In triangle $ABC$, from certain points on the base $AB$, we draw perpendiculars to the sides $AC$ and $BC$. Determine the point $P$ on $AB$ such that the sum of the squares of the perpendiculars drawn from it is minimized.
Let

$$
A P=x, B P=c-x
$$

be the perpendiculars

$$
x \sin \alpha \text { and }(c-x) \sin \beta=\frac{(c-x) b}{a} \sin \alpha
$$

The sum of the squares of the perpendiculars:

$$
\begin{gathered}
y=x^{2} \sin ^{2} \alpha+(c-x)^{2} \frac{b^{2}}{a^{2}} \sin ^{2} \alpha \\
y=\left(1+\frac{b^{2}}{a^{2}}\right) \sin ^{2} \alpha \cdot x^{2}-\frac{2 b^{2} c}{a^{2}} \sin ^{2} \alpha \cdot x+\frac{b^{2} c^{2}}{a^{2}} \sin ^{2} \alpha
\end{gathered}
$$

$y$ is minimized when

$$
x=\frac{b^{2} c}{a^{2}+b^{2}}
$$

and in this case

$$
c-x=\frac{a^{2} c}{a^{2}+b^{2}}
$$

(Pál Sárközy, Pannonhalma.)