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Difference between revisions of "2016 USAMO Problems/Problem 4"

(Created page with "== Problem == Find all functions <math>f:\mathbb{R}\rightarrow \mathbb{R}</math> such that for all real numbers <math>x</math> and <math>y</math>, <cmath>(f(x)+xy)\cdot f(x-3y...")
 
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== Solution ==
 
== Solution ==
  
−
Step 1: Set <math>x = y = 0</math> to obtain <math>f(0) = 0.</math>
+
'''Step 1:''' Set <math>x = y = 0</math> to obtain <math>f(0) = 0.</math>
  
−
Step 2: Set <math>x = 0</math> to obtain <math>f(y)f(-y) = f(y)^2.</math>
+
'''Step 2:''' Set <math>x = 0</math> to obtain <math>f(y)f(-y) = f(y)^2.</math>
  
 
<math>\indent</math> In particular, if <math>f(y) \ne 0</math> then <math>f(y) = f(-y).</math>
 
<math>\indent</math> In particular, if <math>f(y) \ne 0</math> then <math>f(y) = f(-y).</math>
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<math>\indent</math> In addition, replacing <math>y \to -t</math>, it follows that <math>f(t) = 0 \implies f(-t) = 0</math> for all <math>t \in \mathbb{R}.</math>
 
<math>\indent</math> In addition, replacing <math>y \to -t</math>, it follows that <math>f(t) = 0 \implies f(-t) = 0</math> for all <math>t \in \mathbb{R}.</math>
  
−
Step 3: Set <math>x = 3y</math> to obtain <math>\left[f(y) + 3y^2\right]f(8y) = f(4y)^2.</math>
+
'''Step 3:''' Set <math>x = 3y</math> to obtain <math>\left[f(y) + 3y^2\right]f(8y) = f(4y)^2.</math>
  
 
<math>\indent</math> In particular, replacing <math>y \to t/8</math>, it follows that <math>f(t) = 0 \implies f(t/2) = 0</math> for all <math>t \in \mathbb{R}.</math>
 
<math>\indent</math> In particular, replacing <math>y \to t/8</math>, it follows that <math>f(t) = 0 \implies f(t/2) = 0</math> for all <math>t \in \mathbb{R}.</math>
  
−
Step 4: Set <math>y = -x</math> to obtain <math>f(4x)\left[f(x) + f(-x) - 2x^2\right] = 0.</math>
+
'''Step 4:''' Set <math>y = -x</math> to obtain <math>f(4x)\left[f(x) + f(-x) - 2x^2\right] = 0.</math>
  
 
<math>\indent</math> In particular, if <math>f(x) \ne 0</math>, then <math>f(4x) \ne 0</math> by the observation from Step 3, because <math>f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0.</math> Hence, the above equation implies that <math>2x^2 = f(x) + f(-x) = 2f(x)</math>, where the last step follows from the first observation from Step 2.
 
<math>\indent</math> In particular, if <math>f(x) \ne 0</math>, then <math>f(4x) \ne 0</math> by the observation from Step 3, because <math>f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0.</math> Hence, the above equation implies that <math>2x^2 = f(x) + f(-x) = 2f(x)</math>, where the last step follows from the first observation from Step 2.
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<math>\indent</math> Looking back on the equation from Step 3, it follows that <math>f(y) + 3y^2 \ne 0</math> for any nonzero <math>y.</math> Therefore, replacing <math>y \to t/4</math> in this equation, it follows that <math>f(t) = 0 \implies f(2t) = 0.</math>
 
<math>\indent</math> Looking back on the equation from Step 3, it follows that <math>f(y) + 3y^2 \ne 0</math> for any nonzero <math>y.</math> Therefore, replacing <math>y \to t/4</math> in this equation, it follows that <math>f(t) = 0 \implies f(2t) = 0.</math>
  
−
Step 5: If <math>f(a) = f(b) = 0</math>, then <math>f(b - a) = 0.</math>
+
'''Step 5:''' If <math>f(a) = f(b) = 0</math>, then <math>f(b - a) = 0.</math>
  
 
<math>\indent</math> This follows by choosing <math>x, y</math> such that <math>x - 3y = a</math> and <math>3x - y = b.</math> Then <math>x + y = \tfrac{b - a}{2}</math>, so plugging <math>x, y</math> into the given equation, we deduce that <math>f\left(\tfrac{b - a}{2}\right) = 0.</math> Therefore, by the third observation from Step 4, we obtain <math>f(b - a) = 0</math>, as desired.
 
<math>\indent</math> This follows by choosing <math>x, y</math> such that <math>x - 3y = a</math> and <math>3x - y = b.</math> Then <math>x + y = \tfrac{b - a}{2}</math>, so plugging <math>x, y</math> into the given equation, we deduce that <math>f\left(\tfrac{b - a}{2}\right) = 0.</math> Therefore, by the third observation from Step 4, we obtain <math>f(b - a) = 0</math>, as desired.
  
−
Step 6: If <math>f \not\equiv 0</math>, then <math>f(t) = 0 \implies t = 0.</math>
+
'''Step 6:''' If <math>f \not\equiv 0</math>, then <math>f(t) = 0 \implies t = 0.</math>
  
 
<math>\indent</math> Suppose by way of contradiction that there exists an nonzero <math>t</math> with <math>f(t) = 0.</math> Choose <math>x, y</math> such that <math>f(x) \ne 0</math> and <math>x + y = t.</math> The following three facts are crucial:
 
<math>\indent</math> Suppose by way of contradiction that there exists an nonzero <math>t</math> with <math>f(t) = 0.</math> Choose <math>x, y</math> such that <math>f(x) \ne 0</math> and <math>x + y = t.</math> The following three facts are crucial:
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<math>\indent</math> By the second observation from Step 4, these three facts imply that <math>f(y) = y^2</math> and <math>f(x - 3y) = \left(x - 3y\right)^2</math> and <math>f(3x - y) = \left(3x - y\right)^2.</math> By plugging into the given equation, it follows that
 
<math>\indent</math> By the second observation from Step 4, these three facts imply that <math>f(y) = y^2</math> and <math>f(x - 3y) = \left(x - 3y\right)^2</math> and <math>f(3x - y) = \left(3x - y\right)^2.</math> By plugging into the given equation, it follows that
−
\begin{align*}
+
<cmath>\begin{align*}
 
\left(x^2 + xy\right)\left(x - 3y\right)^2 + \left(y^2 + xy\right)\left(3x - y\right)^2 = 0.
 
\left(x^2 + xy\right)\left(x - 3y\right)^2 + \left(y^2 + xy\right)\left(3x - y\right)^2 = 0.
−
\end{align*}But the above expression miraculously factors into <math>\left(x + y\right)^4</math>! This is clearly a contradiction, since <math>t = x + y \ne 0</math> by assumption. This completes Step 6.
+
\end{align*}</cmath> But the above expression miraculously factors into <math>\left(x + y\right)^4</math>! This is clearly a contradiction, since <math>t = x + y \ne 0</math> by assumption. This completes Step 6.
  
−
Step 7: By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.
+
'''Step 7:''' By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:10, 21 April 2016

Problem

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, \[(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.\]

Solution

Step 1: Set $x = y = 0$ to obtain $f(0) = 0.$

Step 2: Set $x = 0$ to obtain $f(y)f(-y) = f(y)^2.$

$\indent$ In particular, if $f(y) \ne 0$ then $f(y) = f(-y).$

$\indent$ In addition, replacing $y \to -t$, it follows that $f(t) = 0 \implies f(-t) = 0$ for all $t \in \mathbb{R}.$

Step 3: Set $x = 3y$ to obtain $\left[f(y) + 3y^2\right]f(8y) = f(4y)^2.$

$\indent$ In particular, replacing $y \to t/8$, it follows that $f(t) = 0 \implies f(t/2) = 0$ for all $t \in \mathbb{R}.$

Step 4: Set $y = -x$ to obtain $f(4x)\left[f(x) + f(-x) - 2x^2\right] = 0.$

$\indent$ In particular, if $f(x) \ne 0$, then $f(4x) \ne 0$ by the observation from Step 3, because $f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0.$ Hence, the above equation implies that $2x^2 = f(x) + f(-x) = 2f(x)$, where the last step follows from the first observation from Step 2.

$\indent$ Therefore, either $f(x) = 0$ or $f(x) = x^2$ for each $x.$

$\indent$ Looking back on the equation from Step 3, it follows that $f(y) + 3y^2 \ne 0$ for any nonzero $y.$ Therefore, replacing $y \to t/4$ in this equation, it follows that $f(t) = 0 \implies f(2t) = 0.$

Step 5: If $f(a) = f(b) = 0$, then $f(b - a) = 0.$

$\indent$ This follows by choosing $x, y$ such that $x - 3y = a$ and $3x - y = b.$ Then $x + y = \tfrac{b - a}{2}$, so plugging $x, y$ into the given equation, we deduce that $f\left(\tfrac{b - a}{2}\right) = 0.$ Therefore, by the third observation from Step 4, we obtain $f(b - a) = 0$, as desired.

Step 6: If $f \not\equiv 0$, then $f(t) = 0 \implies t = 0.$

$\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \ne 0$ and $x + y = t.$ The following three facts are crucial:

$\indent$ 1. $f(y) \ne 0.$ This is because $(x + y) - y = x$, so by Step 5, $f(y) = 0 \implies f(x) = 0$, impossible.

$\indent$ 2. $f(x - 3y) \ne 0.$ This is because $(x + y) - (x - 3y) = 4y$, so by Step 5 and the observation from Step 3, $f(x - 3y) = 0 \implies f(4y) = 0 \implies f(2y) = 0 \implies f(y) = 0$, impossible.

$\indent$ 3. $f(3x - y) \ne 0.$ This is because by the second observation from Step 2, $f(3x - y) = 0 \implies f(y - 3x) = 0.$ Then because $(x + y) - (y - 3x) = 4x$, Step 5 together with the observation from Step 3 yield $f(3x - y) = 0 \implies f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0$, impossible.

$\indent$ By the second observation from Step 4, these three facts imply that $f(y) = y^2$ and $f(x - 3y) = \left(x - 3y\right)^2$ and $f(3x - y) = \left(3x - y\right)^2.$ By plugging into the given equation, it follows that \begin{align*} \left(x^2 + xy\right)\left(x - 3y\right)^2 + \left(y^2 + xy\right)\left(3x - y\right)^2 = 0. \end{align*} But the above expression miraculously factors into $\left(x + y\right)^4$! This is clearly a contradiction, since $t = x + y \ne 0$ by assumption. This completes Step 6.

Step 7: By Step 6 and the second observation from Step 4, the only possible solutions are $f \equiv 0$ and $f(x) = x^2$ for all $x \in \mathbb{R}.$ It's easy to check that both of these work, so we're done.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2016 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
2016 USAJMO (Problems • Resources)
Preceded by
Problem 5 		Last Problem
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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