2. From a point inside an equilateral triangle $ABC$, perpendiculars are dropped to the sides $AB$, $BC$, and $CA$. The lengths of the dropped perpendiculars are $m$, $n$, and $p$, and the length of the side of triangle $ABC$ is $a$.

Determine the ratio of the areas of triangle $ABC$ and the triangle whose vertices are the feet of the perpendiculars.
Solution. Let $P$ be an arbitrary point inside the equilateral triangle $ABC$. The points $K, L$, and $M$ are the feet of the perpendiculars dropped from point $P$ to the sides $AB, BC$, and $CA$ respectively. (diagram) Since

$$
\begin{aligned}
\measuredangle P K B & =\measuredangle P L B=\measuredangle P L C=\measuredangle P M C \\
& =\measuredangle P M A=\measuredangle A K P=90^{\circ}
\end{aligned}
$$

the quadrilaterals $A K P M, B L P K$, and $C L P M$ are cyclic. Therefore,

$\measuredangle K P L=\measuredangle L P M=\measuredangle M P K=120^{\circ}$

(because the angles of triangle $a$ are equal to $n$ and the cyclic nature of the above quadrilaterals). From the equalities $b, d$, and $S$ we get

$$
\begin{aligned}
& P_{\triangle L P K}=\frac{1}{2} m p \sin 120^{\circ}=\frac{\sqrt{3}}{4} m p \\
& P_{\triangle L P M}=\frac{1}{2} m n \sin 120^{\circ}=\frac{\sqrt{3}}{4} m n \\
& P_{\triangle M P K}=\frac{1}{2} n p \sin 120^{\circ}=\frac{\sqrt{3}}{4} n p
\end{aligned}
$$

The area of triangle $K L M$ is equal to

$$
P_{\triangle K L M}=P_{\triangle K P L}+P_{\triangle L P M}+P_{\triangle M P K}=\frac{\sqrt{3}}{4}(p m+m n+n p)
$$

Since the area of triangle $A B C$ is $P_{\triangle A B C}=\frac{\sqrt{3}}{4} a^{2}$, we get

$$
\frac{P_{\triangle A B C}}{P_{\triangle K L M}}=\frac{\frac{\sqrt{3}}{4} a^{2}}{\frac{\sqrt{3}}{4}(p m+m n+n p)}=\frac{a^{2}}{p m+m n+n p}
$$