171. A jeweler's diamond cracked during polishing, as a result, its value decreased by $32 \%$. What part of the diamond broke off if the value of the diamond is proportional to the square of its mass?
Solution. Let a diamond of mass $p$ during polishing split into two parts: of mass $p_{1}$ and mass $p_{2}$. Since the cost of the diamond is proportional to the square of its mass, we have the system of equations:

$$
\left\{\begin{array}{l}
p_{1}+p_{2}=p \\
k p_{1}^{2}+k p_{2}^{2}=k \cdot 0.68 \cdot p^{2}
\end{array}\right.
$$

where $k$ is the proportionality coefficient. After reducing the second equation by $k$, we get:

$$
\left\{\begin{array}{l}
p_{1}+p_{2}=p \\
p_{1}^{2}+p_{2}^{2}=\frac{17}{25} p^{2}
\end{array}\right.
$$

To solve this system, square the first equation:

96

$$
\left\{\begin{array}{l}
p_{1}^{2}+p_{2}^{2}+2 p_{1} p_{2}=p^{2} \\
p_{1}^{2}+p_{2}^{2}=\frac{17}{25} p^{2}
\end{array}\right.
$$

Subtracting the second equation from the first, we get:

$$
\left\{\begin{array}{l}
p_{1}+p_{2}=p \\
p_{1} \cdot p_{2}=\frac{4}{25} p^{2}
\end{array}\right.
$$

from which $p_{1}=\frac{4}{5} p, p_{2}=\frac{1}{5} p$. Therefore, $\frac{1}{5}$ of the diamond was chipped off.

Answer: $\frac{1}{5}$ part.