## Task 4 - 280824

Let $k_{1}$ and $k_{2}$ be two concentric circles with center $M$, whose radii are in the ratio $3:1$. Two diameters $A_{1} B_{1}$ and $C_{1} D_{1}$ of $k_{1}$ intersect $k_{2}$ at points $A_{2}, B_{2}$ and $C_{2}, D_{2}$, respectively, as shown in the figure.

a) Determine the ratio of the areas of the sector $A_{2} M D_{2}$ and the annular sector $D_{2} B_{2} B_{1} D_{1}$, assuming that $\angle A_{1} M D_{1}$ is a right angle!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0785.jpg?height=440&width=377&top_left_y=634&top_left_x=1359)

b) How should the size of the angle $\angle A_{1} M D_{1}$ be chosen so that the area of the sector $A_{2} M D_{2}$ is equal to the area of the annular sector $D_{2} B_{2} B_{1} D_{1}$?
a) Let $A_{1} M=r_{1}$ and $A_{2} M=r_{2}$. Then the area $F_{1}$ of the sector $A_{2} M D_{2}$ is

$$
F_{1}=\frac{r_{2}^{2} \pi \cdot 90^{\circ}}{360^{\circ}}=\frac{r_{2}^{2} \pi}{4}
$$

and the area $F_{2}$ of the annular sector $D_{2} B_{2} B_{1} D_{1}$ is

$$
F_{2}=\frac{r_{1}^{2} \pi \cdot 90^{\circ}}{360^{\circ}}-\frac{r_{2}^{2} \pi}{4}=\frac{\pi}{4}\left(r_{1}^{2}-r_{2}^{2}\right)
$$

Since $r_{1}=3 r_{2}$, it follows that $r_{1}^{2}-r_{2}^{2}=8 r_{2}^{2}$, and thus $F_{2}=2 \pi r_{2}^{2}$. The ratio $F_{1}: F_{2}$ is therefore

$$
F_{1}: F_{2}=\frac{r_{2}^{2} \pi}{4}:\left(2 \pi r_{2}^{2}\right)=1: 8
$$

b) Let $A_{1} M=r_{1} ; A_{2} M=r_{2}$ and $\angle A_{1} M D_{1}=\alpha$. Then the area $F_{3}$ of the sector $A_{2} M D_{2}$ is:

$$
F_{3}=\frac{r_{2}^{2} \pi \cdot \alpha}{360^{\circ}}
$$

Since $\angle B_{1} M D_{1}$ is the supplementary angle of $\angle A_{1} M D_{1}$, its measure is $180^{\circ}-\alpha$. The area $F_{4}$ of the annular sector $D_{2} B_{2} B_{1} D_{1}$ is then:

$$
F_{4}=\frac{r_{1}^{2} \pi\left(180^{\circ}-\alpha\right)}{360^{\circ}}-\frac{r_{2}^{2} \pi\left(180^{\circ}-\alpha\right)}{360^{\circ}}=\frac{\pi\left(180^{\circ}-\alpha\right)}{360^{\circ}}\left(r_{1}^{2}-r_{2}^{2}\right)
$$

Since $r_{1}=3 r_{2}$, it follows that $r_{1}^{2}-r_{2}^{2}=8 r_{2}^{2}$, and thus

$$
F_{4}=\frac{\pi\left(180^{\circ}-\alpha\right)}{360^{\circ}} \cdot 8 r_{2}^{2}
$$

To satisfy the condition $F_{3}=F_{4}$, one must choose

$$
\frac{r_{2}^{2} \pi \alpha}{360^{\circ}}=\frac{\pi\left(180^{\circ}-\alpha\right)}{360} \cdot 8 r_{2}^{2}
$$

From this equation, after multiplying by $360^{\circ}$ and dividing by $\pi \cdot r_{2}^{2}$, we get

$$
\alpha=8 \cdot\left(180^{\circ}-\alpha\right) \quad \Rightarrow \quad \alpha=160^{\circ}
$$

Thus, the angle $\angle A_{1} M D_{1}$ should be $160^{\circ}$.

Solutions of the 2nd Round 1988 adopted from [5]

### 5.30.3 3rd Round 1988, Class 8