A boy has a set of trains and pieces of railroad track. Each piece is a quarter of circle, and by concatenating these pieces, the boy obtained a closed railway. The railway does not intersect itself. In passing through this railway, the train sometimes goes in the clockwise direction, and sometimes in the opposite direction. Prove that the train passes an even number of times through the pieces in the clockwise direction and an even number of times in the counterclockwise direction. Also, prove that the number of pieces is divisible by $4$.

How do we deal with the fact that the rails don't intersect? I tried the following.

Let the train start at the point $(0,0)$ and assume without loss of generality that the radius of the circle from which the quarter circle is made is $1$. Also assume that the first end part of a track is at $(1,1)$. Now consider the circle with radius $1$ with center at the origin. Let the arcs be \begin{align*}&A:(1,0) \to (0,1) \\&B: (0,1) \to (-1,0) \\&C: (-1,0) \to (0,1) \\&D: (0,1) \to (1,0). \end{align*}Notice that we can go from an $A$-track to a $B,C^c$-track; a $B$-track to a $C,D^c$-track; a $C$ track to a $D,A^c$ track; a $D$ track to a $A,B^c$ track (the superscript $c$ denotes the clockwise direction).

  • $\begingroup$ number-theory is for advanced topics. $\endgroup$ – Aryabhata Jun 6 '16 at 20:37

Consider the four points on the circle centre the origin radius 1: $P(1,0),Q(0,1),R(-1,0),S(0,-1)$. A anticlockwise piece can be $A: P\to Q$, or $B:Q\to R$, or $C:R\to S$, or $D:S\to P$. A clockwise piece can be $A':Q\to P$, or $B':R\to Q$, or $C':S\to R$, or $D':P\to S$.

Pieces $C,D,A',B'$ increase the $x$-coordinate by 1 and the other four reduce it by 1. So we have $C+D+A'+B'=C'+D'+A+B$. Similarly, considering the $y$-coordinate, we have $A+D+B'+C'=A'+D'+B+C$. Subtracting, $A-C=A'-C'$ (1) and hence also $B-D=B'-D'$ (2).

Pieces $A$ and $D'$ must be followed by $B$ or $C'$, so $A+D'=B+C'$ (3). Similarly, $A+B'=C'+D$ (4). Subtracting: $D'-B'=B-D$. Comparing with (2), we see that $B=D,B'=D'$. Similarly, $A=C,A'=C'$.

So the number of anticlockwise pieces is $2A+2B$ and the number of clockwise pieces is $2A'+2B'$. Hence both are even.

We may label the initial direction of the train as 0 and the other directions in which it can be heading when entering a piece as 1,2,3,4. wlog the anticlockwise pieces increase the direction by 1 mod 4 and the clockwise pieces reduce it. Its direction after a complete circuit is $A+B+C+D-A'-B'-C'-D'\bmod 4$. But this must be the same as its initial direction so $2A+2B=2A'+2B'\bmod4$. Hence the total number of pieces which is $2A+2B+2A'+2B'$ is $4A+4B=0\bmod 4$.

  • $\begingroup$ Is this true? Let $A^c$ denote the arc $A$ in the opposite direction and similarly for $B,C,D$. Then the train goes over the same number of $A$ arcs as $A^c$ arcs; the same number of $B$ arcs as $B^c$ arcs; the same number of $C$ arcs as $C^c$ arcs; and the same number of $D$ arcs as $D^c$ arcs. $\endgroup$ – user19405892 Jun 6 '16 at 20:52
  • $\begingroup$ No. Take a simple circular track. Then all the pieces are all clockwise (or all anticlockwise). $\endgroup$ – almagest Jun 6 '16 at 20:53
  • $\begingroup$ How about this? The number of $A$ arcs is the same as the number of $C$ arcs and the number of $B$ arcs is the same as the number of $D$ arcs. $\endgroup$ – user19405892 Jun 6 '16 at 20:57
  • $\begingroup$ Yes, that is true. $\endgroup$ – almagest Jun 6 '16 at 21:01
  • $\begingroup$ How do I prove it? $\endgroup$ – user19405892 Jun 6 '16 at 21:01

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