2000 USAMO Problems/Problem 1
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,
n.
n,
n
n > \frac{A+B}{4} - 1.
A
B
f.$ It follows that:
However, by the very convex condition,
This is a contradiction. It follows that there exists no very convex function.
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 |
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