## Task B-1.2.

Two circles $K_{1}$ and $K_{2}$, shaded in color, intersect such that $10 \%$ of the area of circle $K_{1}$ and $60 \%$ of the area of circle $K_{2}$ lie outside their intersection. Calculate the ratio of the radii of the circles $K_{1}$ and $K_{2}$. What is the sum of these radii if the total shaded area is $94 \pi$?

![](https://cdn.mathpix.com/cropped/2024_05_30_ab4e7c3a83d03c059773g-01.jpg?height=377&width=440&top_left_y=1956&top_left_x=1276)
## Solution.

If $10 \%$ of the area of circle $K_{1}$ is outside the common intersection, then the area of the intersection is equal to $90 \%$ of the area of circle $K_{1}$.

If $60 \%$ of the area of circle $K_{2}$ is outside the common intersection, then the area of the intersection is equal to $40 \%$ of the area of circle $K_{2}$.

If we denote the radius of circle $K_{1}$ by $R_{1}$ and the radius of circle $K_{2}$ by $R_{2}$, we get

$$
\frac{90}{100} R_{1}^{2} \pi=\frac{40}{100} R_{2}^{2} \pi
$$

1 point

which simplifies to $\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{4}{9}$ and finally $\frac{R_{1}}{R_{2}}=\frac{2}{3}$.

1 point

The total shaded area can be calculated as $P\left(K_{1}\right)+60 \% P\left(K_{2}\right)$ or $P\left(K_{2}\right)+10 \% P\left(K_{1}\right)$ or $P\left(K_{1}\right)+P\left(K_{1}\right)-P\left(K_{1} \cap K_{2}\right)$.

Thus, we have

$$
P\left(K_{1}\right)+P\left(K_{1}\right)-P\left(K_{1} \cap K_{2}\right)=R_{1}^{2} \pi+R_{2}^{2} \pi-0.90 R_{1}^{2} \pi=94 \pi
$$

It follows that $R_{1}^{2}+R_{2}^{2}-0.90 R_{1}^{2}=94$, or, since $R_{2}^{2}=\frac{9}{4} R_{1}^{2}$,

$R_{1}^{2}+\frac{9}{4} R_{1}^{2}-0.90 R_{1}^{2}=94$,

$\frac{94}{40} R_{1}^{2}=94$, so $R_{1}^{2}=40$.

From this, we get $R_{1}=2 \sqrt{10}, R_{2}=3 \sqrt{10}$

and finally $R_{1}+R_{2}=5 \sqrt{10}$.

Note: The ratio of the radii can also be calculated as follows.

If $10 \%$ of the area of circle $K_{1}$ and $60 \%$ of the area of circle $K_{2}$ are outside the common intersection, then the common intersection occupies $30 \%$ of the total area of $K_{1} \cup K_{2}$.

Now,

$$
\frac{P\left(K_{1}\right)}{P\left(K_{2}\right)}=\frac{(10 \%+30 \%) P\left(K_{1} \cup K_{2}\right)}{(60 \%+30 \%) P\left(K_{1} \cup K_{2}\right)}=\frac{40 \%}{90 \%}=\frac{4}{9}
$$

It follows that $\frac{R_{1}}{R_{2}}=\frac{2}{3}$ (since the squares of the radii are in the same ratio as the areas).

This part is worth 3 points. The rest follows as in the solution above.