25. Solve the inequality

$$
\frac{4^{-|x-6|}}{\sqrt{x^{2}-6 x}+3} \leqslant \frac{2^{1-|x-3|}}{\sqrt{x^{2}+6 x-27}+6} .
$$

In the answer, write the sum of the lengths of the intervals of the solution that belong to the set $[-2020,2018]$. If the inequality has no solutions, write -1.
Answer: 4020.

IV. In triangle $ABC$, the height $BH$ is drawn, point $O$ is the center of the circumscribed circle around it, and the length of its radius is $R$. Find the largest of the angles $\angle BAC, \angle ACB$, expressed in radians, if it is known that $R = (6/5) \cdot BH = 4 \cdot OH$. Round the found value to two decimal places if necessary.

Solution. Let's assume $R = 12$, then $BH = 10, OH = 3$. Thus, $\triangle OBH$ is fully defined, and its area by Heron's formula is

$$
S = \sqrt{\frac{25}{2} \cdot \frac{1}{2} \cdot \frac{5}{2} \cdot \frac{19}{2}} = \frac{5 \sqrt{95}}{4}
$$

Then the height $OH_1 = \frac{2S}{BH} = \frac{2 \cdot 5 \sqrt{95}}{4 \cdot 10} = \frac{\sqrt{95}}{4}$. Therefore, $MH = OH_1 = \frac{\sqrt{95}}{4}$. Further, by the Pythagorean theorem:

$$
\begin{aligned}
& OM^2 = OH^2 - MH^2 = 3^2 - \left(\frac{\sqrt{95}}{4}\right)^2 = \left(\frac{7}{4}\right)^2, \quad OM = \frac{7}{4} \\
& AM^2 = AO^2 - OM^2 = 12^2 - \left(\frac{7}{4}\right)^2 = \frac{41 \cdot 55}{16}, \quad AM = MC = \frac{\sqrt{41 \cdot 55}}{4}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cd42bf56ca8604f2c35fg-10.jpg?height=414&width=628&top_left_y=218&top_left_x=771)

Fig. 1:

Since $AM = MC > MH$, then $\angle ACB > \angle BAC$.

$$
\angle ACB = \operatorname{arctg}\left(\frac{BH}{HC}\right) = \operatorname{arctg}\left(\frac{10 \cdot 4}{\sqrt{41 \cdot 55} - \sqrt{95}}\right) \approx 0.81446 \ldots
$$

Note that in this case, point $O$ lies outside $\triangle ABC$, but all calculations remain valid.

Answer: 0.81.