Variant 134. a) $36+8 \pi$ cm or $23+3 \sqrt{41}+8 \pi$ cm; b) not always.

Evaluation criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 5}$ points - mostly correct solution and correct answer, but there are defects (for example: comparison of numbers in the last part of the solution is not performed or is performed incorrectly; it is not proven that the sum of the lengths of the arcs is equal to the length of the circumference); 10 points - the solution is mostly correct, but contains computational errors; 5 points - only one of the two cases is considered, and the solution and answer (for this case) are completely correct.

## Problem 4 (Variant 131).

When studying the operation of a new type of heat engine working cyclically, it was found that for part of the period it receives heat, with the absolute value of the heat supply power expressed by the law: $P_{1}(t)=P_{0} \frac{\sin \omega t}{100+\sin t^{2}}, 0<t<\frac{\pi}{\omega}$. The gas performs work, developing mechanical power $P_{2}(t)=3 P_{0} \frac{\sin (2 \omega t)}{100+\sin (2 t)^{2}}, 0<t<\frac{\pi}{2 \omega}$. The work done on the gas by external bodies is $2 / 3$ of the magnitude of the work done by the gas. Determine the efficiency of the engine.
Solution. The efficiency of a heat engine is the ratio of useful work to the amount of heat received by the gas per cycle. To find the useful work, it is necessary to find the difference between the work done by the gas ($A_{+}$) and the work done on it ($A_{-}$). Each of these quantities can be found by integrating the power over time or by finding the area of the figure enclosed between the graph of the power dependence on time and the time axis. Similarly, the amount of heat received by the gas ($Q_{+}$) is found.

Note that the function $P_{2}(t)=a P_{1}(b t)$, therefore, since the figure formed by the graph of $P_{2}$ is obtained from the figure corresponding to the function $P_{1}$ by stretching along the vertical axis by a factor of $a$ and compressing along the horizontal axis by a factor of $b$, then $A_{+}=Q_{+} \cdot \frac{a}{b}$. According to the problem, $A_{-}=\beta A_{+}$. Finally, $\eta=\frac{a}{b}(1-\beta)$.

Given in the problem $a=3, b=2, \beta=\frac{2}{3}: \quad \eta=\frac{1}{2}$.

Note that the integral of the function $P_{1}(t)$ cannot be expressed in terms of elementary functions. On the other hand, the integrand is significantly simplified if the second term in the denominator is neglected. This does not result in a significant loss of accuracy, as the second term is limited and two orders of magnitude smaller than the first. After this, the integrals are easily calculated.

Answer: $\frac{1}{2}$.

## Answers of other variants: