Difference between revisions of "2011 USAJMO Problems/Problem 4"
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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>. | <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>: | 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>: | ||
Revision as of 18:19, 1 June 2020
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 , , , be defined as follows: , , and for , is the word formed by writing followed by . Prove that for any , the word formed by writing , , , in succession is a palindrome.
Solution
Let be the reflection function on the set of words, namely for all words , . Then the following property is evident (e.g. by mathematical induction):
, for any words , .
We use mathematical induction to prove the statement of the problem. First, , , are palindromes. Second, suppose , and that the words (, , , ) are all palindromes, i.e. . Now, consider the word :
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.