Geometry + Radical Arithmetic = Headache!

I think I was using the method of Nihar Mahajan . It is as follows:

Let the perpendiculars from $E$ and $G$ to $AP$ be $EH$ and $GI$ and their lengths be $h_1$ and $h$ respectively, and $AP = b$ .

We note that $\angle APE = 30^\circ$ , therefore,

$\dfrac {PH}{EH} = \dfrac {b-2}{h_1} = \cot {30^\circ} = \sqrt{3}$

$\Rightarrow \dfrac {b-2}{h_1} = \dfrac {b-2}{4+4\sin{60^\circ}} = \dfrac {b-2}{4+2\sqrt{3}} = \sqrt{3} \quad \Rightarrow b = 8 + 4\sqrt{3}$

We also note that $\angle PAG = 45^\circ$ , therefore, $AI = GI = h$ , then we have:

$\dfrac {PI}{GI} = \dfrac {AP-AI}{GI} = \dfrac {b-h}{h} = \cot {30^\circ} = \sqrt{3} \quad \Rightarrow b - h = \sqrt{3}h$

$\Rightarrow h = \dfrac {8 + 4\sqrt{3}}{1 + 4\sqrt{3}} = \dfrac {(8 + 4\sqrt{3})(\sqrt{3}-1)}{2} = 2 + 2\sqrt{3}$

The area of $\triangle AGP$ ,

$A=\frac {1}{2} bh = \dfrac {(8 + 4\sqrt{3})(2 + 2\sqrt{3})}{2} = 20 + 12\sqrt{3}$

$\Rightarrow a + b + c = 20 + 12 + 3 = \boxed{35}$

So we know that $CE=4 \quad and \quad \angle CGE=75^{\circ}$

So applying Sine law in $\Delta CGE$

$\displaystyle\frac{CG}{Sin30}=\frac{4}{sin75^{\circ}}$

We get $CG=\frac{4\sqrt2}{\sqrt3+1}$ .

Hence $AG=AC+CG=4\sqrt2 +\frac{4\sqrt2}{\sqrt3+1}$

$=\frac{4\sqrt2(\sqrt3+2)}{\sqrt3+1}$

Now $\angle APG =30^{\circ}\text{ applying sine law in} \Delta APG$

$\displaystyle\frac{AG}{Sin30}=\frac{PG}{sin45^{\circ}}$

We get $PG=\frac{8(\sqrt3+2)}{\sqrt3+1}$

Now $ar(\Delta APG)=\frac{1}{2}\times AG\times PG \times sin\angle AGP$

Substituting values we get $ar(\Delta APG)=20+12\sqrt3$

Hence 20+12+3=35

We know that $CE=4 \quad and \quad \angle CGE=75^{\circ}$ So applying Sine law in $\Delta CGE$ $\dfrac{CG}{4}=\dfrac{sin30^o}{sin75^{\circ}}.~sin75^o=\dfrac{\sqrt6+\sqrt2}{4}.\therefore CG=\dfrac{4\sqrt2}{\sqrt3+1}=2\sqrt2(\sqrt3-1)$ . $\therefore AG=AC+CG=4\sqrt2 +2\sqrt2(\sqrt3-1)$ $=2\sqrt2(\sqrt3+1).$ $\therefore \dfrac{AG}{CG}=\dfrac{2\sqrt2( \sqrt3+1)}{2\sqrt2(\sqrt3-1)} = (\sqrt3+2).~~\therefore~\left(\dfrac{AG}{CG}\right)^2 = 7+4\sqrt3......(1)$ $Area~\Delta GCF=\dfrac{1}{2}* CF*CG *sin\angle FCG\\=\dfrac{1}{2}*4*2\sqrt2(\sqrt3-1)*\dfrac{1}{\sqrt2}=4*\left(\sqrt3-1\right)....(2)\\Area~\Delta AGP=\left (\dfrac{AG}{CG}\right)^2*Area~\Delta GCF=(1)*(2)\\=( 7+4\sqrt3 )*4*\left(\sqrt3-1\right)= 20+12\sqrt3.\\ \Huge \boxed{\color{#D61F06}{ ~~35~~}}$