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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 ==
  
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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 17:10, 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.

These problems are copyrighted © by the Mathematical Association of America, as part of the 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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