Agakhanov N.X.

Vasya drew several circles on a plane and drew all possible common tangents to each pair of these circles. It turned out that the drawn lines contain all the sides of a certain regular 2011-gon. What is the smallest number of circles that Vasya could have drawn?
Let the obtained regular 2011-gon be denoted by $M$, its vertices (in clockwise order) by $X_{1}, X_{2}, \ldots, X_{2011}$, its incircle by $\omega$, and its center by $O$. We will call the lines containing the sides of the polygon the distinguished lines.

Notice that for any five consecutive vertices $A, B, C, D, E$ of the polygon $M$, there exists a circle, different from $\omega$, that is tangent to the lines $A B, B C, C D$, and $D E$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f7c9453eb6f3f37ea7g-26.jpg?height=1241&width=1332&top_left_y=0&top_left_x=349)

Indeed, the vertices $A$ and $E$, as well as $B$ and $D$, are symmetric with respect to the line $CO$. Then the intersection point of the external bisector of angle $A B C$ with the line $CO$ is different from $O$ and is equidistant from the lines $A B, B C, C D$, and $D E$, and thus is the center of the desired circle.

If Vasya draws 503 such circles for the points $(X_{1}, X_{2}, X_{3}, X_{4}, X_{5})$, $(X_{5}, X_{6}, X_{7}, X_{8}, X_{9}), \ldots, (X_{2009}, X_{2010}, X_{2011}, X_{1}, X_{2})$, and also the circle $\omega$, then each distinguished line will be a common tangent to two of the drawn circles. Thus, 504 circles are sufficient.

It remains to prove that there must be at least 504 circles. Each distinguished line must be tangent to at least two circles. The circle $\omega$ is tangent to all 2011 of these lines. Each other circle has no more than four common tangents with $\omega$, meaning it is tangent to no more than four distinguished lines. Thus, if there are $n$ circles, then the total number of tangencies between the circles and the distinguished lines is no more than $2011 + 4(n-1)$; on the other hand, there must be at least $2011 \cdot 2 = 4022$ tangencies. Therefore, $2011 + 4(n-1) \geq 2 \cdot 2011$, from which it follows that $n \geq 2011 / 4 + 1 > 503$.

## Answer

504 circles.