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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 $a+b$ ?

enter image description here


All that I was able to do is: enter image description here

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 : enter image description here

$$\begin{align} \text{Area } \Delta PQR &=\dfrac{3x\ RN}{2}+\dfrac{3x\ PN}{2}+\dfrac{x\ PL}{2}+\dfrac{x\ QL}{2}+ {x\ QM}+{x\ MR} \\ \\ &=\dfrac{3x\ (PN+RN)}{2}+\dfrac{x\ (PL+QL)}{2}+{x\ (QM+MR)}\\ \\ &=\dfrac{3x\ (PR)}{2}+\dfrac{x\ (PQ)}{2}+{x\ (QR)}\\ \\ &=kx(\dfrac{3}{2}+\dfrac{1}{2}+{1})=3kx\\ \end{align}$$ 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 ?

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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.

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  • $\begingroup$ I did not get the same equations as you ($2^{nd},\ 3^{rd}\ \text{and } 4^{th}$). See my working in my (edited) question. $\endgroup$ – K. Rmth Dec 21 '13 at 20:53
  • $\begingroup$ @Rmth: the answer has been completed. let me know if you have any doubt. $\endgroup$ – Suraj M S Dec 21 '13 at 20:55
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    $\begingroup$ you are right i forgot to take the $\frac{1}{2}$ .i will edit it. $\endgroup$ – Suraj M S Dec 21 '13 at 20:58
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    $\begingroup$ I get $\dfrac{a}{b}=\dfrac{11}{36}$ making $a+b=47$ which is the right answer. I hope that's what your answer will turn out to be once you've edited. ;) $\endgroup$ – K. Rmth Dec 21 '13 at 21:14
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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 $\ell$, $m$, $n$.

enter image description here

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.

enter image description here

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

$$\begin{align} |LONP| &= |\triangle PQ^\prime R^\prime| - \frac{1}{2}|\triangle L| - \frac{1}{2}|\triangle N| \\[4pt] &= \frac{4}{9}|\triangle PQR| - \frac{5}{36}|\triangle PQR| \\[4pt] &= \frac{11}{36}|\triangle PQR| \end{align}$$

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    $\begingroup$ Very nice answer, and doubly so since no complicated calculations are involved (this question was from a contest, where calculators are not allowed). :) $\endgroup$ – K. Rmth Dec 23 '13 at 14:00
  • $\begingroup$ Is there any name to the "well-known fact" ?? $\endgroup$ – K. Rmth Dec 23 '13 at 14:01
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    $\begingroup$ @K.Rmth: I don't think the Well-Known Fact has a name, although it would be convenient. (Writing it out every time you want to use it is a pain.) It is well-known, though, so invoking it in a contest setting is reasonable. (The Fact even forms the basis of one of my iPhone games, although that particular game has been put on hiatus for renovations.) Of course, the one-line proof is easy enough to provide if there's some doubt about whether your audience is familiar with the result. $\endgroup$ – Blue Dec 23 '13 at 14:09
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    $\begingroup$ In a deleted answer, @AndrewNguyen noted that the Well-Known Fact does in fact have a name: Viviani's Theorem. $\endgroup$ – Blue Mar 5 '14 at 22:29

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