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2023 USAMO Problems/Problem 2: Difference between revisions

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== Solution 1 ==
== Solution 1 ==
First, let us plug in some special points; specifically, plugging in <math>x=0</math> and <math>x=1</math>, respectively:
<cmath>
\begin{align}
    f(f(0)) &= 2 \\
    f(y + f(1)) &= f(y) + 2
\end{align}
</cmath>
Next, let us find the derivative of this function. First, with (2), we isolate <math>f(y)</math> one one side
<cmath>
\begin{align*}
  f(y) = f(y + f(1)) - 2
\end{align*}
</cmath>
and then take the derivative:
<cmath>
\begin{align*}
    \dfrac{\mathrm{d}f}{\mathrm{d}y}
    &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1)) - 2\right] \\
    &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1))\right] - \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[2\right] \\
    &= f'(y + f(1))\cdot\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\
    &= f'(y + f(1))\cdot(1)\\
    f'(y) &= f'(y + f(1))\\
\end{align*}
</cmath>
With the derivative, we see that the input to the function does not matter: it will return the same result regardless of input, assuming that <math>f(1) \neq 0</math>. We know it is not zero because if it was, then (2) would become <math>f(y) = f(y) + 2</math>, implying that <math>0 = 2</math>, which is not true.
Therefore, the function <math>f'</math> must be a constant, and <math>f</math> must be a linear equation or a constant. We know it is not a constant because if it was, the problem could be reduced to the following:
<cmath>
\begin{align*}
    f(xy + f(x)) &= xf(y) + 2 \\
    f &= xf + 2 \\
    f - xf &= 2 \\
    f(1-x) &= 2 \\
    f &= \dfrac{2}{1-x} \\
\end{align*}
</cmath>
where <math>f</math> is the constant from <math>f(x)</math>. As we see, <math>f</math> would depend on <math>x</math>, making it not a constant function. Thus, <math>f(x)</math> must be linear, meaning we can model it like so:
<cmath>
\begin{align*}
    f(x) = ax + b
\end{align*}
</cmath>
Via (1), we get the following:
<cmath>
\begin{align*}
    f(f(0)) &= 2 \\
    a(a(0) + b) + b &= 2 \\
    ab + b &= 2
\end{align*}
</cmath>
And via (2),
<cmath>
\begin{align*}
    f(y + f(1)) &= f(y) + 2 \\
    a(y + a(1) + b) + b &= ay + b + 2 \\
    ay + a^2 + ab + b &= ay + b + 2 \\
    a^2 + ab &= 2 \\
\end{align*}
</cmath>
Setting these equations equal to each other,
<cmath>
\begin{align*}
    ab + b &= a^2 + ab \\
    b &= a^2 \\
\end{align*}
</cmath>
Therefore,
<cmath>
\begin{align*}
    ab + b &= 2 \\
    a^3 + a^2 &= 2 \\
\end{align*}
</cmath>
There are three solutions to this equation: <math>a = 1</math>, <math>a = -1 + i</math>, and <math>a = -1 - i</math>. Knowing that <math>b = a^2</math>, the respective <math>b</math> values are <math>b = 1</math>, <math>b = -2i</math>, and <math>b = 2i</math>. Thus, <math>f(x)</math> could be the following:
<cmath>
\begin{align*}
    f(x) &= x + 1 \\
    f(x) &= x(-1 + i) - 2i \\
    f(x) &= x(-1 - i) + 2i \\
\end{align*}
</cmath>
Because only the first function maps strictly to positive real numbers, the only solution that works is <math>f(x) = x + 1</math>. <math>\square</math>
~cogsandsquigs
This solution is heavily flawed:
1. <math>f(0)</math> doesn't exist.
2. The function isn't necessarily continuous, let alone differentiable.
3. Even if it was differentiable, the equation 
<cmath>
\begin{align*}
f'(y) &= f'(y + f(1))\\
\end{align*}
</cmath>
doesn't necessarily mean the derivative is constant either. It only implies the derivative is periodic with period <math>f(1)</math>
spencerD.
== Solution 2 ==


Make the following substitutions to the equation:
Make the following substitutions to the equation:
Line 132: Line 10:
2. <math>(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4</math>
2. <math>(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4</math>


3. <math>(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf(1 + \frac{f(1)}{x}) + 2</math>
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>. Solving for the coefficients (in the same way as solution 1), we find that <math>f(x) = x + 1 \forall x > 1</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>.


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.
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
~jkmmm3
==See Also==
{{USAMO newbox|year=2023|num-b=1|num-a=3}}
{{MAA Notice}}

Latest revision as of 12:25, 1 January 2024

Problem 2

Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ such that, for all $x, y \in \mathbb{R}^{+}$,\[f(xy + f(x)) = xf(y) + 2\]

Solution 1

Make the following substitutions to the equation:

1. $(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2$

2. $(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4$

3. $(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf\biggl(1 + \frac{f(1)}{x}\biggr) + 2$

It then follows from (2) and (3) that $f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}$, so we know that this function is linear for $x > 1$. Substitute $f(x) = ax+b$ and solve for $a$ and $b$ in the functional equation; we find that $f(x) = x + 1 \forall x > 1$.

Now, we can let $x > 1$ and $y \le 1$. Since $f(x) = x + 1$, $xy + f(x) > x > 1$, so $f(xy + f(x)) = xy + x + 2 = xf(y) + 2$. It becomes clear then that $f(y) = y + 1$ as well, so $f(x) = x + 1$ 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

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