If the area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths  $2\sqrt{3},5$ and  $\sqrt{37}$, as shown, is $\frac{m\sqrt{3}}{n}$ , (where  $m,n$ are positive integers, m and n are relatively prime) then the value of   $m+n=$

The distance between the origin and any point (x, y) on the line p x + q y=1 is atleast $\frac{1}{\sqrt{p^2+q^2}}$
Let the vertices of the right triangle be $(0,0),(5,0),(0,2\sqrt{3})$ and let $(a,0),(0,b)$ be two of the vertices of the equilateral triangle. Then, the third vertex of the equilateral triangle is  $(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2})$. 
This point must lie on the hypotenuse $\frac{x}{5}+\frac{y}{2\sqrt{3}}=1$i.e., a,b must satisfy $\frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}}=1$.
Which can be simplified to, $\frac{7}{20}a+\frac{11\sqrt{3}}{60}b=1$ .
By the lemma, the minimal value of  $\sqrt{a^2+b^2}$ is $\frac{1}{\sqrt{{\left(\frac{7}{20}\right)}^2+{\left(\frac{11\sqrt{3}}{60}\right)}^2}}=\frac{10\sqrt{3}}{\sqrt{67}}$,
So the minimum area of equilateral triangle is  $\frac{\sqrt{3}}{4}.{\left(\frac{10\sqrt{3}}{\sqrt{67}}\right)}^2=\frac{\sqrt{3}}{4}.\frac{300}{67}=\frac{75\sqrt{3}}{67},$
And hence the answer is 75+67=142