In the interior of triangle $A B C$, a point $P$ is arbitrarily chosen, and its orthogonal projections onto the sides $B C, C A, A B$ are $A^{\prime}, B^{\prime}, C^{\prime}$, respectively. For which position of $P$ will the following expression have the smallest value:

$$
\frac{B C}{P A^{\prime}}+\frac{C A}{P B^{\prime}}+\frac{A B}{P C^{\prime}}
$$
Let $P A^{\prime}=a_{1}, P B^{\prime}=b_{1}, P C^{\prime}=c_{1}$ and $\frac{a}{a_{1}}+\frac{b}{b_{1}}+\frac{c}{c_{1}}=S$. The area of the triangle: $T=\frac{1}{2}\left(a a_{1}+b b_{1}+c c_{1}\right)$.
![](https://cdn.mathpix.com/cropped/2024_05_02_1e73fb21b2a6c698e96dg-1.jpg?height=938&width=718&top_left_y=240&top_left_x=690)

Let's calculate the value of the product $2 T S$:

$$
\left(\frac{a}{a_{1}}+\frac{b}{b_{1}}+\frac{c}{c_{1}}\right) \cdot\left(a a_{1}+b b_{1}+c c_{1}\right)=a^{2}+b^{2}+c^{2}+a b\left(\frac{a_{1}}{b_{1}}+\frac{b_{1}}{a_{1}}\right)+a c\left(\frac{a_{1}}{c_{1}}+\frac{c_{1}}{a_{1}}\right)+b c\left(\frac{b_{1}}{c_{1}}+\frac{c_{1}}{b_{1}}\right)
$$

Apply the inequality $\frac{x}{y}+\frac{y}{x} \geqq 2$ for $(x, y>0)$, equality holds only when $x=y$:

$$
2 T S \geqq a^{2}+b^{2}+c^{2}+2(a b+a c+b c)=(a+b+c)^{2}=K^{2}
$$

The right-hand side is constant: the square of the perimeter. Thus $S \geqq \frac{K^{2}}{2 T}$, and equality holds if and only if $a_{1}=b_{1}=c_{1}$. Therefore, the value of the expression $S$ is the smallest when $P$ is equidistant from all three sides - that is, when $P$ coincides with the center of the inscribed circle.