Difference between revisions of "2000 USAMO Problems/Problem 1"
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However, by the very convex condition, | However, by the very convex condition, | ||
<cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.</cmath> | <cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.</cmath> | ||
− | This is a contradiction. It follows that | + | This is a contradiction. It follows that no very convex function exists. |
== See Also == | == See Also == |
Revision as of 00:57, 5 April 2016
Contents
Problem
Call a real-valued function very convex if
holds for all real numbers and . Prove that no very convex function exists.
Solution 1
Let , and substitute . Then a function is very convex if , or rearranging,
Let , which is the slope of the secant between . Let be arbitrarily small; then it follows that , . Summing these inequalities yields . As (but , so is still arbitrarily small), we have . This implies that in the vicinity of any , the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.
Solution 2
Suppose, for the sake of contradiction, that there exists a very convex function Notice that is convex if and only if is convex, where is a constant. Thus, we may set for convenience.
Suppose that and By the very convex condition, A straightforward induction shows that: for all nonnegative integers Using a similar line of reasoning as above, Therefore, for every nonnegative integer Now, we choose large enough such that This is always possible because and are fixed for any particular function It follows that: However, by the very convex condition, This is a contradiction. It follows that no very convex function exists.
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.