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Difference between revisions of "2011 USAJMO Problems/Problem 4"

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== Solution ==
 
== Solution ==
  
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Let <math>r</math> be the reflection function on the set of words, namely <math>r(a_1\dots a_n) = a_n \dots a_1</math> for all words <math>a_1 \dots \a_n</math>, <math>n\ge 1</math>. Then the following property is evident (e.g. by mathematical induction):
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Let <math>r</math> be the reflection function on the set of words, namely <math>r(a_1\dots a_n) = a_n \dots a_1</math> for all words <math>a_1 \dots a_n</math>, <math>n\ge 1</math>. Then the following property is evident (e.g. by mathematical induction):
  
 
<math> r(w_1 \dots w_k) = r(w_k) \dots r(w_1)</math>, for any words <math>w_1, \dots, w_k</math>, <math>k \ge 1</math>.
 
<math> r(w_1 \dots w_k) = r(w_k) \dots r(w_1)</math>, for any words <math>w_1, \dots, w_k</math>, <math>k \ge 1</math>.
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a, b, ab, bab,
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We use mathematical induction to prove the statement of the problem. First, <math>W_1 = b</math>, <math>W_1W_2 = bab</math>, <math>W_1W_2W_3 = babbab</math>  are palindromes. Second, suppose <math>n\ge 3</math>, and that the words <math>W_1 W_2 \dots W_k</math>  (<math>k = 1</math>, <math>2</math>, <math>\dots</math>, <math>n</math>) are all palindromes, i.e. <math>r(W_1W_2\dots W_k) = W_1W_2\dots W_k</math>. Now, consider the word <math>W_1 W_2 \dots W_{n+1}</math>:
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We use mathematical induction to prove the statement of the problem. First, <math>W_1 = b</math>, <math>W_1W_2 = bab</math>, <math>W_1W_2W_3W_3 = babbab</math>  are palindromes. Second, suppose <math>n\ge 3</math>, and that the words <math>W_1 W_2 \dots W_k</math>  (<math>k = 1</math>, <math>2</math>, <math>\dots</math>, <math>n</math>) are all palindromes, i.e. <math>r(W_1W_2\dots W_k) = W_1W_2\dots W_k</math>. Now, consider the word <math>W_1 W_2 \dots W_{n+1}</math>:
 
  
 
<cmath>r(W_1 W_2 \dots W_{n+1}) = r(W_{n+1}) r(W_1 W_2 \dots W_{n-2}W_{n-1}W_n)</cmath>
 
<cmath>r(W_1 W_2 \dots W_{n+1}) = r(W_{n+1}) r(W_1 W_2 \dots W_{n-2}W_{n-1}W_n)</cmath>
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By the principle of mathematical induction, the statement of the problem is proved. [[User:Lightest|Lightest]] 21:54, 1 April 2012 (EDT)
 
By the principle of mathematical induction, the statement of the problem is proved. [[User:Lightest|Lightest]] 21:54, 1 April 2012 (EDT)
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{{MAA Notice}}
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== Solution 2==

Latest revision as of 20:56, 13 February 2024

Problem

A word is defined as any finite string of letters. A word is a palindrome if it reads the same backwards as forwards. Let a sequence of words $W_0$, $W_1$, $W_2$, $\dots$ be defined as follows: $W_0 = a$, $W_1 = b$, and for $n \ge 2$, $W_n$ is the word formed by writing $W_{n - 2}$ followed by $W_{n - 1}$. Prove that for any $n \ge 1$, the word formed by writing $W_1$, $W_2$, $\dots$, $W_n$ in succession is a palindrome.

Solution

Let $r$ be the reflection function on the set of words, namely $r(a_1\dots a_n) = a_n \dots a_1$ for all words $a_1 \dots a_n$, $n\ge 1$. Then the following property is evident (e.g. by mathematical induction):

$r(w_1 \dots w_k) = r(w_k) \dots r(w_1)$, for any words $w_1, \dots, w_k$, $k \ge 1$.

We use mathematical induction to prove the statement of the problem. First, $W_1 = b$, $W_1W_2 = bab$, $W_1W_2W_3 = babbab$ are palindromes. Second, suppose $n\ge 3$, and that the words $W_1 W_2 \dots W_k$ ($k = 1$, $2$, $\dots$, $n$) are all palindromes, i.e. $r(W_1W_2\dots W_k) = W_1W_2\dots W_k$. Now, consider the word $W_1 W_2 \dots W_{n+1}$:

\[r(W_1 W_2 \dots W_{n+1}) = r(W_{n+1}) r(W_1 W_2 \dots W_{n-2}W_{n-1}W_n)\]

\[= r(W_{n-1}W_n) W_1 W_2 \dots W_{n-2} W_{n-1} W_n\]

\[= r(W_{n-1}W_n) r(W_1 W_2 \dots W_{n-2}) W_{n+1}\]

\[= r(W_1 W_2 \dots W_{n-2} W_{n-1}W_n) W_{n+1}\]

\[= W_1W_2\dots W_n W_{n+1}.\]

By the principle of mathematical induction, the statement of the problem is proved. Lightest 21:54, 1 April 2012 (EDT) The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Solution 2

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