Problem 1201

14.12. We will prove that if segments $B_{1} C_{1}$ and $B_{2} C_{2}$ are drawn through point $O$, with endpoints $B_{1}$ and $B_{2}$ lying on side $A B$ of the given angle

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Fig. 101

$B A C$, and endpoints $C_{1}$ and $C_{2}$ lying on side $A C$, and if

$$\angle A O B_{2}>\angle A O \dot{B}_{1} \geqslant 90^{\circ}$$

(Fig. 101), then

$$\frac{1}{B_{1} O}+\frac{1}{C_{1} O}>\frac{1}{B_{2} O}+\frac{1}{C_{2} O}$$

Let $B_{3}$ and $C_{3}$ be the points of intersection of line $B_{1} C_{1}$ with the lines passing through points $B_{2}$ and $C_{2}$, respectively, parallel to line $A O$. Since

$$\angle B_{2} B_{3} O=\angle A O B_{3} \geqslant 90^{\circ}$$

then $O B_{2}>O B_{3}$ and

$$\frac{1}{B_{1} O}-\frac{1}{B_{2} O}=\frac{B_{2} O-B_{1} O}{B_{1} O \cdot B_{2} O}>\frac{B_{3} O-B_{1} O}{B_{1} O \cdot B_{2} O}=\frac{B_{1} B_{3}}{B_{1} O \cdot B_{2} O}=\frac{B_{2} B_{3}}{A O \cdot B_{2} O}$$

(because $\triangle B_{1} B_{3} B_{2} \sim \triangle B_{1} O A$). Similarly, we obtain

$$\frac{1}{C_{1} O}-\frac{1}{C_{2} O}>-\frac{C_{2} C_{3}}{A O \cdot C_{2} O}$$

Adding the resulting inequalities and considering the similarity of triangles $B_{2} B_{3} O$ and $C_{2} C_{3} O$, we have

$$\begin{aligned} \frac{1}{B_{1} O}+\frac{1}{C_{1} O}-\left(\frac{1}{B_{2} O}+\frac{1}{C_{2} O}\right)>\frac{B_{2} B_{3}}{A O \cdot B_{2} O}-\frac{C_{2} C_{3}}{A O \cdot C_{2} O} & = \\ & =\frac{1}{A O}\left(\frac{B_{2} B_{3}}{B_{2} O}-\frac{C_{2} C_{3}}{C_{2} O}\right)=0 . \end{aligned}$$

If the line passing through point $O$ is perpendicular to line $A O$ and intersects rays $A B$ and $A C$ at points $B_{1}$ and $C_{1}$, respectively, then segment $B_{1} C_{1}$ is the desired one. Indeed, for any other segment $B_{2} C_{2}$ passing through point $O$, one of the inequalities holds

$$\angle A O B_{2}>90^{\circ} \text { or } \angle A O C_{2}>90^{\circ}$$

(for definiteness, we can assume that points $B_{2}$ and $C_{2}$ lie on rays $A B$ and $A C$, respectively, and the first inequality is satisfied). Then the above inequality holds. If the line passing through point $O$ perpendicular to line $A$ does not intersect one of the sides of angle $B A C$, say side $A C$, then the desired segment cannot be constructed. Suppose that the desired segment is some segment $B_{2} C_{2}$, with endpoints $B_{2}$ and $C_{2}$ lying on rays $A B$ and $A C$, respectively. Then

$$\angle A O B_{2}>90^{\circ},$$

and if we choose point $C_{1}$ on the extension of segment $A C_{2}$ beyond point $C_{2}$ and draw another segment $C_{1} B_{1}$ through point $O$, then again

$$90^{\circ}<\angle A O B_{1}<\angle A O B_{2}$$

and by the above inequality, segment $B_{2} C_{2}$ does not satisfy the required condition.