Balti Tee 1991 · Ülesanne 20

Geomeetria

Consider two points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the graph of the function $y=\frac{1}{x}$ such that $0<x_{1}<x_{2}$ and $|A B|=2 \cdot|O A|(O$ is the reference point, i.e., $O(0,0))$ . Let $C$ be the midpoint of the segment $A B$ . Prove that the angle between the $x$ -axis and the ray $O A$ is equal to three times the angle between $x$ -axis and the ray $O C$ .

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Lahendused

Lahendus

Solution:

We have $A\left(x_{1}, \frac{1}{x_{1}}\right)$ , $B\left(x_{2}, \frac{1}{x_{2}}\right)$ and $C\left(\frac{x_{1}+x_{2}}{2}, \frac{1}{2 x_{1}}+\frac{1}{2 x_{2}}\right)$ . Computing the coordinates of $\bar{v}=|O C| \cdot \overline{A C}+|A C| \cdot \overline{O C}$ we find that the vector $\bar{v}$ —and hence also the bisector of the angle $\angle O C A$ —is parallel to the $x$ -axis. Since $|O A|=|A C|$ this yields $\angle A O C=\angle A C O=2 \cdot \angle C O x$ (see Figure 3) and $\angle A O x=\angle A O C+\angle C O x=3 \cdot \angle C O x$ .

Figure 3

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AoPS · Baltic Way 1991