Probability of a point in a region of a triangle

An uniformly random chosen point $P$ in the triangle has the following parametrization:

$\displaystyle P=\{(x,y) \in \mathbb{R^2} \hspace{5pt} | \hspace{5pt} \frac{4}{5}y \leq x \leq 12-\frac{2}{5}y, \hspace{5pt} 0 \leq y \leq 10\}$ .

Point $P$ has to be closer to $B$ than to the other vertices, so $\overline{PB} \leq \overline{PA}$ and $\overline{PB} \leq \overline{PC}$ :

$\displaystyle \begin{cases} (x-12)^2+y^2 \leq x^2+y^2 \\ (x-12)^2+y^2 \leq (x-8)^2+(y-10)^2 \end{cases} \hspace{8pt} \Longrightarrow \hspace{8pt} \begin{cases} x \geq 6 \\ \displaystyle y \leq \frac{1}{5} (2x+5) \end{cases}$

The previous result describes a portion that we'll call $A$ of the $xy$ plane:

$\displaystyle A=\{(x,y) \in \mathbb{R^2} \hspace{5pt} | \hspace{5pt} x \geq 6, \hspace{5pt} y \leq \frac{1}{5} (2x+5)\}$ .

The portion of the region described by $P$ we're interested in is $S$ , such that

$\displaystyle S=P \cap A = S_1 \cup S_2$

where

$\displaystyle S_1=\{(x,y) \in \mathbb{R^2} \hspace{5pt} | \hspace{5pt} 6 \leq x \leq 12-\frac{2}{5}y, \hspace{5pt} 0 \leq y \leq \frac{17}{5}\}$ ;

$\displaystyle S_2=\{(x,y) \in \mathbb{R^2} \hspace{5pt} | \hspace{5pt} \frac{5}{2} (y-1) \leq x \leq 12-\frac{2}{5}y, \hspace{5pt} \frac{17}{5} \leq y \leq 5\}$ .

Than

$\displaystyle Area(S)=Area(S_1)+Area(S_2)=\int_{0}^{\frac{17}{5}} \int_{6}^{12-\frac{2}{5}y} dxdy+\int_{\frac{17}{5}}^{5} \int_{\frac{5}{2}(y-1)}^{12-\frac{2}{5}y} dxdy=\frac{109}{5}$ ;

$\displaystyle Area(P)=\frac{10 \cdot 12}{2}=60$ ;

Eventually

$\displaystyle \mathbb{P}(S)=\frac{Area(S)}{Area(P)}=\frac{109}{5\cdot 60}=\frac{109}{300}=\frac{p}{q} \hspace{5pt} \Longrightarrow \hspace{5pt} p+q=\boxed{409}$