Difference between revisions of "2023 USAMO Problems/Problem 2"
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Let <math>\mathbb{R}^{+}</math> be the set of positive real numbers. Find all functions <math>f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}</math> such that, for all <math>x, y \in \mathbb{R}^{+}</math>,<cmath>f(xy + f(x)) = xf(y) + 2</cmath> | Let <math>\mathbb{R}^{+}</math> be the set of positive real numbers. Find all functions <math>f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}</math> such that, for all <math>x, y \in \mathbb{R}^{+}</math>,<cmath>f(xy + f(x)) = xf(y) + 2</cmath> | ||
− | == Solution == | + | == Solution 1 == |
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− | + | Make the following substitutions to the equation: | |
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− | + | 1. <math>(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2</math> | |
− | < | + | 2. <math>(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4</math> |
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− | + | 3. <math>(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf\biggl(1 + \frac{f(1)}{x}\biggr) + 2</math> | |
− | < | + | It then follows from (2) and (3) that <math>f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}</math>, so we know that this function is linear for <math>x > 1</math>. Substitute <math>f(x) = ax+b</math> and solve for <math>a</math> and <math>b</math> in the functional equation; we find that <math>f(x) = x + 1 \forall x > 1</math>. |
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− | + | Now, we can let <math>x > 1</math> and <math>y \le 1</math>. Since <math>f(x) = x + 1</math>, <math>xy + f(x) > x > 1</math>, so <math>f(xy + f(x)) = xy + x + 2 = xf(y) + 2</math>. It becomes clear then that <math>f(y) = y + 1</math> as well, so <math>f(x) = x + 1</math> is the only solution to the functional equation. | |
− | + | ~jkmmm3 | |
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− | + | ==See Also== | |
− | + | {{USAMO newbox|year=2023|num-b=1|num-a=3}} | |
− | + | {{MAA Notice}} | |
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Latest revision as of 12:25, 1 January 2024
Problem 2
Let be the set of positive real numbers. Find all functions such that, for all ,
Solution 1
Make the following substitutions to the equation:
1.
2.
3.
It then follows from (2) and (3) that , so we know that this function is linear for . Substitute and solve for and in the functional equation; we find that .
Now, we can let and . Since , , so . It becomes clear then that as well, so is the only solution to the functional equation.
~jkmmm3
See Also
2023 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.