Question

Let circumcentre of a triangle $ABC$ be $O (0,0)$ and orthocentre of triangles $OBC , OCA$ and $OAB$ are $A , B$ and $C$ respectively. If $OA =\frac{3}{2}$ and area of $\triangle ABC =\frac{ a \sqrt{3}}{ b }$ where $a , b$ are respectively prime numbers, then find the value of $(2 a-3 b)$.

Answer 1

O is the circumcentre as well as orthocentre of the triangle.

$\therefore \triangle ABC$ is equilateral triangle.
$OA =\frac{3}{2} \Rightarrow AD =\frac{3}{2}+\frac{3}{4}=\frac{9}{4} $
$\Delta=\frac{( AD )^2}{\sqrt{3}}=\frac{81}{16 \sqrt{3}}=\frac{27 \sqrt{3}}{16} \equiv \frac{ a \sqrt{3}}{ b }$
$\therefore 2 a -3 b =54-48=6$