Difference between revisions of "1974 USAMO Problems/Problem 1"
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<cmath> 0 = \lvert (a-b) + (b-c) + (c-a) \rvert = \lvert 2(a-b) + (c-a) \rvert \ge 2 \lvert a-b \rvert - \lvert c-a \rvert = \lvert a-b \rvert , </cmath> | <cmath> 0 = \lvert (a-b) + (b-c) + (c-a) \rvert = \lvert 2(a-b) + (c-a) \rvert \ge 2 \lvert a-b \rvert - \lvert c-a \rvert = \lvert a-b \rvert , </cmath> | ||
so <math>a=b= P(a)</math>, and <math>c= P(b) = P(a) = b</math>, so <math>a</math>, <math>b</math>, and <math>c</math> are equal, as desired. <math>\blacksquare</math> | so <math>a=b= P(a)</math>, and <math>c= P(b) = P(a) = b</math>, so <math>a</math>, <math>b</math>, and <math>c</math> are equal, as desired. <math>\blacksquare</math> | ||
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Revision as of 15:08, 17 September 2012
Problem
Let , , and denote three distinct integers, and let denote a polynomial having all integral coefficients. Show that it is impossible that , , and .
Solution
It suffices to show that if are integers such that , , and , then .
We note that so the quanitities must be equal in absolute value. In fact, two of them, say and , must be equal. Then so , and , so , , and are equal, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1974 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |