find the maximum possible area of $\triangle{ABC}$

Let $ABC$ be of triangle with $\angle BAC = 60^\circ$ . Let $P$ be a point in its interior so that $PA=1, PB=2$ and $PC=3$. Find the maximum area of triangle $ABC$.

I took reflection of point $P$ about the three sides of triangle and joined them to vertices of triangle. Thus I got a hexagon having area double of triangle, having one angle $120$ and sides $1,1,2,2,3,3$. We have to maximize area of this hexagon. For that, I used some trigonometry but it went very complicated and I couldn't get the solution.

Solution 1:

Let $\theta=\measuredangle PAB$ in the triangle you specify. Then $\measuredangle PAC=60°-\theta$.

By the law of cosines,

$$2^2=1^2+c^2-2\cdot 1\cdot c\cdot \cos\theta$$ $$3^2=1^2+b^2-2\cdot 1\cdot b\cdot \cos(60°-\theta)$$

Solving those equations for $b$ and $c$,

$$c=\cos\theta+\sqrt{\cos^2\theta+3}$$ $$b=\cos(60°-\theta)+\sqrt{\cos^2(60°-\theta)+8}$$

Since the triangle's area is $\frac 12bc\sin A$ and $A=60°$, putting them all together we get

$$Area=\frac{\sqrt 3}{4}\left(\cos(60°-\theta)+\sqrt{\cos^2(60°-\theta)+8}\right)\left(\cos\theta+\sqrt{\cos^2\theta+3}\right)$$

where $0°<\theta<60°$. This looks like a bear to maximize analytically, so I did it numerically with a graph and got

$$Area_{max}\approx 4.66441363567 \quad\text{at}\quad \theta\approx 23.413°$$

I could get only five significant digits for $\theta$ but the maximum area should be accurate to all shown digits. I checked this with a construction in Geogebra and it checks. WolframAlpha timed out while trying to find an exact maximum value of the triangle's area from that my formula.

Solution 2:

Let $\mathcal{A}$ be the area of $\triangle ABC$. Let $\theta$ and $\phi$ be the angles $\angle PAC$ and $\angle BAP$ respectively.
We have $\theta + \phi = \angle BAC = \frac{\pi}{3}$. As functions of $\theta$ and $\phi$, the side lengths $b$, $c$ and area $\mathcal{A}$ are:

$$ \begin{cases} c(\theta) &= \cos\theta + \sqrt{2^2-\sin^2\theta}\\ b(\phi) &= \cos\phi + \sqrt{3^2-\sin^2\phi} \end{cases} \quad\text{ and }\quad \mathcal{A}(\theta) = \frac{\sqrt{3}}{4} c(\theta)b\left(\frac{\pi}{3}-\theta\right) $$ In order for $\mathcal{A}(\theta)$ to achieve maximum has a particular $\theta$, we need

$$\frac{d\mathcal{A}}{d\theta} = 0 \iff \frac{1}{\mathcal{A}}\frac{d\mathcal{A}}{d\theta} = 0 \iff \frac{1}{c}\frac{dc}{d\theta} - \frac{1}{b}\frac{db}{d\phi} = 0 \iff \frac{\sin\theta}{\sqrt{2^2-\sin^2\theta}} - \frac{\sin\phi}{\sqrt{3^2-\sin^2\phi}} = 0$$ This implies $$\frac{\sin\theta}{2} = \frac{\sin\phi}{3} = \frac13 \sin\left(\frac{\pi}{3} - \theta\right) = \frac13 \left(\frac{\sqrt{3}}{2}\cos\theta - \frac12\sin\theta\right) \iff 4\sin\theta = \sqrt{3}\cos\theta$$ and hence $$\theta = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 0.4086378550975924 \;\;( \approx 23.41322444637054^\circ )$$

Furthermore, we have $\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3} = \frac{\sqrt{3}}{2\sqrt{19}}\;$. Substitute this into the expression for side lengths and area, we get $$ \begin{cases} c &= \frac{4+\sqrt{73}}{\sqrt{19}}\\ b &= \frac{7+3\sqrt{73}}{2\sqrt{19}} \end{cases} \quad\implies\quad \mathcal{A} = \frac{\sqrt{3}}{8}(13+\sqrt{73}) \approx 4.664413635668018 $$ Please note that the condition $\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3}$ is equivalent to $\angle ABP = \angle ACP$. If one can figure out why these two angles equal to each other when $\mathcal{A}$ is maximized, one should be able to derive all the result here w/o using any calculus.