Task B-4.4. Three congruent ellipses with semi-axes $a$ and $b(a>b)$ are arranged so that each pair touches, and their major axes lie on the sides of an equilateral triangle (figure). Two concentric circles, the larger with radius $R$ and the smaller with radius $r(R \neq r)$, touch all three ellipses. Express $R$ and $r$ in terms of $a$ and $b$.

![](https://cdn.mathpix.com/cropped/2024_05_30_38fd7d226823545bc90fg-18.jpg?height=404&width=393&top_left_y=801&top_left_x=844)
## Solution.

Let the axes of the ellipse $c$ coincide with the coordinate axes, and the center with the origin.

Then its equation is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the equation of the circle $k_{1}$ is $x^{2}+(y-p)^{2}=R^{2}$. It is obvious that the line $A S$ is a tangent to the ellipse $c$. Its slope is $k=\operatorname{tg} 30^{\circ}=\frac{\sqrt{3}}{3}$, and the intercept on the $y$-axis is $l=p$.

Therefore, it has the equation

$$
y=\frac{\sqrt{3}}{3} x+p
$$

From the condition of tangency $a^{2} k^{2}+b^{2}=l^{2}$ of the line and the ellipse, we get

$$
a^{2}\left(\frac{\sqrt{3}}{3}\right)^{2}+b^{2}=p^{2} \quad \Rightarrow \quad p^{2}=\frac{a^{2}+3 b^{2}}{3}
$$

Then $r=p-b$, that is,

$$
r=\sqrt{\frac{a^{2}+3 b^{2}}{3}}-b
$$

The circle $k_{1}$ touches the ellipse $c$ at points $D_{1}$ and $D_{2}$ whose ordinates are equal.

From the system of equations formed by the equation of the circle $k_{1}$ and the ellipse $c$

$$
\left\{\begin{array}{l}
x^{2}+(y-p)^{2}=R^{2} \\
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_38fd7d226823545bc90fg-19.jpg?height=713&width=770&top_left_y=213&top_left_x=648)

we get

$$
\frac{R^{2}-(y-p)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$

From here, after simplification, we obtain

$$
\left(a^{2}-b^{2}\right) y^{2}+2 b^{2} p y+b^{2}\left(R^{2}-p^{2}-a^{2}\right)=0
$$

This equation must have only one solution, so its discriminant is zero, that is,

$$
\begin{aligned}
& \left(2 b^{2} p\right)^{2}-4\left(a^{2}-b^{2}\right) b^{2}\left(R^{2}-p^{2}-a^{2}\right)=0 \\
& \Rightarrow R^{2}=\frac{a^{2}\left(p^{2}+a^{2}-b^{2}\right)}{a^{2}-b^{2}}
\end{aligned}
$$

Substituting $(*)$ into this expression, after simplification, we get $R=\frac{2 a^{2}}{\sqrt{3\left(a^{2}-b^{2}\right)}}$.