12.140. In what ratio does the point $O$, from which all three sides of the isosceles triangle $ABC$ are seen at the same angle $(\angle AOB = \angle BOC = \angle COA)$, divide the height of the isosceles triangle $ABC$ if the angle at the base of the triangle is $\alpha\left(\alpha>\frac{\pi}{6}\right)$?
## Solution.

Let $B D$ be the height, bisector, and median of $\triangle A B C$ (Fig.12.13).

$\angle A O B=\angle B O C=\angle A O C=\frac{2 \pi}{3}$.

$\angle A O K=\frac{1}{2} \angle A O C=\frac{\pi}{3}$.

Let $O K=1$, then from $\triangle A D O\left(\angle A D O=\frac{\pi}{2}\right)$ :

$$
A D=O D \operatorname{tg} \angle A O D=\operatorname{tg} \frac{\pi}{3}=\sqrt{3}
$$

From $\triangle A D B\left(\angle A D B=\frac{\pi}{2}\right): B D=A D \operatorname{tg} \angle B A D=\sqrt{3} \operatorname{tg} \alpha$.

$\frac{B O}{O D}=\frac{B D-O D}{O D}=\frac{B D}{O D}-1=\sqrt{3} \operatorname{tg} \alpha-1=\sqrt{3}\left(\operatorname{tg} \alpha-\frac{1}{\sqrt{3}}\right)=\sqrt{3}\left(\operatorname{tg} \alpha-\operatorname{tg} \frac{\pi}{6}\right)=$ $=\frac{\sqrt{3} \sin \left(\alpha-\frac{\pi}{6}\right)}{\cos \alpha \cos \frac{\pi}{6}}=\frac{2 \sin \left(\alpha-\frac{\pi}{6}\right)}{\cos \alpha}$.

Answer: $\frac{2 \sin \left(\alpha-\frac{\pi}{6}\right)}{\cos \alpha}$, measured from the vertex.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0901.jpg?height=533&width=537&top_left_y=79&top_left_x=125)

Fig. 12.13

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0901.jpg?height=523&width=492&top_left_y=90&top_left_x=735)

Fig. 12.14