Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Could someone write a solution for this nice IMO problem ?

You can find solutions here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=110250#p110250

We begin with some definitions. For any set of points $X$, let $a(X)$ denote the area of $X$. For $1 \leq i \leq n$ and for any point $X$ on $C_i$, let $t_i(X)$ be the line tangent to $C_i$ at $X$. For $1 \leq i, j \leq n$, $i \neq j$, let the common internal tangents of $C_i$ and $C_j$ intersect at $A_{i,j}, D_{i,j}$, respectively, and let their common internal tangents intersect at $B_{i,j}$, $C_{i,j}$, respectively, so that $A_{i,j}, B_{i,j}, C_{i,j}, D_{i,j}$ appear on $C_i$ in that order clockwise along the circumference of $C_i$. Let $T_{i,j}$ be the region $(\triangle O_i A_{i,j} B_{i,j}) \cup (\triangle O_i C_{i,j} D_{i,j})$. Lemma 1: $a(T_{i, j}) = \frac{2}{O_i O_j}$. Proof: In this lemma, we will refer to $A_{i,j}, B_{i,j}, C_{i,j}$, and $D_{i,j}$ by $A, B, C$, and $D$, respectively. Furthermore, let $t_i(B)$ intersect $O_i O_j$ at $P$, let it intersect $t_i(A)$ at $T$, and let $t_i(A)$ intersect $C_j$ at $F$. (See Fig 1.) Since \[ \angle BO_i A = \pi - \angle ATB = \angle FTP = \angle TPO_i, \] we have \[ [O_i AB] = \frac{1^2 \sin \angle B O_i A}{2} = \frac{\sin \angle TPO_i}{2} = \frac{\frac{1}{\frac{O_i O_j}{2}}}{2} = \frac{1}{O_i O_j}. \] By symmetry, $[O_i CD] = \frac{1}{O_i O_j}$ as well, so $a(T_{i,j}) = [O_i AB] + [O_i CD] = \frac{2}{O_i O_j}$. as desired. Lemma 2: If $1 \leq i, j, k, l \leq n$, $i \neq j$,$k \neq l$, and $(i, j) \neq (k, l)$, $T_{i,j} \cap T_{k,l} = \emptyset$. Proof: $T_{i,j} \subseteq C_i$ and $T_{k,l} \subseteq C_k$; hence, if $i \neq k$, since $C_i \neq C_k$ the result follows immediately, so we may assume that $i = k$. Let $X$ be any point on minor arc $AB$. As $X$ varies from $A_{i,j}$ to $B_{i,j}$, $t_i(X)$ must always meet $C_j$; similarly, as $X$ varies from $C_{i,j}$ to $D_{i,j}$, $t_i(X)$ must always meet $C_j$ (see Fig 2). The same holds when we vary $X$ from $A_{i,l}$ to $B_{i,l}$ and from $C_{i,l}$ to $D_{i,l}$. Because no line intersects 3 of our unit circles, minor arcs $A_{i,j} B_{i,j}$, $C_{i,j}D_{i,j}$, $A_{i,l} B_{i,l}$, and $C_{i,l} D_{i,l}$ must be pairwise disjoint; it follows immediately that $T_{i,j}$ and $T_{i,l}$ must also be disjoint. Without loss of generality, let $O_1, O_2 \ldots, O_m$ be the convex hull of $O_1, O_2, \ldots, O_n$, with $O_1, O_2, \ldots O_m$ appearing in clockwise order. Let the common external tangent to $C_{i-1}$ and $C_i$ lying outside of $O_1 O_2 \ldots O_m$ intersect $C_i$ at $X_i$, let the common external tangent to $C_i$ and $C_{i+1}$ lying outside of $O_1 O_2 \ldots O_m$ intersect $C_i$ at $Y_i$, let these two external tangents intersect at $Z_i$, and let $\theta_i = X_i Z_i Y_i$ (all indices are taken modulo $m$) (see Fig 3.) For any $1 \leq i \leq m$, it is clear that for any $j \neq i$, the tangency points to $C_i$ of the common tangents of $C_i$ and $C_j$ all lie within major arc $X_i Y_i$. Thus, $T_{i,j}$ always lies in the sector $X_i Y_i$ farther from $Z_i$, which has area $\frac{\pi + \theta_i}{2}$. Hence, by lemmas 1 and 2, \begin{align*} \sum_{1 \leq j \leq n, j \neq i} \frac{1}{O_i O_j} = \frac{1}{2} \sum_{1 \leq j \leq n, j \neq i} a(T_{i,j}) = \frac{1}{2} a \left( \bigcup_{1 \leq j \leq n, j \neq i} T_{i,j} \right) \leq \frac{\pi + \theta_i}{4}. \tag{1} \end{align*} For any $i > m$, we have \begin{align*} \sum_{1 \leq j \leq n, j \neq i} \frac{1}{O_i O_j} = \frac{1}{2} \sum_{1 \leq j \leq n, j \neq i} a(T_{i,j}) = \frac{1}{2} a \left( \bigcup_{1 \leq j \leq n, j \neq i} T_{i,j} \right) \leq \frac{\pi}{2}. \tag{2} \end{align*} Adding up all the equations obtained from (1) and (2), we get that \begin{align*} \sum_{1 \leq i < j \leq n} \frac{1}{O_i O_j} &= \frac{1}{2} \sum_{1 \leq i, j \leq n, i \neq j} \frac{1}{O_i O_j} \\ &= \frac{1}{2} \sum_{1 \leq i \leq m} \sum_{1 \leq j \leq n, j \neq i} \frac{1}{O_i O_j} + \frac{1}{2} \sum_{m+1 \leq i \leq n} \sum_{1 \leq j \leq n, j \neq i} \frac{1}{O_i O_j} \\ &\leq \frac{1}{2} \sum_{1 \leq i \leq m} \frac{\pi + \theta_i}{4} + \frac{1}{2} \sum_{m+1 \leq i \leq n} \frac{\pi}{2} \\ &= \frac{1}{2} \left( \frac{m \pi}{4} + \frac{\pi(m-2)}{4} \right) + \frac{(n-m) \pi}{4} \\ &= \frac{(n-1) \pi}{4}, \end{align*} as desired.

Attachments:

Let $O_{k_1},O_{k_2},\ldots,O_{k_m}$ ($m\ge3$) be the vertices (in clockwise order) of the convex hull of $O_1,O_2,\ldots,O_n$. Lemma 1: For $i\notin\{k_1,\ldots,k_m\}$, $\sum_{1\le j\ne i\le n}\frac{4}{O_iO_j}\le2\pi$. Proof: For each $j\ne i$, consider the four half angles (each of measure $\arcsin(1/O_iO_j)$) formed by line $O_iO_j$ and the two tangent lines to $(O_j)$ passing through $O_i$; clearly no two of these angles overlap. Summing over all the half angles and using the inequality $\arcsin{x}\ge x$, we get the desired inequality.$\blacksquare$ Lemma 2: For $1\le r\le m$ (indices of $k_s$ taken modulo $m$), $\sum_{1\le j\ne k_r\le n}\frac{2}{O_{k_r}O_j}-\frac{1}{O_{k_r}O_{k_{r-1}}}-\frac{1}{O_{k_r}O_{k_{r+1}}}\le\angle{O_{k_{r-1}}O_{k_r}O_{k_{r+1}}}$. Proof: For each $j\ne k_r$, consider (as earlier) the two half angles (each of measure $\arcsin(1/O_{k_r}O_j)$) formed by ray $O_{k_r}O_j$ and the two tangents from $O_{k_r}$ to $(O_j)$; clearly no two of these angles overlap. Summing over the half angles lying in between $O_{k_r}O_{k_{r-1}}$ and $O_{k_r}O_{k_{r+1}}$ (so each angle appears twice except for $j=O_{k_{r-1}}$ and $j=O_{k_{r+1}}$, whose corresponding angles each appear once) and using the inequality $\arcsin{x}\ge x$, we get the desired inequality.$\blacksquare$ Lemma 3: For $1\le r\le m$, $\frac{2}{O_{k_r}O_{k_{r-1}}}+\frac{2}{O_{k_r}O_{k_{r+1}}}\le \pi-\angle{O_{k_{r-1}}O_{k_r}O_{k_{r+1}}}$. Proof: For convenience, set $u = k_{r-1}$, $v = k_r$, $w = k_{r+1}$. Keeping the clockwise orientation in mind, let the two internal tangents of $(O_u),(O_v)$ meet at $P$ (the midpoint of $O_uO_v$) and $\ell=PT$ denote the one to the right of $O_u$ and left of $O_v$, where $T\in(O_v)$. Considering the set of lines through $P$ passing through both $(O_u)$ and $(O_v)$, it's clear that $(O_w)$ contains no points on the same side of $\ell$ as $O_v$, i.e. the right of $\ell$. Now setting $Q$ to be the insimilicenter of $(O_v)$ and $(O_w)$ (i.e. the midpoint of $O_vO_w$, since $(O_v)$ and $(O_w)$ have the same radius) and noting that $(O_v)$ is tangent to $\ell$ at $T$, we must have $Q$ to the left of $\ell$ (otherwise, $O_w$ will either be to the right of $\ell$ or closer to $\ell$ than $O_v$). Thus $2/O_{k_r}O_{k_{r-1}}\le\arcsin(2/O_{k_r}O_{k_{r-1}})=\angle{O_iPT}<\angle{O_iPQ}$. But similarly, we have $2/O_{k_r}O_{k_{r+1}}<\angle{O_iQP}$, so we're done by adding these two inequalities.$\blacksquare$ Now summing up half of lemma 1 over all $n-m$ indices $i\notin\{k_1,\ldots,k_m\}$ in addition to lemma 2 and half of lemma 3 over all $m$ indices $r\in\{1,\ldots,m\}$, we get \begin{align*} \sum_{1\le i<j\le n}\frac{2}{O_iO_j}+\frac{2}{O_iO_j}&\le\pi(n-m)+\sum_{r=1}^{m}\frac{\pi+\angle{O_{k_{r-1}}O_{k_r}O_{k_{r+1}}}}{2} \\ &\le \pi(n-m)+\frac{\pi(m)}{2}+\frac{\pi(m-2)}{2}=\pi(n-1), \end{align*}as desired.

Someone probably needs to check this solution, since I weaken the condition a significant amount, and somehow still get bound. Denote $d(X, AB)$ to be the distance from $X$ to the line $AB$. We pretty much will only use $d(O_i, O_jO_k) \ge 2$ (*) for all the stuff below. Let the convex hull of the centers be the points $H_1, H_2, \dots, H_k$ in that order. (a $k$-gon). All angles in radians of course. We will use the inequality $\sin{x} \le x$. Lemma 1: $\sum_{i \not= j}\frac{1}{H_iH_j} \le \frac{1}{2} \cdot \frac{n-1}{n-2} \cdot \angle H_{j-1}H_jH_{j+1}$ for any $j$ (indices modulo $k$ of course) Proof: So take $H_j$ and order the remaining $n-1$ vertices based on angle, so when we sweep from $H_{j-1}$ to $H_{j+1}$ we hit all the middle vertices in order. Call these remaining $n-1$ vertices $O_{j_1}, O_{j_2}, \dots, O_{j_{n-1}}$ (so $O_{j_1} = H_{j-1}$ and $O_{j_{n-1}} = H_{j+1}$). This way we form $n-2$ angles. For each of the $n-1$ lengths $H_jO_{j_p}$ for some $p$, by simple trig and (*) it is $\ge \max(2/\sin{\alpha_\ell}, 2/\sin{\alpha_r})$ where $\alpha_\ell, \alpha_r$ are the angles to the left and right (if they exist). Summing over all of these, it is easy to get that ($\alpha_i$ is the optimal angle for the side) $\sum_{i \not= j}\frac{1}{H_iH_j} \le \frac{1}{2} \sum{\sin \alpha_i} \le \frac{n-1}{n-2} \cdot \frac{1}{2} \cdot \angle H_{j-1}H_jH_{j+1}$ since $\sin{\alpha_i} \le \alpha_i$ and $\sum{\alpha_i} \le \frac{n-1}{n-2} \cdot \angle H_{j-1}H_jH_{j+1}$ (the extra factor since we have more sides than angles). Now we turn our attention to points inside the convex hull. For any point $O_j$ inside (or on) the convex hull, we have Lemma 2: $\sum_{i \not= j} \frac{1}{O_iO_j} \le \frac{n-1}{n-2} \cdot \frac{1}{2} \cdot \pi$. Proof: Trying to repeat the same thing as above, we notice a slight issue: we would get a total angle when sweeping of $2\pi$, so our RHS would actually be twice what I wrote in Lemma 2. To resolve this, relax our condition even more to the following: for any points $O_i, O_k \not= O_j$, $d(O_i, O_kO_j) \ge 2$ (**), i.e. we only look at lines through our $O_j$ vertex. Now, note that if we reflect a point $O_i$ over $O_j$ to a point $O_i'$ then the new set of points still satisfies (**) (it's easy to see that $d(O_i, O_kO_j) = d(O_i', O_kO_j)$ and $d(O_i, O_kO_j) = d(O_i, O_k'O_j)$) Now pick the maximum angle of the form $O_iO_jO_k \le \pi$ and reflect some of the remaining points over $O_j$ so they all lie in this angle. Now repeat the process in Lemma 1 to finish. (Note that Lemma 1 only used (**) actually) Summing over everything finally, we get for Lemma 1 and Lemma 2, since the sum of angles of a convex k-gon is $\pi(k-2)$ and $n-k$ points inside, we get $\frac{n-1}{n-2} \cdot \frac{1}{2} \cdot (\pi(k-2) + \pi(n-k)) = \frac{\pi(n-1)}{2}$. But we added every edge twice, so divide by $2$ to finish.

Sorry for LaTeX; this was mostly copy/pasted from a my solution I sent the above poster (lol). Note the $n=3$ case is trivial. For any segment, label it with the minimal angle that segment makes with any other segment (the acute angle or right angle). If the sum of these labelings is at most $(n-1)pi/2$, we are happy because every segment has a length at most half its labeling. Let $f(a,b)$ be the labeling we give $O_aO_b$. Fix a, say $a=n$. Look at $f(n,1)$, $f(n,2)$,... $f(n,n-1)$. Their sum is clearly bounded above by the sum over all the lines through $O_n$ of the angle between that line and the line through $O_n$ immediately clockwise of it, which is at most $pi$. So the sum is at most $pi$. So averaging over everything the average of $f(a,b)$ is at most $\frac {pi} {n-1}$ and the sum is at most half of $n$ times $pi$, meaning the LHS is at most a quarter of $n$ times pi. Unfortunately, the LHS is ever slightly so less. But without the silly sin(theta)<theta bound, instead we get a bound of $n(n-1)/2 * sin(pi/(n-1))$ which is at most $pi*(n-1)/2$ iff $sin(pi/(n-1))$ is at most $pi/n$. So I will just sharpen the bound slightly, using the convex hull. If $O_n$ is on the convex hull, and its adjacent sides form an angle $R<pi$, then $f(n,1)+f(n,2)+...f(n,n-1)$ is bounded above by $R$, more or less. I say more or less because of the fact that it is actually $R(1+1/(n-2))$, because the first $k$ counterclockwise lines the angle is at most what is immediately clockwise of that, and for the rest it is at most what is immediately counterclockwise. This counts all the $n-2$ pieces of the R angle except a piece which it double counts and averaging gives this. So now, sum over all $j$ the sum of $f(i,j)$ over $i \neq j$. This is twice the sum $f(i,j)$ over $i \neq j$. This is also what? It is NOT just at most $pi*(n)$, but rather at most $pi(n)-2pi+R/(n-2)$ because the sum of the R terms in total is 2pi less than the sum of a bunch of pi's. The sum over hull of ( pi- angle of convex hull) is 2*pi. Now, sum $R/(n-2)$ is at most $pi$, since the sum of $R$ is at most $(n-2)pi$. So this is at most $pi(n-1)$. Thus, sum $f(i,j)$ is at most $pi(n-1)/2$, and the LHS is at most $pi(n-1)/4$, as desired.

First, one observation. Notice that the distance from $A$ to $BC$ (where $A$, $B$ and $C$ are centers of some circles) is at least 2. So by trigonometry it is easy to get that $1/BC \le sinABC/2 < \angle ABC/2$. So, for any segment $AB$ defined by our centers, if $\alpha(AB)$ is defined as the smallest possible angle $\angle CBA$ or $\angle CAB$, we have $1/AB < \alpha(AB)/2$. From here, one is tempted to start assigning angles. Unfortunately, it is hard to control the angles, because the segments that contain a certain center form a $2\pi$ angle, so we get a $(n-1)\pi/2$ bound. To solve this, the following idea is required: For a given center $A$, we can reflect some of the other centers across $A$ so that they all lie on the same side of a certain line through $A$. This doesn't change the angles nor the segment lengths containing $A$. So, if $B_i$ are the centers distinct from $A$, we get $\sum 1/AB_i < \frac{1}{2} * \sum \alpha(AB_i) \le (n-1)\pi/2(n-2)$. To prove the second bound: the angle defined "clockwise" by segment $AB_i$, and then alter the angle defined to specific segments so that the smallest angle $\angle B_iAB_j$ is repeated. We get a sum of at most $\pi + (\pi/(n-2))$. So we get a bound of $\frac{\pi n(n-1)}{4(n-2)}$ To fix this (it's still slightly more than the required bound), one can consider the convex hull and the segments containing points on the convex hull, and gets the required bound easily.

This is probably the same solution, but with a probabilistic interpretation. On the space of all lines intersecting a unit circle $C$ define the probability measure $\frac{1}{2 \pi}d\theta dr$, where $r\in[0,1]$ and $\theta\in[0, 2\pi)$ are the polar coordinates of the perpendicular $OP$ from the center of $C$ to the line $l$. The probability of a line to intersect both the circle $C$ and $C'$ is the integral of the length of the intersection of their projections on the line through $OP$, and can be computed as $$ \frac 1 {2\pi}\int_0^{2\pi} (2-OO'|cos(\alpha)|)^+ d\alpha:= f\left(\frac 1 {OO'}\right),\, \alpha:=\measuredangle POO' $$It is not hard to compute $f(b)=\frac 2 {b \pi}( 2 b \arcsin [2 b]+\sqrt{1 - 4 b^2}-1)\geq \frac 4 {\pi} b.$ Thus, from the condition of the problem we have the probability of lines intersecting the circle $C_1$ and one of the circles $C_2, \ldots, C_n$ is $$ \sum_{k>1} f\left( \frac {1} {O_1 O_k} \right)\leq 1. $$So $$ \sum_{k>1} \frac 1 {O_1 O_k}\leq \frac{\pi} 4. $$Writing similar inequalities for all other circles and summing up we get \[ \sum\limits^{}_{1\leq i<j\leq n}{1\over O_iO_j}\leq{n \pi\over 8}<{(n-1)\pi\over 4}. \]

Thanks to Yang Liu for helping me finish this solution (and thus making it basically the same as his...). For brevity, let $d_{ij}$ be the length of $O_{ij}$ and let $\angle(ijk)$ be shorthand for $\angle O_i O_j O_k$ (or its measure in radians). First, we eliminate the circles completely and reduce the problem to angles using the following fact (which is in part motivated by the mysterious presence of $\pi$ on right-hand side, and also brings $d_{ij}^{-1}$ into the picture). Lemma: For any indices $i$, $j$, $m$ we have the inequalities \[ \angle (imj) \ge \max \left( \frac{2}{d_{mi}}, \frac{2}{d_{mj}} \right) \quad\text{and}\quad \pi - \angle (imj) \ge \max \left( \frac{2}{d_{mi}}, \frac{2}{d_{mj}} \right). \] Proof. We first prove the former line. Consider the altitude from $O_i$ to $O_m O_j$. The altitude must have length at least $2$, otherwise its perpendicular bisector passes intersects all of $C_i$ , $C_m$, $C_j$. Thus \[ 2 \le d_{mi} \sin \angle(imj) \le \angle(imj) \]proving the first line. The second line follows by considering the external angle formed by lines $O_m O_i$ and $O_m O_j$ instead of the internal one. $\blacksquare$ Our idea now is for any index $m$ we will make an estimate on $\sum_{\substack{1 \le i \le n \\ i \ne b}} \frac{1}{d_{bi}}$ for each index $b$. If the centers formed a convex polygon, this would be much simpler, but because we do not have this assumption some more care is needed. Claim: Suppose $O_a$, $O_b$, $O_c$ are consecutive vertices of the convex hull. Then \[ \frac{n-1}{n-2} \measuredangle(abc) \ge \frac{2}{d_{1b}} + \frac{2}{d_{2b}} + \dots + \frac{2}{d_{nb}} \]where the term $\frac{2}{d_{bb}}$ does not appear (obviously). Proof. WLOG let's suppose $(a,b,c) = (2,1,n)$ and that rotating ray $O_2 O_1$ hits $O_3$, $O_4$, \dots, $O_n$ in that order. We have \begin{align*} \frac{2}{d_{12}} &\le \angle(213) \\ \frac{2}{d_{13}} &\le \min \left\{ \angle(213), \angle(314) \right\} \\ \frac{2}{d_{14}} &\le \min \left\{ \angle(314), \angle(415) \right\} \\ &\vdots \\ \frac{2}{d_{1(n-1)}} &\le \min \left\{ \angle((n-2)1(n-1)), \angle((n-1)1n) \right\} \\ \frac{2}{d_{1n}} &\le \angle\left( (n-1) 1 n \right). \end{align*}Of the $n-1$ distinct angles appearing on the right-hand side, we let $\kappa$ denote the smallest of them. We have $\kappa \le \frac{1}{n-2} \angle(21n)$ by pigeonhole principle. Then we pick the minimums on the right-hand side in the unique way such that summing gives \begin{align*} \sum_{i=2}^n \frac{2}{d_{1i}} &\ge \left( \angle(213)+\angle(314)+\dots+\angle( (n-1)1n ) \right) + \kappa \\ &\ge \angle(21n) + \frac{1}{n-2} \angle(21n) = \frac{n-1}{n-2} \angle(21n) \end{align*}as desired. $\blacksquare$ Next we show a bound that works for any center, even if it does not lie on the convex hull $\mathcal H$. Claim: For any index $b$ we have \[ \frac{n-1}{n-2} \pi \ge \frac{2}{d_{1b}} + \frac{2}{d_{2b}} + \dots + \frac{2}{d_{nb}} \]where the term $\frac{2}{d_{bb}}$ does not appear (obviously). Proof. This is the same argument as in the previous proof, with the modification that because $O_b$ could lie inside the convex hull now, our rotation argument should use lines instead of rays (in order for the angle to be $\pi$ rather than $2\pi$). This is why the first lemma is stated with two cases; we need it now. Again WLOG $b=1$. Consider line $O_{1} O_2$ (rather than just the ray) and imagine rotating it counterclockwise through $O_2$; suppose that this line passes through $O_3$, $O_4$, \dots, $O_{n}$ in that order before returning to $O_{2}$ again. We let $\measuredangle (i1j) \in \{ \angle (i1j), \pi-\angle(i1j) \} \in [0, \pi)$ be the counterclockwise rotations obtained in this way, so that \[ \measuredangle(21n) = \measuredangle(213) + \measuredangle(314) + + \dots + \measuredangle((n-1)1n). \](This is not ``directed angles'', but related.) Then we get bounds \begin{align*} \frac{2}{d_{12}} &\le \measuredangle(213) \\ \frac{2}{d_{13}} &\le \min \left\{ \measuredangle(213), \measuredangle(314) \right\} \\ &\vdots \\ \frac{2}{d_{1(n-1)}} &\le \min \left\{ \measuredangle((n-2)1(n-1)), \measuredangle((n-1)1n) \right\} \\ \frac{2}{d_{1n}} &\le \measuredangle\left\{ (n-1) 1 n \right\} \end{align*}as in the last proof, and so as before we get \[ \sum_{i=1}^n \frac{2}{d_{1i}} \le \frac{n-1}{n-2} \measuredangle(21n) \]which is certainly less than $\frac{n-1}{n-2} \pi$. $\blacksquare$ Now suppose there were $r$ vertices in the convex hull. If we sum the first claim across all $b$ on the hull, and the second across all $b$ not on the hull (inside it), we get \begin{align*} \sum_{1 \le i<j \le n} \frac{2}{d_{ij}} &= \frac{1}{2} \sum_b \sum_{i \ne b} \frac{2}{d_{bi}} \\ &\le \frac{1}{2} \cdot \frac{n-1}{n-2} \left( (r-2)\pi + (n-r)\pi \right) \\ &= \frac{(n-1)\pi}{4} \end{align*}as needed (with $(r-2)\pi$ being the sum of angles in the hull).

Draw two parallel external tangents between every pair of unit circles, call them "tubes" joining pairs of circles. We will only use the condition that no tube intersects another circle. Consider any circle $C_1$ and the others $C_2,...,C_n$. Let $d_i = O_1O_i (2\le i \le n)$ and we will prove that $\sum \frac 1{d_i}$ is small enough. Consider angles between tubes mod $\pi$ and denote these by $\theta_1,...,\theta_{n-1}$. For now we will erase all tubes between $O_iO_j (2\le i , j \le n)$ and only consider the case where tubes $O_1O_i$ are present. Now the configuration satisfies no tube intersecting another circle iff when one of the circles is flipped over $O_1$ the configuration still has this property. Hence we can flip all of them to get them onto the same side. There exists $\theta_i$ among all the $\theta$'s having radian measure $\ge \frac{\pi}{n-1}$ by Pigeonhole so with a suitable flip (we keep all circles between tubes with "acute" angle $\theta_i$ between them in the "obtuse" region) all the angles will have sums $\le \frac{n-2}{n-1}\pi$. Note that $d_j/2 \le \frac{1}{\sin \theta_j}$ and use Jensen with $\sin \theta < \theta, \theta = \sum_{j\ne i} \theta_j \frac{1}{n-1}$ to finish.

For convenience, we denote $\angle O_iO_jO_k$ by $\angle(ijk)$. We begin with the following lemma, which allows us to capture the mysterious presence of $\pi$ and the $\tfrac{1}{O_iO_j}$. Lemma: For any distinct indices $i,j$, if $\theta_{ij}$ is the angle between the two common internal tangents of $C_i, C_j$, we have $$\frac{1}{O_iO_j} \leq \frac{\theta_{ij}}{4}$$Proof: Obvious, (recall that $\sin\theta \leq \theta$) $$\frac{1}{O_iO_j} = \frac{\sin\frac{\theta_{ij}}{2}}{2} \leq \frac{\theta_{ij}}{4} \quad\blacksquare.$$ The observation that we will repeatedly use afterward is that if we draw two nonparallel external tangents from either $O_i$ or $O_j$ to $n-2$ other circles, then, up to a configuration issues, the angle between these two lines are larger or equal to $\theta_{ij}$. The configuration issues will be clear in the context. Now we prove the following two main claims. Claim 1: $\theta_{n1} + \theta_{n2} + \hdots + \theta_{n(n-1)} \leq 2\pi$. Proof: Refer to the figure below. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); real x = 30; real y = 20; pair O = (0,0); pair A = 6*dir(x); pair B = (7,0); pair C = 5*dir(180-y); filldraw(arc(O,1.5,-y,x)--O--cycle,lightgreen,green); draw(circle(O,1)); draw(circle(A,1),dashed); draw(circle(B,1)); draw(circle(C,1),dashed); draw(shift(dir(90-y))*(C--8*dir(-y)),red+linewidth(0.5)); draw(shift(dir(x-90))*(O--7*dir(x)),red+linewidth(0.5)); draw(O--1.2*A^^O--1.2*B^^O--(-1.6*C),linewidth(1)); draw(O--1.3*C); dot("$O_n$",O,S); dot("$O_1$",A,dir(120)); dot("$O_2$",B,N); dot("$O_3$",C,N); label("$\alpha_1$",2*dir(x/2)); label("$\alpha_2$",2*dir(-y/2)); [/asy][/asy] Suppose that we draw a line $\ell_i = O_nO_i$ for any $i=1,\hdots,n-1$ and read through these lines in clockwise order. WLOG, we may assume that we get $O_nO_1, O_nO_2,\hdots,O_nO_{n-1}$ in this order. Let $\alpha_i$ denote the angle between $\ell_i$ and $\ell_{i+1}$ that does not contain any other lines (indices modulo $n-1$). Then, from the observation above, we get \begin{align*} \theta_{n1} &\leq \alpha_{n-1} + \alpha_1\\ \theta_{n2} &\leq \alpha_1 + \alpha_2 \\ \theta_{n3} &\leq \alpha_2 + \alpha_3 \\ &\vdots \\ \theta_{n(n-1)} & \leq \alpha_{n-2} + \alpha_{n-1}. \end{align*}Adding up all these inequalities gives the result. $\blacksquare$ Unfortunately, summing up similar inequalities gives the bound of $\tfrac{n\pi}{4}$. So we need to strengthen the bound. We will be more precise when $O_i$ is on the convex hull. WLOG, let the convex hull of $O_1,O_2,\hdots,O_n$ be the polygon $O_1O_2\hdots O_m$. Claim 2: $\theta_{12} + \theta_{23} + \hdots + \theta_{(m-1)m} + \theta_{m1} \leq 2\pi$. Proof: Refer to the figure below (in the case $m=6$). [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = (2.3,2.4); pair B = (-1,2.75); pair C = (-3,0.5); pair D = (-1.7,-2.2); pair E = (3,-3.1); pair F = (3.8,0.3); real r = 0.55; markscalefactor=0.1; filldraw(anglemark(F,B,A),palered,red); filldraw(anglemark(B,A,C),palered,red); markscalefactor=0.05; filldraw(anglemark(C,B,D)^^anglemark(E,A,F) ^^anglemark(A,C,B)^^anglemark(B,D,C)^^anglemark(D,C,E) ^^anglemark(E,D,F)^^anglemark(C,E,D)^^anglemark(F,E,A) ^^anglemark(D,F,E)^^anglemark(A,F,B),lightgreen,green); draw(circle(A,r)^^circle(B,r)^^circle(C,r)^^circle(D,r)^^circle(E,r)^^circle(F,r),linewidth(1)); draw(shift(rotate(90,0.0)*(r*(A-C)/abs(A-C)))*(A--C), red+linewidth(0.5)); draw(shift(rotate(-90,0.0)*(r*(B-F)/abs(B-F)))*(B--F), red+linewidth(0.5)); draw(A--B--C--D--E--F--cycle,linewidth(1.2)); draw(A--C--E--cycle^^B--F--D--cycle,dashed); dot("$O_1$",A,dir(60)); dot("$O_2$",B,dir(110)); dot("$O_3$",C,W); dot("$O_4$",D,0.6*SW); dot("$O_5$",E,0.8*S); dot("$O_6$",F,dir(0)); [/asy][/asy] From the key observation, we have \begin{align*} \theta_{12} &\leq \angle(m21) + \angle(312) \\ \theta_{23} &\leq \angle(132) + \angle(423) \\ \theta_{34} &\leq \angle(243) + \angle(534) \\ &\vdots \\ \theta_{(m-1)m} &\leq \angle((m-2)m(m-1)) + \angle(1(m-1)m)\\ \theta_{m1} &\leq \angle((m-1)1m) + \angle(2m1) \end{align*}One can easily show that the sum of RHS is $2\pi$, so adding up all these inequalities gives the result. $\blacksquare$ In the Claim 1's bound, we have bounded $\theta_{12}$ to \begin{align*} \theta_{12} &\leq (\pi - \angle(21m)) + \angle(213) \\ \theta_{21} &\leq (\pi - \angle(123)) + \angle(m21) \end{align*}Therefore, the correction term (of $\sum\limits_{1\leq i<j\leq n}\theta_{ij}$) is (important: indices here are in modulo $m$) \begin{align*}&\quad 2\pi - \sum_{\text{cyc}} \frac{\angle(213) + \angle(m21) + (\pi - \angle(213)) + (\pi - \angle(m21))}{2} \\ &= 2\pi - \frac 12\sum_{\text{cyc}} (\angle(213) + \angle(m21)) - \sum_{cyc} (\pi - \angle(123) \\ &= 2\pi - \pi - 2\pi \\ &= -\pi \end{align*}Therefore, instead of having $\sum \theta_{ij} \leq n\pi$, we will have $\sum \theta_{ij} \leq (n-1)\pi$. This gives the desired bound.

Suppose there exist three points $O_i$, $O_j$, and $O_k$ such that $d(O_i,O_jO_k)\le 2$. Then, draw the $O_i$-midline of $\triangle O_iO_jO_k$ and we will find that since the centers of $C_i$, $C_j$, and $C_k$ are distance at most $1$ from this line, the line intersects all three circles, contradiction. Therefore, we must have \[d(O_i,O_jO_k)>2\]for all $1\le i,j,k\le n$. $~$ Now, note that $\sin(x)<x$ for all $x$. We have for all $O_i,O_j,O_k$ that $O_iO_j\sin(angle O_iO_jO_k)=d(O_i,O_jO_k)>2$ and this implies \[\frac{1}{O_iO_j}<\frac{\sin(\angle O_iO_jO_k)}{2}<\frac{\angle O_iO_jO_k}{2}\]where the angle is in radians and less than $\tfrac{\pi}{2}$ (If it is greater, simply subtract it from $\pi$). Define \[s(O_j)=\sum_{\stackrel{1\le i\le n}{i\neq j}}{\frac{1}{O_iO_j}}\]It is easy to see that the LHS of the desired inequality is $\tfrac12 (s(O_1)+s(O_2)+\dots+s(O_n))$. $~$ Fix $j$. Note that if we reflect some points across $O_j$, the set of lines $O_iO_j$ stay the same, therefore $d(O_i,O_jO_k)>2$ for all $i,k\neq j$ and so the bound on $O_iO_j$ still holds, and the distance $O_iO_j$ also doesn't change. Hence we can reflect points across $O_j$ until all points are on the same half-plane defined by a line going through $O_j$. At this point, let the rest of the points with $P_1,P_2,\dots,P_{n-1}$ where $P_1,P_2,\dots,P_{n-1}$ are arranged in angular order with respect to $O_j$. Now note that \[s(O_j)=\sum_{i=1}^{n-1}{\frac{1}{O_jP_i}}<\sum_{i=1}^{n-1}{\frac{\measuredangle P_iO_jP_{i+1}}{2}}=\frac{\pi}{2}\]where the angles are directed and $P_n=P_1$. $~$ If $O_{i_1},O_{i_2},\dots,O_{i_k}$ are the convex hull in that order, then we do not do the step of reflections, and we have \[s(O_{i_j})=\sum_{i=1}^{n-1}{\frac{1}{O_{i_j}P_i}}<\sum_{i=1}^{n-1}{\frac{\measuredangle P_iO_{i_j}P_{i+1}}{2}}=\frac{\angle O_{i_{j-1}}O_{i_j}{O_{i_{j+1}}}}{2}\cdot \frac{n-1}{n-2}\]where we replace $\measuredangle P_{n-1}O_{i_j}P_{1}$ with the smallest of the other angles. $~$ Adding everything up, we have that \begin{align*} \frac{1}{2}\sum_{i=1}^{n}{O_j} &= \frac{n-1}{n-2}\sum_{j=1}^{k}{\frac{\angle O_{i_{j-1}}O_{i_j}{O_{i_{j+1}}}}{4}}+(n-k)\frac{\pi}{4} \\ &= \frac{n-1}{n-2} (k-2) \frac{\pi}{4} + (n-k) \frac{\pi}{4} \\ &\le (k-1) \frac{\pi}{4} + (n-k) \frac{\pi}{4} \\ &= \frac{(n-1)\pi}{4} \end{align*}where we used the fact that the sum of the angles in a $k$-gon are $(k-2)\pi$ and that $k<n$ implies $\tfrac{n-1}{n-2}<\tfrac{k-1}{k-2}$ and we're done.