13.379 The price of a diamond is proportional to the square of its mass. A diamond with a mass of $p$ carats was broken into two parts, after which its value decreased by a factor of $k$. Find the mass of the parts into which the diamond was broken. Prove that the greatest loss in the value of the diamond occurs when both parts are equal in mass.
Solution. Let $x$ and $p-x$ be the masses of the parts into which the diamond was divided. Then, according to the problem, we have:

$$
\frac{p^{2}}{x^{2}+(p-x)^{2}}=k \Rightarrow 2 k x^{2}-2 p k x+p^{2} k-p^{2} \Rightarrow
$$

$\Rightarrow x=\frac{p k \pm \sqrt{2 k p^{2}-p^{2} k^{2}}}{2 k}$, and $\left\{\begin{array}{l}2 k-k^{2} \geq 0, \\ p k-\sqrt{2 k p^{2}-p^{2} k^{2}} \geq 0\end{array} \Leftrightarrow 1 \leq k \leq 2\right.$.

Obviously, the greatest loss in value will occur at the smallest value of $x^{2}+(p-x)^{2}$. The minimum of this function is achieved at $x=\frac{p}{2}$.

Answer: the masses of the parts are $\frac{p k \pm \sqrt{2 k p^{2}-p^{2} k^{2}}}{2 k}$, where $1 \leq k \leq 2$.