Problem 1. Three sisters divided a certain amount of money, with the first sister taking $\frac{1}{5}$ of the money, the second sister taking $\frac{5}{8}$ of the money, and the remainder going to the third sister. However, the third sister then gave $\frac{3}{4}$ of her share to the first sister. What fraction of the total amount of money did the first sister receive?
Solution. According to the condition of the problem, the first sister took $\frac{1}{5} x$, the second $\frac{5}{8} x$, and the third sister took $x-\frac{1}{5} x-\frac{5}{8} x=\frac{7}{40} x$.

The third sister gave the first sister $\frac{3}{4} \cdot \frac{7}{40} x=\frac{21}{160} x$.

The first sister received in total $\frac{1}{5} x+\frac{21}{160} x=\frac{53}{160} x$, which is $\frac{53}{160}$ of the money.

In a right-angled triangle $ABC$ with the right angle at vertex $C$, point $M$ is given on side $AB$, and point $N$ is given on side $AC$, such that $\bar{i}: \bar{C}=\bar{\imath}: \ddot{i} \bar{\Lambda}=\overline{M N}=\overline{A N}$. Calculate the angles of triangle $ABC$.

Solution. Triangles $ANM$, $NMC$, and $BCM$ are isosceles, so

$$
\alpha=\angle CAM=\angle NMA, \quad \beta=\angle CBM=\angle CMB, \quad \angle MCN=\angle CNM
$$

Angle $\angle MNC$ is an exterior angle for triangle $AMN$, so $\angle MNC=2 \alpha$. Then

$$
\angle MCN=2 \alpha \text{ and } \angle CMN=180^{\circ}-4 \alpha .
$$

For angle $\angle BMC$ we get

![](https://cdn.mathpix.com/cropped/2024_06_05_4fa80d9916883368a7a6g-5.jpg?height=246&width=446&top_left_y=533&top_left_x=1015)

$$
\angle BMC=180^{\circ}-\left(180^{\circ}-4 \alpha\right)-\alpha=3 \alpha
$$

i.e., $\angle CBM=3 \alpha$. But $\angle CBM=\beta$, so we get $\alpha+3 \alpha=90^{\circ}, \alpha=22.5^{\circ}$.

For $\beta$ we get $\beta=3 \alpha=67.5^{\circ}$.