11.7. Vasya drew several circles on the plane and drew all possible common tangents to each pair of these circles. It turned out that the drawn lines contain all sides of a certain regular 2011-gon. What is the smallest number of circles Vasya could have drawn?

(N. Agakhanov)

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# Answer: 504.

Solution. Let the obtained regular 2011-gon be denoted by $M$, its vertices (in a clockwise direction) by $X_{1}, X_{2}, \ldots, X_{2011}$, its inscribed circle by $\omega$, and its center by $O$. We will call the lines containing the sides of the polygon the selected lines.

![](https://cdn.mathpix.com/cropped/2024_05_06_6cb139fb80eb844cc174g-3.jpg?height=359&width=392&top_left_y=300&top_left_x=899)

Figure 5

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 $D E$ (see Figure 5). Indeed, the vertices $A$ and $E$, as well as $B$ and $D$, are symmetric with respect to the line $C O$. Then the intersection point of the external bisector of angle $A B C$ with the line $C O$ 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. Now, if Vasya draws 503 such circles for the points $\left(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\right), \left(X_{5}, X_{6}, X_{7}, X_{8}, X_{9}\right), \ldots, \left(X_{2009}, X_{2010}, X_{2011}, X_{1}, X_{2}\right)$, and also the circle $\omega$, then any selected 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 selected line must be tangent to at least two circles. The circle $\omega$ is tangent to all 2011 of these lines. Any other circle has no more than four common tangents with $\omega$; hence, it is tangent to no more than four selected lines. Therefore, if there are $n$ circles, then the total number of tangencies between circles and selected lines is no more than $2011 + 4(n-1)$; on the other hand, there must be at least $2011 \cdot 2 = 4022$ tangencies. Thus, $2011 + 4(n-1) \geq 2 \cdot 2011$, from which it follows that $n \geq 2011 / 4 + 1 > 503$, which is what we needed to prove.

Comment. Only the answer - 0 points.

Only shown that 504 circles are sufficient - 2 points.

Only shown that 503 circles are insufficient - 4 points.