The 16th Japanese Mathematical Olympiad - The First Round

Take an arbitrary route. For each face, count the number of the edges on the face which are on the route. Since they can be neither less than $3$ nor more than $4$, and they add up to $2\times 20=40$, there exists exactly $4$ faces with $4$ of their edge on the route.

Now, assume that the ant passed consecutive four edges of a face as below.

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Since the ant did not pass edge $AB$, the ant passed the edges $BC$ and $AK$. Also, since all the vertices must be passed just once, looking at vertices $E$, $G$ and $I$, it follows that the route contains edges $DE$, $EF$, $FG$, $GH$, $HI$ and $IJ$. By the same reason, looking at vertices $O$ and $P$, it follows that it contains $NO$, $OP$ and $PQ$. There are two ways to complete the route: one is to take the edges $KL$, $LM$, $MN$, $CD$ and $JQ$; the other is to take $CL$, $LM$, $MQ$, $KJ$ and $DN$.

Now we can compute how many ways are there to make a route. There are $12$ ways to choose the first face, $5$ ways to choose the lacked edge, $2$ ways to complete the route, and $2$ ways to decide the direction. Since each route appears exactly $4$ times (note that there are $4$ faces of which $4$ edges are contained in a route), there are $12\times 5\times 2\times 2÷4=60$ routes.