IMO 2002 · Problem 6

\(n>2\) circles of radius \(1\) are drawn in the plane so that no line meets more than two of the circles. Their centers are \(O_1,O_2,\cdots,O_n\). Show that \[ \sum_{i<j}1/O_iO_j\le (n-1)\pi/4. \]

Official text of the problem. Official problem paper (PDF)

Solution

Write \(d_{ij}=O_iO_j\). First, \(d_{ij}>2\) for every pair. Indeed, if \(d_{ij}\le2\), choose a third center \(O_k\) and a line containing both \(O_k\) and the midpoint \(M\) of \(O_iO_j\). Its distance from each of \(O_i,O_j\) is at most \(d_{ij}/2\le1\), so it meets all three circles, a contradiction.

We count directions of tangent lines. For \(0\le\theta<2\pi\), put \[ u_\theta=(\cos\theta,\sin\theta), \] and let \[ L_i(\theta)=\{X:(X-O_i)\cdot u_\theta=1\}. \] This is the tangent to the \(i\)-th circle with outward unit normal \(u_\theta\). For sets of directions, write \(|S|\) for their angular length, measured in radians.

For \(i\ne j\), define \[ A_{ij}=\{\theta:L_i(\theta)\text{ meets the }j\text{-th circle}\}. \] If \(O_j-O_i=d_{ij}(\cos\alpha,\sin\alpha)\), the distance from \(O_j\) to \(L_i(\theta)\) is \[ \left|d_{ij}\cos(\theta-\alpha)-1\right|. \] Consequently, \[ \theta\in A_{ij} \quad\Longleftrightarrow\quad 0\le\cos(\theta-\alpha)\le\frac{2}{d_{ij}}. \] Because \(d_{ij}>2\), these directions form two arcs, each of angular length \(\arcsin(2/d_{ij})\). Thus \[ |A_{ij}|=2\arcsin\frac{2}{d_{ij}}. \tag{1} \] For each fixed \(i\), the sets \(A_{ij}\), \(j\ne i\), are pairwise disjoint: a direction belonging to two of them would give a tangent line meeting three circles.

There are also directions in which an outermost circle has a tangent meeting no other circle. More precisely, define \[ B_i=\{\theta:O_i\cdot u_\theta>O_j\cdot u_\theta \text{ for every }j\ne i\}. \] For \(\theta\in B_i\), every other center satisfies \[ (O_j-O_i)\cdot u_\theta<0, \] so its distance from \(L_i(\theta)\) is greater than \(1\). Hence \(B_i\) is disjoint from every \(A_{ij}\).

The sets \(B_i\) are pairwise disjoint and together contain all directions except possibly finitely many. Indeed, failure of a unique maximum among the projections \(O_i\cdot u_\theta\) requires \[ (O_i-O_j)\cdot u_\theta=0 \] for some pair \(i\ne j\); each pair gives only two directions. Therefore \[ \sum_{i=1}^n|B_i|=2\pi. \tag{2} \]

For each \(i\), disjointness gives \[ |B_i|+\sum_{j\ne i}|A_{ij}|\le2\pi. \] Summing over \(i\) and using (1) and (2), we obtain \[ 2\pi+4\sum_{i<j}\arcsin\frac{2}{d_{ij}}\le2\pi n. \] Thus \[ \sum_{i<j}\arcsin\frac{2}{d_{ij}} \le\frac{(n-1)\pi}{2}. \] Finally, the elementary inequality \(\sin t\le t\) implies \(\arcsin x\ge x\) for \(0\le x\le1\). It follows that \[ \boxed{\displaystyle \sum_{i<j}\frac1{O_iO_j} \le\frac12\sum_{i<j}\arcsin\frac{2}{d_{ij}} \le\frac{(n-1)\pi}{4}.} \]

The solution was checked against Evan Chen's published IMO solution notes.