Let $P$ be an arbitrary interior point of the equilateral triangle $ABC$. Let the feet of the perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$ be $D$, $E$, and $F$ respectively. Determine the value of the ratio

$$
\frac{P D+P E+P F}{B D+C E+A F}
$$
Let's denote the length of the side of the triangle by $a$. We will show that the numerator and the denominator of the fraction in question depend only on $a$.

The area of triangle $ABC$ can be written in two ways (Fig. 1):

$$
t_{A B C}=t_{A P B}+t_{B P C}+t_{C P A}
$$

which is

$$
\frac{a^{2} \sqrt{3}}{4}=\frac{a \cdot P F}{2}+\frac{a \cdot P D}{2}+\frac{a \cdot P E}{2}
$$

From this, the numerator of the fraction is $P D+P E+P F=\frac{a \sqrt{3}}{2}$.

Draw lines through $P$ parallel to the sides of triangle $ABC$, and denote the intersection points of these lines with the sides of the triangle as shown in Fig. 2. The quadrilaterals $A B K L, B C G H$, and $C A J I$ are symmetric trapezoids, so $A L=B K, B H=C G$, and $A J=C I$. The triangles $P J H, P K I$, and $P G L$ have all angles of $60^{\circ}$, so they are equilateral triangles, in which $P F, P D$, and $P E$ are perpendicular bisectors. Based on this, the perimeter of triangle $ABC$ can be written as:

$$
\begin{gathered}
3 a=(A J+J F+F H+H B)+(B K+K D+D I+I C)+(C G+G E+E L+L A)= \\
=[(A J+J F)+(I C+F H)]+[(B K+K D)+(A L+D I)]+[(C G+G E)+(H B+E L)]= \\
=2(A F+B D+C E)
\end{gathered}
$$

Thus, $A F+B D+C E=\frac{3 a}{2}$.

Therefore, the value of the desired fraction is:

$$
\frac{P D+P E+P F}{B D+C E+A F}=\frac{\frac{a \sqrt{3}}{2}}{\frac{3 a}{2}}=\frac{\sqrt{3}}{3}
$$

Péter Csordás (Kecskemét, Zrínyi I. Secondary School, 8th grade)
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