Geometry With Calculus?

Let $\angle PBC = \alpha$ , $\angle ACB = ABC = \theta$ and $AB=AC=3a$ $\implies CP=a$ and $AP=2a$ .

Using sine rule on $\triangle PBC$ : $\dfrac {\sin \alpha}a = \dfrac {\sin \theta}1$ $\implies {\color{#3D99F6} \sin \alpha = a \sin \theta}$ .

Using sine rule on $\triangle ABP$ :

$\begin{aligned} \frac {\sin (\theta-\alpha)}{2a} & = \frac {\sin (180^\circ - 2\theta)}1 \\ \sin \theta \cos \alpha - \sin \alpha \cos \theta & = 2a \sin 2 \theta \\ & = 4 {\color{#3D99F6} a \sin \theta} \cos \theta & \small \color{#3D99F6} \text{Note that } \sin \alpha = a \sin \theta \\ & = 4 {\color{#3D99F6} \sin \alpha} \cos \theta \\ \sin \theta \cos \alpha & = 5 \sin \alpha \cos \theta \\ \implies \tan \theta & = 5 \tan \alpha \end{aligned}$

Let $BC = b$ , then $b = \cos \alpha + \dfrac {\sin \alpha}{\tan \theta} = \cos \alpha + \dfrac {\sin \alpha}{5 \tan \alpha} = \dfrac 65 \cos \alpha$

Let the height of $\triangle ABC$ be $h$ . Then the area of $\triangle ABC$ is:

$\begin{aligned} A & = \frac {bh}2 = \frac b2 \cdot \frac b2 \tan \theta = \frac {5b^2}4 \tan \alpha = \frac 54 \left( \dfrac 65 \cos \alpha \right)^2 \tan \alpha = \frac 95 \sin \alpha \cos \alpha \end{aligned}$

$\implies A = \dfrac 9{10} \sin 2\alpha$ and $A_{max} = \dfrac 9{10} = \boxed{0.9}$ , when $\sin 2 \alpha = 1$ .

We use the Cartesian Coordinate System. Take $A= (0,0) , C = (3a,3b)$ , where $a,b > 0$ and $B= (-3a,3b)$ . By section formula, $P= (2a,2b)$ . Now,

Area of $\Delta ABC = \frac{1}{2} \times$ height $\times BC = \frac{1}{2} \times 3b \times 6a = 9ab$ .

Also, $1= BP^{2} = (5a)^{2} + b^{2} \geq 2(5a)(b) = 10ab$ , so $\Delta ABC = 9ab \leq 0.9$ , with equality iff $5a=b (= \frac{1}{\sqrt{2}})$

Let $\angle B = \angle C = a \implies \angle A = 180 - 2a$

Let $x = CP \implies AP = 2x \ \text{and} \ AB = 3x$

By law of cosines in ∆ABP,

$\begin{aligned} 9x^2 + 4x^2 - 2(3x)(2x)( \cos(180-2a)) & = 1 \\ 13x^2 - 12(- \cos 2a)(x^2) & = 1 \\ 13x^2 + 12(\cos 2a)(x^2) & = 1 \\ 13 + 12 \cos 2a & = \cfrac{1}{x^2} \\ 12 \cos 2a & = \cfrac{1 - 13x^2}{x^2} \\ \cos 2a & = \cfrac{1 - 13x^2}{12x^2} = \cfrac{x}{r} \\ \sin 2a & = \cfrac{y}{r} = \cfrac{\sqrt{144x^4 - (1 - 26x^2 + 169x^4)}}{12x^2} \\ \sin 2a & = \cfrac{\sqrt{-1 + 26x^2 - 25x^4}}{12x^2} \end{aligned}$

Now, Area of ∆ABC = $\cfrac{1}{2}(AB)(AC)(\sin A) = 0.5 \times 3x \times 3x \times \sin (180-2a)$

$= 0.5\times (9x^2) \times (\sin 2a) \\ = 0.5 \times (9x^2) \times \cfrac{\sqrt{-1 + 26x^2 - 25x^4}}{12x^2} \\ = \cfrac{9}{24} \left(\sqrt{-1 + 26x^2 - 25x^4} \right) \\ = \cfrac{9}{24} \left(\sqrt{-25 \left(x^2 - \cfrac{13}{25} \right)^2 + \cfrac{144}{25}} \right)$

Max Area of ∆ABC occurs when $-25 \left(x^2 - \cfrac{13}{25} \right)^2 = 0$

$\begin{aligned} \implies \text{Max area of} \ \triangle ABC & = \cfrac{9}{24} \left(\sqrt{0 + \cfrac{144}{25}} \right) \\ & = \cfrac{9}{24} \times \cfrac{12}{5} \\ & = \cfrac{9}{10} \end{aligned}$ .