Three Chords & The Truth

$\angle APB=\angle BPC \Rightarrow \widehat{AB}=\widehat{BC} \Rightarrow \overline{AB}=\overline {BC}$

Let, $\overline{AB}=\overline {BC}=x$ .

Then, by Ptolemy's Theorem,

$\overline{PA}\cdot \overline{CB}+\overline{PC} \cdot \overline{AB}= \overline{PB}\cdot \overline{AC}$
$\Rightarrow 5x+3x= 8\overline{AC}$
$\Rightarrow 8x=8\overline{AC}$
$\Rightarrow x=\overline{AC}$
$\therefore \overline{AB}=\overline{BC}=\overline{AC}$

So, $\triangle ABC$ is equilateral.
$\therefore \angle APB=\angle ACB=60^\circ$

Now, applying Cosine-Rule to $\triangle APB$ , we get,

$AB^2$ $=PB^2+PA^2-2\cdot PB\cdot PA\cdot \cos{\angle APB}$
$= 8^2+5^2-2.8.5\cos{60^\circ}$
$=49$
$\therefore \overline{AB}=7$ .

Now, $\overline{AB}=2R\cdot\sin{60^\circ}\Rightarrow R=\frac{7}{\sqrt{3}}$

$\therefore$ Area of $\Gamma=\pi R^2=\pi\cdot(\frac{7}{\sqrt{3}})^2=\frac{49}{3}\pi$

$\therefore a+b=49+3=\boxed{52}$

The question asks to find $R^2$ where $R$ is the radius. image

In the figure, $\angle OCP = \angle OPC = \theta$ and $\angle OPA =\angle OAP = \alpha$ .

Now,

$2Rcos(\theta)= 3$

$2Rcos(\alpha)=5$

$2Rcos(\frac{\theta- \alpha}{2})=8$

So,

$2cos^2(\frac{\theta- \alpha}{2})-1=cos(\theta- \alpha)=2(\frac{4}{R})^2-1$

$\implies cos(\theta)cos(\alpha)+sin(\theta)sin(\alpha)=\frac{32}{R^2}-1$

$\implies \frac{3}{2R} \frac{5}{2R} + \frac{\sqrt{4R^2-9}}{2R} \frac{\sqrt{4R^2-25}}{2R}=\frac{32}{R^2}-1$

$\implies \sqrt{(4R^2-9)(4R^2-25)} = 113-4R^2 \implies R^2=\frac{12544}{768} = \frac{49}{3}$

$\implies \pi R^2 = \pi \frac{49}{3}$

Hence, $49+3= \boxed{52}$

Since angle APB = angle BPC, line AB = line BC. Let angle APB = x. Using cosine rule on triangle PAB and PBC, we get AB^2 = 73 - 48cosx = 89 - 80cosx. Solving it, we get x = 60 degrees and AB = 7. Using the extended sine rule on triangle PAB, we get 7/sin(60) = 2R where R is the circumradius of triangle PAB. Finally, we get a/b = R^2 = 49/3. a+b = 52.