4. Among all points $P$ lying on the sides $A B, B C$ or $C A$ of the right triangle $A B C\left(\measuredangle C=90^{\circ}\right)$, find the one for which the sum $\overline{A P}+\overline{B P}+\overline{C P}$ is the smallest.
Solution. If point $P$ is on the leg $BC$, then $\overline{BP} + \overline{CP} = a$. Then $\overline{AP} + a$ is the smallest when $\overline{AP} = b$. Similarly, if $P$ is on the leg $AC$, the sum has the smallest value when it is equal to the sum of the legs $a + b$. Let $P$ be on the hypotenuse $AC$. Then $\overline{BP} + \overline{AP} = c$, and $\overline{CP} + c$ has the smallest value if $\overline{CP} = h_c$. It remains to compare the sums $a + b$ and $c + h_c$. From $c^2 = a^2 + b^2$ it follows that $c^2 + h_c^2 > a^2 + b^2$. Since $c h_C = a b$, we get

$$
\begin{aligned}
c^2 + 2 c h_c + h_c^2 & > a^2 + 2 a b + b^2 \\
(c + h_c)^2 & > (a + b)^2
\end{aligned}
$$

and therefore $a + b < c + h_c$. Thus, the smallest value of the sum $\overline{AP} + \overline{BP} + \overline{CP}$ is $a + b$, and then $P \equiv C$.

## III year