B4. A square with a side length of $\sqrt{2} \mathrm{~cm}$ is circumscribed by a circle. Over each side of the square, a semicircle is drawn that lies outside the square. What is the total area of the figures bounded by the drawn semicircles and the circumscribed circle?

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-05.jpg?height=422&width=423&top_left_y=874&top_left_x=1482)

## PROBLEMS FOR THE THIRD YEAR

Before you are two sets of problems. Problems 1 to 6 of the first set are solved by selecting the correct answer from the five proposed answers on this sheet and writing it in the table under the corresponding number. Only one answer is correct. A correct answer will be awarded two points, while a wrong answer will result in the deduction of one point.

Problems 1 to 4 of the second set are solved on the provided paper. The solution to each of these problems will be evaluated on a scale of 0 to 6 points. Do not sign the sheets where you will solve the problems; write only your code. Write your solution clearly and legibly with a pen.

The time allowed for solving is 90 minutes.

THE NATIONAL COMPETITION COMMITTEE WISHES YOU GREAT SUCCESS.

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-06.jpg?height=166&width=485&top_left_y=1002&top_left_x=800)

## PART I
B4. The radius of the circle circumscribed around the square is equal to half the length of the diagonal: $R=\frac{d}{2}=\frac{1}{2} \sqrt{2} \cdot \sqrt{2}=1$. The radius of the semicircle above the side of the square is $r=\frac{a}{2}=\frac{\sqrt{2}}{2}$. The sum of the areas of the figures bounded by the drawn semicircles and the circumscribed circle is obtained by subtracting the area of the square circumscribed circle from the sum of the areas of the square and the semicircles above its sides. Thus, we have $S=a^{2}+4 \cdot \frac{1}{2} \cdot \pi r^{2}-\pi R^{2}=\sqrt{2}^{2}+2 \cdot \pi \cdot \frac{1}{2}-\pi=2 \mathrm{~cm}^{2}$.

The expression for the radius of the circle circumscribed around the square $R=\frac{d}{2}=\frac{1}{2} \sqrt{2} \cdot \sqrt{2}=1 \mathrm{~cm} \ldots \ldots \ldots \ldots \ldots 1$ point The area of this circle: $S_{1}=\pi \mathrm{~cm}^{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

The expression for the radius of the semicircle above the side of the square: $r=\frac{a}{2}=\frac{\sqrt{2}}{2} \mathrm{~cm} \ldots \ldots \ldots \ldots \ldots \ldots .1$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-13.jpg?height=57&width=1639&top_left_y=822&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-13.jpg?height=60&width=1642&top_left_y=869&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-13.jpg?height=60&width=1641&top_left_y=918&top_left_x=276)

## Third Year

I. PART

| Task | A1 | A2 | A3 | A4 | A5 | A6 |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: |
| Answer | C | B | B | A | D | C |