In triangle ABC, with AB<AC, let H be the orthocenter and O be the circumcentre. Given that mid point of OH lies on BC, BC=1 and the perimeter of $\Delta ABC$ is 6 units then the area of  $\Delta \mathcal{l}e$ABC is  $\frac{a}{b}$sq. units  $(GCD(a,b)=1)$ then a + b is equal to _______

$O\to$ outside the triangle ABC
$H\to$ outside the triangle ABC
 Mid point of OH lies on BC
ABC is obtuse angled triangle 
consider $O\left(\overline{O}\right),A\left(\overline{a}\right),B\left(\overline{b}\right),C\left(\overline{c}\right)$  
$H\to \overline{a}+\overline{b}+\overline{c}\left|\overline{a}\right| =\left|\overline{b}\right|=\left|\overline{c}\right|=R$

mid point of  $OH=\frac{\overline{a}+\overline{b}+\overline{c}}{2}$
$\overline{b},\frac{\overline{a}+\overline{b}+\overline{c}}{2},\overline{c}$ are collinear
 $\Rightarrow \frac{1}{2}|(\frac{\overline{a}+\overline{b}+\overline{c}}{2}-\overline{b})\times (\overline{c}-\overline{b})|=0$
$\Rightarrow |\overline{a}\times \overline{b}-\overline{a}\times \overline{c}|=0\Rightarrow \overline{a}\times \overline{b}=\overline{a}\times \overline{c}$

consider  $\overline{)AOB}=\theta \Rightarrow \overline{)AOC}=\pi -\theta$  
 sine rule in  $\Delta \mathcal{l}eAOB:\frac{R}{\sin \left(\frac{\pi -\theta }{2}\right)}=\frac{AB}{\sin \theta }$
  $AB=\frac{R\sin \theta }{\cos \left(\frac{\theta }{2}\right)}=2R\sin \left(\frac{\theta }{2}\right)$
sine rule in  $\Delta \mathcal{l}eBOC :\frac{R}{\sin \theta }=\frac{BC}{\sin (\pi -2\theta )}$
$R=\frac{1}{2\cos \theta }$

sine rule in $\Delta \mathcal{l}eABC:\frac{AC}{\sin \left(\frac{\pi +\theta }{2}\right)}=2R$

$AC=2R\cos \left(\frac{\theta }{2}\right)$