Finding the ratio of areas produced by perpendiculars from the $3$ sides of an equilateral triangle. 
A point O is inside an equilateral triangle $PQR$ and the perpendiculars $OL,OM,\text{and } ON$ are drawn to the sides $PQ,QR,\text{and } RP$ respectively. The ratios of lengths of the perpendiculars $OL:OM:ON \text{ is } 1:2:3$.
If  $\ \dfrac{\text{area of }LONP}{\text{area of }\Delta PQR}=\dfrac{a}{b}, \quad$ where $a$ 
  and $b$ are integers with no common factors,
what is the value of $a+b$ ?



All that I was able to do is:

Area $LONP=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP||ON|=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP|\ 3|OL|=\frac{1}{2} |OL|\ \left[\ |PL|+3|NP|\ \right]$
Area $PQR=\frac{1}{2} |PR||PQ|\sin 60^o=\frac{\sqrt 3}{4} |PR||PQ|=\frac{\sqrt 3}{4} |NP+RN||PL+LQ|=\frac{\sqrt 3}{4} \left[\ |NP|+|PL|+|RN|+|LQ| \ \right]$
Area $\Delta LON=\frac{1}{2} |ON||OL|\sin 120=\frac{\sqrt 3}{4} |OL|\ 3|OL|=\frac{3\sqrt 3}{4} |OL|^2$

$\mathbf{EDIT : }$Following Suraj M.S 's answer :

$$\begin{align}
\text{Area } \Delta PQR &=\dfrac{3x\ RN}{2}+\dfrac{3x\ PN}{2}+\dfrac{x\ PL}{2}+\dfrac{x\ QL}{2}+ {x\ QM}+{x\ MR} \\
\\
&=\dfrac{3x\ (PN+RN)}{2}+\dfrac{x\ (PL+QL)}{2}+{x\ (QM+MR)}\\
\\
&=\dfrac{3x\ (PR)}{2}+\dfrac{x\ (PQ)}{2}+{x\ (QR)}\\
\\
&=kx(\dfrac{3}{2}+\dfrac{1}{2}+{1})=3kx\\
\end{align}$$
Area $\Delta PQR=\dfrac{1}{2} k^2 \sin 60^o=\dfrac{k^2 \sqrt 3}{4} \implies x=\dfrac{k}{4\sqrt 3}$

$\mathbf{Question: }$How do I now find the area of $LONP$ in terms of $x'$s and/or $k'$s only ?

 A: Here's a different approach.


Well-known Fact. The sum of the distances from an interior point to the sides of an equilateral triangle is equal to the height of that triangle.

Proof (in case you've never seen it). Using our current triangle, writing $s$ for its side-length, and $h$ for its height:
$$\frac{1}{2} s h = |\triangle PQR| = |\triangle OPQ| + |\triangle OQR| + \triangle ORP| = \frac{1}{2} s(|\overline{OL}|+|\overline{OM}|+|\overline{ON}|)$$ 

From the Fact, and the given proportionality condition, we have
$$|\overline{OL}| : |\overline{OM}| : |\overline{ON}| : h \;=\; 1 : 2 : 3 : (1+2+3) \;=\; 1:2:3:6 \quad (\star)$$
Now, through $O$, draw lines parallel to the sides of the triangle to create three new equilateral triangles of which $\overline{OL}$, $\overline{OM}$, $\overline{ON}$ are altitudes; call these $\triangle L$, $\triangle M$, $\triangle N$, with side-lengths $\ell$, $m$, $n$.

Observe that $(\star)$ implies
$$\begin{align}
\ell : m : n : s \;&=\; 1 : 2 : 3 : 6\\[4pt]
|\triangle L| : |\triangle M| : |\triangle N| : |\triangle PQR| \;&=\; 1 : 4 : 9 : 36
\end{align}$$
The lines also create a parallelogram with diagonal $\overline{OP}$ (and others with diagonals $\overline{OQ}$ and $\overline{OR}$); call it $\square P$.
Then 
$$\begin{align}
|LONP| &= |\square P| + \frac{1}{2}|\triangle L| + \frac{1}{2}|\triangle N| \\[4pt]
&= n |\overline{OL}| + \frac{1}{2} \cdot \frac{1}{36}|\triangle PQR| + \frac{1}{2}\cdot\frac{9}{36}|\triangle PQR| \\[4pt]
&= \frac{3}{6}s \cdot \frac{1}{6} h + \frac{5}{36}|\triangle PQR| \\[4pt]
&= \frac{1}{6} |\triangle PQR| + \frac{5}{36}|\triangle PQR| \\[4pt]
&= \frac{11}{36} |\triangle PQR|
\end{align}$$
So, $\frac{a}{b}=\frac{11}{36}$, whereupon $a+b=47$.

Here's a slicker route to the area ratio. Let $\overline{Q^\prime R^\prime}$ be the added segment through $O$ parallel to $\overline{QR}$, and let $P^\prime$ be the foot of the perpendicular from $P$ to that segment.

Then
$$\begin{align}
|\overline{PP^\prime}| + |\overline{OM}| = h \quad &\implies \quad |\overline{PP^\prime}|:h = (6-2):6 = 2:3 \quad \\[4pt]
&\implies \quad |\triangle PQ^\prime R^\prime| :|\triangle PQR| = 4:9
\end{align}$$
so that
$$\begin{align}
|LONP| &= |\triangle PQ^\prime R^\prime| - \frac{1}{2}|\triangle L| - \frac{1}{2}|\triangle N| \\[4pt]
&= \frac{4}{9}|\triangle PQR| - \frac{5}{36}|\triangle PQR| \\[4pt]
&= \frac{11}{36}|\triangle PQR|
\end{align}$$
A: HINT:
Let $OL, OM, ON$ be $x, 2x, 3x$ respectively . Now you will observe 
$$area(\triangle PQR) = area(\triangle ONR) +area(\triangle ORM)+area(\triangle OMQ)+area(\triangle OQL)+area(\triangle OPL)+area(\triangle OPN)$$
since these are right angled and assuming $k$ be the side of the triangle you will get the resultant area as 
$$area(\triangle PQR)=3kx$$
equating it with the usual area
$$3kx=\frac{\sqrt{3}k^2}{2}$$
we get 
$$x=\frac{\sqrt{3}k}{6}$$
with the help of $x$ you will get $OL, OM, ON$.
Now let $\angle NPO=\theta$ then $\angle OPL=60^\circ-\theta$. with the help of the sides find $\frac{\sin \theta}{\sin (60^\circ-\theta)}$
you get
$$ \frac{\sin\theta}{\sin (60^\circ-\theta)}=\frac{ON}{OL}=3$$
solving for $\theta$
$$\tan \theta=\frac{3\sqrt{3}}{5}$$
with the help of $\tan$ you get $NP=\frac{5}{6}k$
now further you can get $PL$ by interchanging the angles $\theta$ and $60^\circ-\theta$ and once more solving  you get the new $\theta$ as 
$$\tan \theta=\frac{\sqrt{3}}{7}$$
using the same method of finding $NP$ find $PL$.
$$PL=\frac{7}{6}k$$
now we have found all our needed unknowns.
$$a=area(\triangle NOP)+area(\triangle POL)$$
$$=\frac{1}{2}(\frac{\sqrt{3}k}{2}.\frac{5k}{6}+ \frac{\sqrt{3}k}{6}.\frac{7k}{6} )$$
$$a=\frac{11\sqrt{3}k^2}{72}$$
also
$$b=\frac{\sqrt{3}k^2}{2}$$
solving you get $$\frac{a}{b}=\frac{11}{36}$$
$$a+b=\left(\frac{a}{b}+1\right)b$$
which solves to 
$$a+b=\frac{47\sqrt{3}k^2}{72}$$
where $k$ is the side of the equilateral triangle.
