Given a triangle $ABC$. Construct two circles inside the triangle that touch each other, one of which touches the sides meeting at $A$, and the other touches the sides meeting at $B$, and the ratio of their radii is $1:2$.

What can we say about the angles of a triangle in which such a pair of circles cannot be constructed?
We carry out the construction using a similarity transformation. With an arbitrary radius $r_{1}^{*}$, we take a circle $k_{1}^{*}$ tangent to the side segment $AB$ at $E$ inside the triangle, construct a circle $k_{2}^{*}$ with radius $2 r_{1}^{*}$ that is tangent to $k_{1}^{*}$ and the ray $EB$ - if necessary, take a new $k_{1}^{*}$ such that the tangency point $F$ of $k_{2}^{*}$ also lies on the segment $AB$ - then among the tangents of $k_{1}^{*}$ parallel to $AC$, we choose the one $t_{1}^{*}$ that intersects the ray $EA$, and among the tangents of $k_{2}^{*}$ parallel to $BC$, we choose the one that intersects the ray $FB$. Let the intersection points be $A^{*}, B^{*}$, and the intersection point of the drawn tangents be $C^{*}$, so in the triangle $A^{*} B^{*} C^{*}=H^{*}$, $k_{1}^{*}, k_{2}^{*}$ meet the conditions of the problem, and $H^{*}$ and the given triangle $A B C=H$ are centrally similar at the intersection point $O$ of the lines $AB$ and $CC^{*}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4b8153d2217d7cceefb8g-1.jpg?height=746&width=1119&top_left_y=478&top_left_x=492)

Now, by enlarging $k_{1}^{*}$ and $k_{2}^{*}$ from $O$ in the ratio $OA : OA^{*}$, we obtain the desired pair of circles $k_{1}, k_{2}$. The proof of the correctness of the construction is left to the reader.

The pair of circles $k_{1}^{*}, k_{2}^{*}$ - even with the given constraints - can be constructed in any triangle $ABC$, and $k_{1}^{*}$ is indeed inside $H^{*}$, so $k_{1}$ will also be inside $H$. The circle $k_{2}$ will be inside $H$ if and only if $k_{2}^{*}$ is inside $H^{*}$. This condition is that $t_{1}^{*}$ does not intersect $k_{2}^{*}$, at most it touches it. This is equivalent to $t_{1}^{*}$ making at least as large an angle with $AB$ as the second common tangent $t$ of $k_{1}^{*}$ and $k_{2}^{*}$. Denoting this angle by $\varphi$ and the projection of $O_{1}^{*}$ on $O_{2}^{*} F$ by $G$,

$$
\sin \frac{\varphi}{2}=\frac{G O_{2}^{*}}{O_{1}^{*} O_{2}^{*}}=\frac{1}{3}
$$

$\varphi=38^{\circ} 57^{\prime}$, and the condition for the solution is: $\alpha \geq \varphi$.

With this, we have completed the solution.