Difference between revisions of "2009 OIM Problems/Problem 1"

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== Solution ==
 
== Solution ==
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WLOG first travel from <math>x_1</math> to <math>x_2</math>. If we reverse back to <math>x_1</math> immediately, then we must continue in that direction until <math>x_2</math> (since otherwise there is no way to return to those bridges). This pattern continues for all <math>x_i</math>.
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Notice that at every pair of bridges, we can always choose to take one or the other, and the remaining bridge will be forced upon us; thus there are <math>2^n</math> ways to choose between each pair of bridges. Next, we can reverse at any <math>x_i</math> for <math>i>1</math>, and if we continue all the way to <math>x_1</math> clockwise, we can either continue forwards or reverse. Thus there are <math>n+1</math> cases if we initially head clockwise. This number is doubled since we can initially travel from <math>x_1</math> to <math>x_n</math> instead, so in total we have <math>2\cdot (n+1)\cdot2^n=\boxed{2^{n+1}(n+1)}</math> choices.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
[[OIM Problems and Solutions]]
 
[[OIM Problems and Solutions]]

Latest revision as of 11:59, 20 March 2025

Problem

Let $n$ be a natural number greater than 2. Suppose that $n$ islands are located in a circle and that between each two neighboring islands there are two bridges, with the islands $x_1, x_2, \cdots , x_n$ in order of the clock hands. Starting on island $x_1$, in how many ways can the $2n$ bridges be crossed by passing over each bridge exactly once?

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

WLOG first travel from $x_1$ to $x_2$. If we reverse back to $x_1$ immediately, then we must continue in that direction until $x_2$ (since otherwise there is no way to return to those bridges). This pattern continues for all $x_i$.

Notice that at every pair of bridges, we can always choose to take one or the other, and the remaining bridge will be forced upon us; thus there are $2^n$ ways to choose between each pair of bridges. Next, we can reverse at any $x_i$ for $i>1$, and if we continue all the way to $x_1$ clockwise, we can either continue forwards or reverse. Thus there are $n+1$ cases if we initially head clockwise. This number is doubled since we can initially travel from $x_1$ to $x_n$ instead, so in total we have $2\cdot (n+1)\cdot2^n=\boxed{2^{n+1}(n+1)}$ choices.

~ eevee9406

See also

OIM Problems and Solutions