Difference between revisions of "2016 USAMO Problems/Problem 4"
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+ | ==Solution 2== | ||
+ | Step 1: <math>x=y=0 \implies f(0)=0</math> | ||
+ | |||
+ | Step 2: <math>x=0 \implies f(y)f(-y)=f(y)^{2}</math>. Now, assume <math>y \not = 0</math>. Then, if <math>f(y)=0</math>, we substitute in <math>-y</math> to get <math>f(y)f(-y)=f(-y)^{2}</math>, or <math>f(y)=f(-y)=0</math>. Otherwise, we divide both sides by <math>f(y)</math> to get <math>f(y)=f(-y)</math>. If <math>y=0</math>, we obviously have <math>f(0)=f(0)</math>. Thus, the function is even. | ||
+ | . | ||
+ | Step 3: <math>y=-x \implies 2f(4x)(f(x)-x^{2})=0</math>. Thus, <math>\forall x</math>, we have <math>f(4x)=0</math> or <math>f(x)=x^{2}</math>. | ||
+ | |||
+ | Step 4: We now assume <math>f(x) \not = 0</math>, <math>x\not = 0</math>. We have <math>f(\frac{x}{4})=\frac{x^{2}}{16}</math>. Now, setting <math>x=y=\frac{x}{4}</math>, we have <math>f(\frac{x}{2}=\frac{x^{2}}{4}</math> or <math>f(\frac{x}{2})=0</math>. The former implies that <math>f(x)=0</math> or <math>x^{2}</math>. The latter implies that <math>f(x)=0</math> or <math>f(x)=\frac{x^{2}}{2}</math>. Assume the latter. <math>y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}</math>. Clearly, this implies that <math>f(x)</math> is negative for some <math>m</math>. Now, we have <math>f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0</math>, which is a contradiction. Thus, <math>\forall x</math><math>f(x)=0</math> or <math>f(x)=x^{2}</math>. | ||
+ | |||
+ | Step 5: We now assume <math>f(x)=0</math>, <math>f(y)=y^{2}</math> for some <math>x,y \not = 0</math>. Let <math>m</math> be sufficiently large integer, let <math>z=|4^{m}x|</math> and take the absolute value of <math>y</math>(since the function is even). Choose <math>c</math> such that <math>3z-c=y</math>. Note that we have <math>\frac{c}{z}</math>~<math>3</math> and <math>\frac{y}{z}</math>~<math>0</math>. Note that <math>f(z)=0</math>. Now, <math>x=z, y=c \implies</math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <math>(z+c)^{4}</math>~<math>256z^{4}</math>. Now if <math>f(z-3c)=0</math>, the second term of the LHS/RHS clearly ~0 as <math>m \to \infty</math>. if <math>f(z-3c)=0</math>, then we have LHS/RHS ~ <math>0</math>, otherwise, we have LHS/RHS~<math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}</math>~<math>\frac{3}{4}</math>, a contradiction, as we're clearly not dividing by <math>0</math>, and we should have LHS/RHS=1. | ||
==See also== | ==See also== |
Revision as of 23:25, 24 April 2016
Contents
Problem
Find all functions such that for all real numbers
and
,
Solution
Step 1: Set to obtain
Step 2: Set to obtain
In particular, if
then
In addition, replacing
, it follows that
for all
Step 3: Set to obtain
In particular, replacing
, it follows that
for all
Step 4: Set to obtain
In particular, if
, then
by the observation from Step 3, because
Hence, the above equation implies that
, where the last step follows from the first observation from Step 2.
Therefore, either
or
for each
Looking back on the equation from Step 3, it follows that
for any nonzero
Therefore, replacing
in this equation, it follows that
Step 5: If , then
This follows by choosing
such that
and
Then
, so plugging
into the given equation, we deduce that
Therefore, by the third observation from Step 4, we obtain
, as desired.
Step 6: If , then
Suppose by way of contradiction that there exists an nonzero
with
Choose
such that
and
The following three facts are crucial:
1.
This is because
, so by Step 5,
, impossible.
2.
This is because
, so by Step 5 and the observation from Step 3,
, impossible.
3.
This is because by the second observation from Step 2,
Then because
, Step 5 together with the observation from Step 3 yield
, impossible.
By the second observation from Step 4, these three facts imply that
and
and
By plugging into the given equation, it follows that
But the above expression miraculously factors into
! This is clearly a contradiction, since
by assumption. This completes Step 6.
Step 7: By Step 6 and the second observation from Step 4, the only possible solutions are and
for all
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.
Solution 2
Step 1:
Step 2: . Now, assume
. Then, if
, we substitute in
to get
, or
. Otherwise, we divide both sides by
to get
. If
, we obviously have
. Thus, the function is even.
.
Step 3:
. Thus,
, we have
or
.
Step 4: We now assume ,
. We have
. Now, setting
, we have
or
. The former implies that
or
. The latter implies that
or
. Assume the latter.
. Clearly, this implies that
is negative for some
. Now, we have
, which is a contradiction. Thus,
or
.
Step 5: We now assume ,
for some
. Let
be sufficiently large integer, let
and take the absolute value of
(since the function is even). Choose
such that
. Note that we have
~
and
~
. Note that
. Now,
LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to
~
. Now if
, the second term of the LHS/RHS clearly ~0 as
. if
, then we have LHS/RHS ~
, otherwise, we have LHS/RHS~
~
, a contradiction, as we're clearly not dividing by
, and we should have LHS/RHS=1.
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 |