{3}\right)\\ y = -\frac{4}{193}\left(25 \sqrt{3} + 36\right) \end{cases} \end{array} \right.$$ The second case doesn't fit as $y<0$, then we have $$S_{ABC}=\frac{\sqrt{3}}{4a^2}=\frac{1}{4} \left(36 + 25 \sqrt{3}\right)$$ Although there are two solutions given in [Crux Vol. 1, No. 7][1] (pp. 64-66), I wonder is there more neat geometrical solution?

Answer 1

Rotate The whole figure about $B$ anticlockwise by $\pi/3$ radians and assume $P$ moves to $P'$ then $PP'B$ is equilateral and since $C$ moves to $A$ , $APP'$ is a $3-4-5$ triangle.

Now you easily calculate $AB$ from $AP'B$ by cosine rule.

Answer 2

Set the triangle with $A(0,0)$, $B(0,z)$ and $ C(z/2, z\sqrt{3}/2)$, where $z$ is the side length (this is the same orientation of the figure in the OP). Call $x$ and $y$ the coordinates of $P$. Now you can solve the problem nearly in one step by solving the following system in three variables, where the equations express the distances of $P$ from the vertices:

$$\displaystyle\left\{ \begin{array}{ll} (x^2+y^2)=9\\ (z-x)^2+y^2=16\\ (z/2-x)^2+(z\sqrt{3}/2-y)^2=25 \end{array} \right. $$

Solving the system and taking the only solution where $x$, $y$, and $z$ are all positive directly leads, after some simplification of the radicals, to

$$z = (14 \sqrt{3} - 3) \sqrt{\frac{1}{579} (31 + 16 \sqrt{3})}\\ = \sqrt{25+12\sqrt{3}}$$

as confirmed by WA here and here.

From this we easily obtain

$$S= \frac{\sqrt{3}}{4} (25+12\sqrt{3})\\ =\frac{1}{4} (36 + 25 \sqrt{3}) $$