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

Revision as of 21:19, 3 January 2019

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 1

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.

Alternative Solution 1

From steps 1 and 2 of Solution 1 we have that $f(0)=0$ , and $f(-y)f(y)=(f(y))^2$ , so if $f(y) \ne 0$ , then $f(y)=f(-y)$ . Furthermore, setting $y=-x$ gives us $(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0$ . The LHS can be factored as $(f(x)+f(-x)-2x^2) \cdot f(4x) = 0$ . Therefore, if $f(4x) \ne 0$ , then we have $f(x)+f(-x)-2x^2=0$ , or $f(x)+f(-x)=2x^2$ . However, since from step 2 we have that $f(x)=f(-x)$ assuming $f(x) \ne 0$ , the equation becomes $2f(x)=2x^2$ , so $f(x)=0, x^2$ are the only solutions.

Solution 2

Step 1: $x=y=0 \implies f(0)=0$

Step 2: $x=0 \implies f(y)f(-y)=f(y)^{2}$ . Now, assume $y \not = 0$ . Then, if $f(y)=0$ , we substitute in $-y$ to get $f(y)f(-y)=f(-y)^{2}$ , or $f(y)=f(-y)=0$ . Otherwise, we divide both sides by $f(y)$ to get $f(y)=f(-y)$ . If $y=0$ , we obviously have $f(0)=f(0)$ . Thus, the function is even. . Step 3: $y=-x \implies 2f(4x)(f(x)-x^{2})=0$ . Thus, $\forall x$ , we have $f(4x)=0$ or $f(x)=x^{2}$ .

Step 4: We now assume $f(x) \not = 0$ , $x\not = 0$ . We have $f(\frac{x}{4})=\frac{x^{2}}{16}$ . Now, setting $x=y=\frac{x}{4}$ , we have $f(\frac{x}{2}=\frac{x^{2}}{4}$ or $f(\frac{x}{2})=0$ . The former implies that $f(x)=0$ or $x^{2}$ . The latter implies that $f(x)=0$ or $f(x)=\frac{x^{2}}{2}$ . Assume the latter. $y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}$ . Clearly, this implies that $f(x)$ is negative for some $m$ . Now, we have $f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0$ , which is a contradiction. Thus, $\forall x$ $f(x)=0$ or $f(x)=x^{2}$ .

Step 5: We now assume $f(x)=0$ , $f(y)=y^{2}$ for some $x,y \not = 0$ . Let $m$ be sufficiently large integer, let $z=|4^{m}x|$ and take the absolute value of $y$ (since the function is even). Choose $c$ such that $3z-c=y$ . Note that we have $\frac{c}{z}$ ~ $3$ and $\frac{y}{z}$ ~ $0$ . Note that $f(z)=0$ . Now, $x=z, y=c \implies$ LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to $(z+c)^{4}$ ~ $256z^{4}$ . Now if $f(z-3c)=0$ , the second term of the LHS/RHS clearly ~0 as $m \to \infty$ . if $f(z-3c)=0$ , then we have LHS/RHS ~ $0$ , otherwise, we have LHS/RHS~ $\frac{8^{2}\cdot 3z^{4}}{256z^{4}}$ ~ $\frac{3}{4}$ , a contradiction, as we're clearly not dividing by $0$ , and we should have LHS/RHS=1.

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

@@ Line 46: / Line 46: @@
 ==Alternative Solution 1==
-From steps 1 and 2 we have that, setting <math>x=y=0</math>, <math>f(0)=0</math>, and, setting <math>x=0</math>, <math>f(-y)f(y)=(f(y))^2</math>, so if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. Therefore, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)-2x^2=0</math>, or <math>f(x)+f(-x)=2x^2</math>. However, since from step 2 we have that <math>f(x)=f(-x)</math> assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.
+From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y)f(y)=(f(y))^2</math>, so if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. Therefore, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)-2x^2=0</math>, or <math>f(x)+f(-x)=2x^2</math>. However, since from step 2 we have that <math>f(x)=f(-x)</math> assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.
 ==Solution 2==
 Step 1: <math>x=y=0 \implies f(0)=0</math>

2016 USAMO (Problems • Resources)
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2016 USAJMO (Problems • Resources)
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All USAJMO Problems and Solutions

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