B4. Let $k$ be a circle with radius $R$. On the diameter $AB$ are points $C$ and $D$ such that $|AC|=|CD|=|DB|$. Over $AC$ and $AD$ are semicircles drawn on one side of the segment $AB$, and over $DB$ and $CB$ on the other side (see figure). Express the area of the shaded figure in terms of $R$.

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-09.jpg?height=377&width=463&top_left_y=1382&top_left_x=1456)

## Solutions to Problems and Scoring

A contestant who arrives at the solution by any correct method (even if the scoring does not provide for it) receives all possible points.

For a correct method, each procedure that

- sensibly takes into account the wording of the problem,
- leads to the solution of the problem,
- is mathematically correct and complete.

A contestant who has only partially solved the problem, from otherwise correct solving procedures but does not show the way to the final solution of the problem, cannot receive more than half of the possible points.

## First Year

I. PART

| Problem | A1 | A2 | A3 | A4 | A5 | A6 |
| :--- | ---: | ---: | ---: | ---: | ---: | :---: |
| Answer | D | E | B | C | B | B |
B4. The area of the shaded figure from the problem is equal to the area of the shaded figure on the right, which is the difference between the area of a circle with radius $\frac{2}{3} R$ and the area of a circle with radius $\frac{1}{3} R$: $S=\pi\left(\frac{4}{9} R^{2}-\frac{1}{9} R^{2}\right)=\frac{\pi}{3} R^{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-16.jpg?height=69&width=1137&top_left_y=1082&top_left_x=274)
The relationship $r_{2}=\frac{2}{3} R \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots . \ldots$ point Knowledge of the formula for the area of a circle ......................................................... Conclusion $S=\pi r_{2}^{2}-\pi r_{1}^{2}=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-16.jpg?height=391&width=463&top_left_y=884&top_left_x=1439)
$=\pi\left(\frac{4}{9} R^{2}-\frac{1}{9} R^{2}\right)=\frac{\pi}{3} R^{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point