We are given a regular icosahedron having three red vertices. To show that it has a vertex that has at least two red neighbours we can use an argument by contradiction. We suppose that no vertex has more than one red neighbour and reach a contradiction thus showing that this statement must be false.

If no vertex has more than one red neighbour then each red vertex has to be surrounded by 5 vertices that are not red otherwise one of those vertices would have two red neighbours. We know that an icosahedron has 12 vertices, 20 faces and 30 edges and we can argue that this leads to a contradiction.

Solution to Proximity by Tony Cardell and John Lesieutre, State College Area High School, PA, USA, both ages 14.

Following the hint: We give a proof by contradiction. Suppose no vertex has more than one red neighbor. If you look at an cosahedron, you'll notice that there are five neighbors to every vertex. Five non-red vertices must then surround every red vertex (otherwise a vertex would have two red neighbors). However, we have three red vertices, each requiring five non-red vertices, so we need a total of fifteen non-red vertices. The fact that an icosahedron has 12 vertices renders this an impossibility, therefore there must be at least one vertex with at least two red neighbors.