There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn. Find all possible numbers of tangent segments when he stops drawing. (Australia)
First, consider a particular arrangement of circles $C_{1}, C_{2}, \ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $\left(i^{2}, 0\right)$ and radius $i / 2$ works. Then the only tangent segments that can be drawn are between adjacent circles $C_{i}$ and $C_{i+1}$, and exactly three segments can be drawn for each pair. So Luciano will draw exactly $3(n-1)$ segments in this case. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-72.jpg?height=258&width=1335&top_left_y=1059&top_left_x=355) For the general case, start from a final configuration (that is, an arrangement of circles and segments in which no further segments can be drawn). The idea of the solution is to continuously resize and move the circles around the plane, one by one (in particular, making sure we never have 4 circles with a common tangent line), and show that the number of segments drawn remains constant as the picture changes. This way, we can reduce any circle/segment configuration to the particular one mentioned above, and the final number of segments must remain at $3 n-3$. Some preliminary considerations: look at all possible tangent segments joining any two circles. A segment that is tangent to a circle $A$ can do so in two possible orientations - it may come out of $A$ in clockwise or counterclockwise orientation. Two segments touching the same circle with the same orientation will never intersect each other. Each pair $(A, B)$ of circles has 4 choices of tangent segments, which can be identified by their orientations - for example, $(A+, B-)$ would be the segment which comes out of $A$ in clockwise orientation and comes out of $B$ in counterclockwise orientation. In total, we have $2 n(n-1)$ possible segments, disregarding intersections. Now we pick a circle $C$ and start to continuously move and resize it, maintaining all existing tangent segments according to their identifications, including those involving $C$. We can keep our choice of tangent segments until the configuration reaches a transition. We lose nothing if we assume that $C$ is kept at least $\varepsilon$ units away from any other circle, where $\varepsilon$ is a positive, fixed constant; therefore at a transition either: (1) a currently drawn tangent segment $t$ suddenly becomes obstructed; or (2) a currently absent tangent segment $t$ suddenly becomes unobstructed and available. Claim. A transition can only occur when three circles $C_{1}, C_{2}, C_{3}$ are tangent to a common line $\ell$ containing $t$, in a way such that the three tangent segments lying on $\ell$ (joining the three circles pairwise) are not obstructed by any other circles or tangent segments (other than $C_{1}, C_{2}, C_{3}$ ). Proof. Since (2) is effectively the reverse of (1), it suffices to prove the claim for (1). Suppose $t$ has suddenly become obstructed, and let us consider two cases. Case 1: $t$ becomes obstructed by a circle ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-73.jpg?height=318&width=1437&top_left_y=292&top_left_x=315) Then the new circle becomes the third circle tangent to $\ell$, and no other circles or tangent segments are obstructing $t$.