1998 All-Russian Olympiad
yeah you're right,the official problem is the following one:



there are 1998 cities in Russia, each being connected (in both directions) by flights to three other cities. any city can be reached by any other city by a sequence of flights. the KGB plans to close off 200 cities, no two joined by a single flight. show that this can be done so that any open city can be reached from any other open city by a sequence of flights only passing through open cities.





Solution.we begin with some terminology. define a trigraph to be a connected undirected graph in which every vertex has degree at most $ 3$. a trivalent vertex of such a graph is a vertex of degree $ 3$. in this wording, the problem becomes: we have a trigraph $ G$ with $ 1998$ vertices, all of which are trivalent. we want to remove $ 200$ vertices, no two of which are adjacent, such that the remaining vertices stay connected.

we remove the vertices one at a time. suppose we have deleted $ k$ of the $ 1998$ vertices, no two of which are adjacent, such the trigraph $ G'$ induced by the remaining vertices is connected. we will show that if $ K < 200$, we can always delete a trivalent vertex of $ G'$ such that the graph remains connected. this vertex cannot be adjacent in $ G$ to any of the other $ k$ deleted vertices, because then its degree in $ G'$ would be less than $ 3$. hence repeating this $ 200$ times gives us the desired set of vertices.



Lemma. let $ G$ be a trigraph such that the removal of any trivalent vertex disconnects $ G$. then $ G$ is planer. moreover $ G$ can be drawn in such a way that every vertex lies on the "outside" face; in other words, for any point $ P$ outside some bounded set, each vertex $ v$ of $ G$ can be joined to $ P$ by a curve which does not intersect any edges of $ G$ (except at $ v$).



Proof. we induct on the number of trivalent vertices of $ G$. if $ G$ does not contain any trivalent vertices, then $ G$ must be a path or a cycle and the claim is obvious. so suppose $ G$ contains $ n\geq 1$ trivalent vertices and that every trigraph with fewer trivalent vertices can be drawn as described. if $ G$ is a tree the claim is obvious, so suppose $ G$ contains a cycle; let $ v_1,\ldots ,v_k$ $ (k\geq 3)$ be a minimal cycle. let $ S = \{v_1,\ldots ,v_k\}$, and let $ T = \{i\mid v_i \textrm{ is trivalent }\}$. ($ T$ cannot be empty, because then no $ v_i$ would be connected to a vertex of degree $ 3$.) for each $ i\in T$, let $ w_i$ denote the third vertex which is adjacent to $ v_i$ (other than $ v_{i - 1}$ and $ v_{i + 1}$), and let $ S_i$ be the set of vertices in $ G$ which can be reached from $ w_i$ without passing though $ S$. (for $ i\not\in T$, let $ S_i = \emptyset$.) we claim that the sets $ S,S_1,\ldots ,S_k$ partition the vertices of $ G$. first, note that if $ v$ is a vertex of $ G$ not in $ S$ then there is a shortest path joining $ v$ to a vertex $ v_i$ of $ S$; the penultimate vertex on this path must be $ w_i$, so $ v\in S_i$. now suppose that $ v\in S_i\cap S_j$ for some $ i\neq j$; then $ v_i$ and $ v_j$ are trivalent and there exist paths $ w_i\to v$, $ w_j\to v$ which do not pass though $ S$. we will show that there exists a path from every vertex of $ G - \{v_i\}$ to $ v$ which does not pass though $ v_i$. for $ k\neq i$ there is a path $ v_k\to v_j\to w_j\to v$; if $ w\in S_k$ for $ k\neq i$, then there is a path $ w\to w_k\to v_k\to v$; if $ w\in S_i$, there is a path $ w\to w_i\to v$. since $ S\cup S_1\cup\ldots\cup S_k = G$, we have shown that the graph obtained from $ G$ by deleting $ v_i$ is connected,a contradiction, as $ v_i$ is trivalent. therefore $ S_i\cap S_j = \emptyset$ for $ i\neq j$. obviously $ S_i\cap S$ is empty for all $ i$; hence $ S,S_1,\ldots ,S_k$ partition the vertices of $ G$. let $ G',G_1,\ldots ,G_k$ be the induced subgraphs of $ S,S_1,\ldots ,S_k$ in $ G$, respectively. by construction, the only edges in $ G$ which are not in one of the graphs $ G',G_1,\ldots ,G_k$ are the edges $ v_iw_i$ for $ i\in T$. now $ G_i$ is a trigraph with fewer than $ n$ trivalent vertices, since at least one of the $ n$ trivalent vertices in $ G$ is in $ S$. hence by the inductive hypothesis, we can draw each $ G_i$ in the plane in such a way that every vertex lies on the outside face. since $ v_1,\ldots ,v_k$ was a minimal cycle, there are no "extra" edges between these vertices, so the graph $ G'$ is a $ k$-cycle. now place the vertices of $ S$ at the vertices of a small regular $ k$-gon far from all the graphs $ G_i$; then we can draw a curve joining each pair $ v_i,w_i$. it is east to check that this gives us a drawing of $ G$ with the desired properties. $ \square$



now suppose we have removed $ k$ vertices from $ G$, no two of which are adjacent, such that the trigraph $ G'$ induced by the remaining vertices is connected, and suppose that removing any trivalent vertex of $ G'$ disconnects the graph; we must show $ k\geq 200$. by the lemma, $ G'$ is planner. we will call a dace other than the outside one a "proper face". let $ F$ be the number of proper faces of $ G'$; since $ G'$ has $ 1998 - k$ vertices and $ 2997 - 3k$ edges,



$ F\geq 1 - (1998 - k) + (2997 - 3k) = 1000 - 2k$.



we now show that no two proper faces can share a vertex. observe that each vertex belongs to at most as many faces as its degree; thus vertices of degree $ 1$ lie only on the outside face. no two proper faces can intersect in a vertex of degree $ 2$, or that vertex would not lie on the outside face, contracting the lemma. if two proper daces intersected in a trivalent vertex $ v$, each face would give a path between two of $ v$'s neighbors, so removing $ v$ would not disconnect the graph, by an argument similar to that of the lemma.

since each proper face contains at least $ 3$ vertices and no two share a vertex, we have $ 3F\leq 1998 - k$. combining this with the previous inequality gives



$ 3000 - 6k\leq 3F\leq 1998 - k$



so $ 1002\leq 5k$ and $ k\geq 200$, as desired.