Difference between revisions of "2000 USAMO Problems/Problem 1"
Lopkiloinm (talk | contribs) (→Solution 3) |
Lopkiloinm (talk | contribs) (→Solution 3) |
||
Line 35: | Line 35: | ||
=== Solution 3 === | === Solution 3 === | ||
− | + | We consider the function | |
− | <cmath> | + | <cmath> |
+ | f(x, y) = \frac{x + y}{2} + |x - y|, | ||
+ | </cmath> | ||
+ | defined on <math>\mathbb{R}^2</math>. Define the graph of <math>f</math> as the manifold | ||
+ | <cmath> | ||
+ | \mathcal{M} = \{ (x, y, z) \in \mathbb{R}^3 : z = f(x, y) \}. | ||
+ | </cmath> | ||
+ | |||
+ | Claim: The manifold <math>\mathcal{M}</math> is not the graph of a function <math>\mathbb{R} \to \mathbb{R}^2</math>, i.e., it cannot be inverted to express <math>(x, y)</math> as a function of <math>z</math>. | ||
− | We | + | Proof: |
+ | We first observe that <math>f</math> is symmetric: | ||
<cmath> | <cmath> | ||
− | f(x, y) = | + | f(x, y) = f(y, x), |
+ | </cmath> | ||
+ | and convex, as it is the sum of a linear function and the convex function <math>|x - y|</math>. | ||
+ | |||
+ | Consider the behavior of <math>f</math> along the line <math>x = y</math>. For <math>x > y</math>, we have | ||
+ | <cmath> | ||
+ | f(x, y) = \frac{x + y}{2} + (x - y) = \frac{3x - y}{2}, | ||
+ | </cmath> | ||
+ | while for <math>x < y</math>, we have | ||
+ | <cmath> | ||
+ | f(x, y) = \frac{x + y}{2} + (y - x) = \frac{3y - x}{2}. | ||
+ | </cmath> | ||
+ | The partial derivatives with respect to <math>x</math> and <math>y</math> do not match along <math>x = y</math>, creating a non-differentiable crease. Specifically, | ||
+ | <cmath> | ||
+ | \nabla f(x, y) = | ||
\begin{cases} | \begin{cases} | ||
− | \frac{ | + | \left( \frac{3}{2}, -\frac{1}{2} \right), & x > y, \\ |
− | \frac{ | + | \left( -\frac{1}{2}, \frac{3}{2} \right), & x < y. |
− | \end{cases} | + | \end{cases} |
− | </cmath> | + | </cmath> |
+ | At <math>x = y</math>, the gradient does not exist; the directional derivatives have a jump discontinuity, forming a kink along the line <math>x = y</math>. | ||
+ | |||
+ | Suppose for contradiction that <math>\mathcal{M}</math> is the graph of a function <math>g: \mathbb{R} \to \mathbb{R}^2</math> such that <math>g(z) = (x, y)</math> with <math>f(x, y) = z</math>. Then each level set | ||
+ | <cmath> | ||
+ | L_z = \{ (x, y) \in \mathbb{R}^2 : f(x, y) = z \} | ||
+ | </cmath> | ||
+ | must be the image of a unique point <math>g(z) \in \mathbb{R}^2</math>. | ||
+ | |||
+ | However, since <math>f(x, y) = f(y, x)</math>, the level sets are symmetric across the line <math>x = y</math>. Furthermore, near any <math>z</math> for which the level set intersects the kink <math>x = y</math>, the gradient directions from either side differ, and the level set contains multiple points mapping to the same height <math>z</math>. Therefore, the preimage <math>f^{-1}(z)</math> contains at least two distinct points, violating the definition of a function <math>z \mapsto (x, y)</math>. | ||
+ | |||
+ | Hence, <math>\mathcal{M}</math> cannot be the graph of a function <math>\mathbb{R} \to \mathbb{R}^2</math>. The surface folds along the kink, failing the vertical line test in the direction of <math>z</math>. | ||
+ | |||
+ | <cmath> | ||
+ | \boxed{\text{Therefore, } \mathcal{M} \text{ is a non-functional manifold.}} \quad \blacksquare | ||
+ | </cmath> | ||
Revision as of 03:31, 30 June 2025
Problem
Call a real-valued function very convex if
holds for all real numbers and
. Prove that no very convex function exists.
Solution
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.
Solution 3
We consider the function
defined on
. Define the graph of
as the manifold
Claim: The manifold is not the graph of a function
, i.e., it cannot be inverted to express
as a function of
.
Proof:
We first observe that is symmetric:
and convex, as it is the sum of a linear function and the convex function
.
Consider the behavior of along the line
. For
, we have
while for
, we have
The partial derivatives with respect to
and
do not match along
, creating a non-differentiable crease. Specifically,
At
, the gradient does not exist; the directional derivatives have a jump discontinuity, forming a kink along the line
.
Suppose for contradiction that is the graph of a function
such that
with
. Then each level set
must be the image of a unique point
.
However, since , the level sets are symmetric across the line
. Furthermore, near any
for which the level set intersects the kink
, the gradient directions from either side differ, and the level set contains multiple points mapping to the same height
. Therefore, the preimage
contains at least two distinct points, violating the definition of a function
.
Hence, cannot be the graph of a function
. The surface folds along the kink, failing the vertical line test in the direction of
.
~Lopkiloinm
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.