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

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A word is defined as a finite sequence of letters.  A ''palindrome'' is a word that reads the same forwards and backwards.  Consider the sequence of words W0,W1,W2... such that W0 = a, W1 = b, and for all n >= 2, Wn is the word formed by writing W(n-2) followed by W(n-1).  Prove that for all n >= 1, the word formed by writing W1,W2,...,Wn in sequence is a palindrome.
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== Problem ==
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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 <math>W_0</math>, <math>W_1</math>, <math>W_2</math>, <math>\dots</math> be defined as follows: <math>W_0 = a</math>, <math>W_1 = b</math>, and for <math>n \ge 2</math>, <math>W_n</math> is the word formed by writing <math>W_{n - 2}</math> followed by <math>W_{n - 1}</math>.  Prove that for any <math>n \ge 1</math>, the word formed by writing <math>W_1</math>, <math>W_2</math>, <math>\dots</math>, <math>W_n</math> in succession is a palindrome.
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== 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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<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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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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<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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<cmath>= r(W_{n-1}W_n) W_1 W_2 \dots W_{n-2} W_{n-1} W_n</cmath>
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<cmath>= r(W_{n-1}W_n) r(W_1 W_2 \dots W_{n-2}) W_{n+1}</cmath>
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<cmath>= r(W_1 W_2 \dots W_{n-2} W_{n-1}W_n) W_{n+1}</cmath>
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<cmath>= W_1W_2\dots W_n W_{n+1}.</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)
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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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