22. A pharmacist needs to prepare a mixture from three components $A$, $B$, and $C$ with the following composition: $a \% \quad A, b \% B, c \% C$, but he does not have pure substances $A, B, C$. At his disposal are three different mixtures of these substances with the following compositions:
1) $a_{1} \% A, b_{1} \% \quad B, c_{1} \% C$;
2) $a_{2} \% A, b_{2} \% \quad B, c_{2} \% \quad C$;
3) $a_{3} \% A, b_{3} \% B, c_{3} \% \quad C$.

How can he prepare the required mixture? Solve this problem using the Gibbs triangle. You can illustrate the solution with a specific case, when

$$
\begin{array}{rlr}
a_{1}=10, & b_{1}=30, & c_{1}=60 \\
a_{2}=20, & b_{2}=60, & c_{2}=20 \\
a_{3}=80, & b_{3}=10, & c_{3}=10 \\
a=50, & b=30, & c=20
\end{array}
$$

(P. 221.)
22. Let the four considered mixtures be represented by points $K_{1}, K_{2}, K_{3}, K$ (Fig. 180). The problem reduces to representing the material point $(K, 1)$ as a combination of three material points with masses $K_{1}, K_{2}, K_{3}$. Let the masses (to be found) of these three material points be $m_{1}, m_{2}, m_{3}$, respectively. Draw a line through $K$ and each vertex of the triangle $K_{1} K_{2} K_{3}$ and mark the point where it intersects the opposite side; we obtain three points $L_{1}, L_{2}, L_{3}$. Since

![](https://cdn.mathpix.com/cropped/2024_05_21_27f86caa754616304e40g-223.jpg?height=881&width=983&top_left_y=466&top_left_x=137)

Fig. 180.

$K$ is the center of gravity of the three material points $\left(K_{1}, m_{1}\right), \left(K_{2}, m_{2}\right)$, and $\left(K_{3}, m_{3}\right)$, it follows that

$$
\frac{m_{1}}{m_{2}+m_{3}}=\frac{L_{1} K}{K K_{1}}
$$

Forming the corresponding proportion, we get

$$
\frac{m_{1}}{m_{1}+m_{2}+m_{3}}=\frac{L_{1} K}{L_{1} K_{1}}
$$

Since $m_{1}+m_{2}+m_{3}=1$, we have

$$
m_{1}=\frac{L_{1} K}{L_{1} K_{1}}
$$

Similarly,

$$
m_{2}=\frac{L_{2} K}{L_{2} K_{2}} \text { and } m_{3}=\frac{L_{3} K}{L_{3} K_{3}}.
$$

The lengths of the six segments involved in the last three fractions can be found from the drawing. The numbers $m_{1}, m_{2}$, and $m_{3}$ show the concentrations with which the three given mixtures should be combined to form the required mixture. In our specific example, using Fig. 180, we find that $m_{1}=\frac{3}{23} \approx 0.13, m_{2}=\frac{8}{23} \approx 0.35$, and $m_{3}=\frac{12}{23} \approx 0.52$. In other words, the pharmacist needs to take 13 g of the first mixture, 35 g of the second mixture, and 52 g of the third mixture to make 100 g of the required mixture.

Problem 22 has no solution if point $K$ is outside the triangle $K_{1} K_{2} K_{3}$; the required mixture cannot be formed in this case.

The case where $\triangle K_{1} K_{2} K_{3}$ degenerates into a segment is no exception. However, for the problem to be solvable, point $K$ must lie on this segment. In the last case, the problem has infinitely many solutions.