A target board's 18 sectors are delimited by three concentric circles and three lines passing through the center, as shown in the figure. The areas of sectors marked with the same number are equal, the area of the 2 is half that of the 3. How many times larger is the area of the 4 sector compared to the 1 sector?

![](https://cdn.mathpix.com/cropped/2024_05_02_d1189564410d90884283g-1.jpg?height=276&width=287&top_left_y=304&top_left_x=908)
Solution. We know that the area of a sector can be written in the form $\frac{r^{2} \varphi}{2}$, where $r$ is the radius of the circle forming the sector, and $\varphi$ is the central angle of the sector measured in radians.

Let the central angle of the I-type sector marked with 1, 2, 3 be $\alpha$, and the angle of the II-type sector marked with 2, 3, 4 be $\beta$. The radii of the concentric circles are $r_{1}<r_{2}<r_{3}$. Let's write the areas of the sectors marked with the same number from I and II.

For the areas marked with 2:

$$
t_{2}=\frac{r_{1}^{2} \beta}{2}=\frac{r_{2}^{2} \alpha}{2}-\frac{r_{1}^{2} \alpha}{2}=\frac{\alpha\left(r_{2}^{2}-r_{1}^{2}\right)}{2}
$$

Simplifying, we get
a)

$$
r_{1}^{2} \beta=\alpha\left(r_{2}^{2}-r_{1}^{2}\right)
$$

For the areas marked with 3:

$$
t_{3}=\frac{r_{2}^{2} \beta}{2}-\frac{r_{1}^{2} \beta}{2}=\frac{r_{3}^{2} \alpha}{2}-\frac{r_{2}^{2} \alpha}{2}
$$

From this
b)

$$
\beta\left(r_{2}^{2}-r_{1}^{2}\right)=\alpha\left(r_{3}^{2}-r_{2}^{2}\right)
$$

Furthermore, we know that the area of the sector marked with 2 is half the area of the sector marked with 3, that is

$$
\frac{r_{2}^{2} \beta}{2}-\frac{r_{1}^{2} \beta}{2}=2 \cdot \frac{r_{1}^{2} \beta}{2}=r_{1}^{2} \beta
$$

c)

$$
\beta\left(r_{2}^{2}-r_{1}^{2}\right)=2 r_{1}^{2} \beta
$$

If we substitute the value of $r_{1}^{2} \beta=\alpha\left(r_{2}^{2}-r_{1}^{2}\right)$ from equation a) into the right side of equation c), we get

$$
\beta\left(r_{2}^{2}-r_{1}^{2}\right)=2 \alpha\left(r_{2}^{2}-r_{1}^{2}\right)
$$

and from this $\beta=2 \alpha$.

Let's write the ratio of the areas $t_{4}$ and $t_{1}$:

$$
\frac{t_{4}}{t_{1}}=\frac{\beta\left(r_{3}^{2}-r_{2}^{2}\right)}{\alpha r_{1}^{2}}
$$

Substituting $2 \alpha$ for $\beta$ from the previous result:

$$
\frac{t_{4}}{t_{1}}=\frac{2 \alpha\left(r_{3}^{2}-r_{2}^{2}\right)}{\alpha r_{1}^{2}}
$$

From equation b):

$$
r_{3}^{2}-r_{2}^{2}=\frac{\beta\left(r_{2}^{2}-r_{1}^{2}\right)}{\alpha}=\frac{2 \alpha\left(r_{2}^{2}-r_{1}^{2}\right)}{\alpha}=2\left(r_{2}^{2}-r_{1}^{2}\right)
$$

From equation c) $r_{2}^{2}-r_{1}^{2}=2 r_{1}^{2}$. Substituting these:

$$
\frac{t_{4}}{t_{1}}=\frac{4\left(r_{2}^{2}-r_{1}^{2}\right)}{r_{1}^{2}}=\frac{8 r_{1}^{2}}{r_{1}^{2}}=8
$$

Thus, the area of $t_{4}$ is 8 times the area of $t_{1}$.