Difference between revisions of "2007 AIME II Problems/Problem 14"
(add prob, no solution yet) |
m (→Solution 1) |
||
(22 intermediate revisions by 14 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math>f(x)</math> be a [[polynomial]] with real [[coefficient]]s such that <math> | + | Let <math>f(x)</math> be a [[polynomial]] with real [[coefficient]]s such that <math>f(0) = 1,</math> <math>f(2)+f(3)=125,</math> and for all <math>x</math>, <math>f(x)f(2x^{2})=f(2x^{3}+x).</math> Find <math>f(5).</math> |
− | == Solution == | + | == Official Solution (MAA)== |
− | {{ | + | If the leading term of <math>f(x)</math> is <math>ax^m</math>, then the leading term of <math>f(x)f(2x^2) = ax^m \cdot a(2x^2)^m = 2^ma^2x^{3m}</math>, and the leading term of <math>f(2x^3 + x) = 2^max^{3m}</math>. Hence <math>2^ma^2 = 2^ma</math>, and <math>a = 1</math>. Because <math>f(0) = 1</math>, the product of all the roots |
+ | of <math>f(x)</math> is <math>\pm 1</math>. If <math>f(\lambda) = 0</math>, then <math>f(2\lambda^3 + \lambda) = 0</math>. Assume that there exists a root <math>\lambda</math> with <math>|\lambda| \neq 1</math>. Then there must be such a root <math>\lambda_1</math> with <math>|\lambda_1| > 1</math>. Then <cmath>|2\lambda^3 + \lambda| \ge 2|\lambda|^3 - |\lambda| > 2|\lambda| - |\lambda| = |\lambda|.</cmath> But then <math>f(x)</math> would have infinitely many roots, given by <math>\lambda_{k+1} = 2\lambda_k^3 + \lambda_k</math>, for <math>k \ge 1</math>. | ||
+ | Therefore <math>|\lambda| = 1</math> for all of the roots of the polynomial. Thus <math>\lambda\overline{\lambda}=1</math>, and <math>(2\lambda^3 + \lambda)\overline{(2\lambda^3 + \lambda)} = 1</math>. | ||
+ | Solving these equations simultaneously for <math>\lambda = a + bi</math> yields <math>a = 0</math>, <math>b^2 = 1</math>, and so <math>\lambda^2 = -1</math>. Because the polynomial has real coefficients, the polynomial must have the form <math>f(x) = (1 + x^2)^n</math> for some integer <math>n \ge 1</math>. The condition <math>f(2) + f(3) = 125</math> implies <math>n = 2</math>, giving <math>f(5) = \boxed{676}</math>. | ||
+ | |||
+ | == Solution 1== | ||
+ | Let <math>r</math> be a root of <math>f(x)</math>. Then we have <math>f(r)f(2r^2)=f(2r^3+r)</math>; since <math>r</math> is a root, we have <math>f(r)=0</math>; therefore <math>2r^3+r</math> is also a root. Thus, if <math>r</math> is real and non-zero, <math>|2r^3+r|>r</math>, so <math>f(x)</math> has infinitely many roots. Since <math>f(x)</math> is a polynomial (thus of finite degree) and <math>f(0)</math> is nonzero, <math>f(x)</math> has no real roots. | ||
+ | |||
+ | Note that <math>f(x)</math> is not constant. We then find two complex roots: <math>r = \pm i</math>. We find that <math>f(i)f(-2) = f(-i)</math>, and that <math>f(-i)f(-2) = f(i)</math>. This means that <math>f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0</math>. Thus, <math>\pm i</math> are roots of the polynomial, and so <math>(x - i)(x + i) = x^2 + 1</math> will be a factor of the polynomial. (Note: This requires the assumption that <math>f(-2)\neq1</math>. Clearly, <math>f(-2)\neq-1</math>, because that would imply the existence of a real root.) | ||
+ | |||
+ | The polynomial is thus in the form of <math>f(x) = (x^2 + 1)g(x)</math>. Substituting into the given expression, we have | ||
+ | |||
+ | <cmath>(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)</cmath> | ||
+ | <cmath>(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)</cmath> | ||
+ | |||
+ | Thus either <math>4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)</math> is 0 for any <math>x</math>, or <math>g(x)</math> satisfies the same constraints as <math>f(x)</math>. Continuing, by infinite descent, <math>f(x) = (x^2 + 1)^n</math> for some <math>n</math>. | ||
+ | |||
+ | Since <math>f(2)+f(3)=125=5^n+10^n</math> for some <math>n</math>, we have <math>n=2</math>; so <math>f(5) = \boxed{676}</math>. | ||
+ | |||
+ | |||
+ | Comment: | ||
+ | The answer is clearly correct, but the proof has a gap, i.e. there is no reason that <math>f(-2)\neq1</math>. Since <math>f(x)</math> has no real roots, the degree must be even. Consider <math>g(x)= f(x)/f(-x)</math>. Then since <math>f</math> is non-zero, <math>g(x)=g(2x^3+x)</math>. Now the function <math>2x^3+x</math> applied repeatedly from some real starting value of x becomes arbitrarily large, and the limit of <math>g(x)</math> as <math>|x|</math> approaches infinity is 1, so <math>g(x)</math>=1 for all x, or <math>f(x)=f(-x)</math>. Then <math>f(x)=h(x^2+1)</math> for some polynomial <math>h(x)</math>, and <math>h(x^2+1)h(4x^4+1)=h(4x^6+4x^4+x^2+1) = h((x^2+1)(4x^4+1))</math>. Now suppose h has degree m. It is clearly monic. Assume that the next highest non-zero coefficient in h is k. Then, subtracting <math>((x^2+1)(4x^4+1))^m</math> from both sides of the equation yields a polynomial equality with degree <math>4m+2k</math> on the left and degree <math>6k</math> on the right, a contradiction. So <math>h(x)=x^m</math>, and <math>f(x)=(1+x^2)^m</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>r</math> be a root of <math>f(x).</math> This means that <math>f(r)f(2r^2)=f(2r^3+r).</math> In other words, <math>2r^3+r</math> is a root of <math>f(x)</math> too. Since <math>f(x)</math> can't have infinitely many roots, <cmath>Q(x)=P(P(\dotsb P(P(r)) \dotsb))</cmath> is cyclic, where <math>P(x)=2x^3+x.</math> Now, we will do casework. | ||
+ | |||
+ | Case 1: <math>\deg f\geq1</math> | ||
+ | |||
+ | Subcase 1: <math>|r|>1</math> | ||
+ | |||
+ | This means that <cmath>|2r^3+r|\geq|2r^3|-|r|=|r|(2|r|^2-1)>|r|(2\cdot1^2-1)=|r|.</cmath> It follows that <math>|2r^3+r|>|r|</math> for all <math>r.</math> This implies that <math>Q(r)</math> can't be cyclic. Thus, it is impossible for <math>|r|>1</math> to be true. | ||
+ | |||
+ | Subcase 2: <math>|r|<1</math> | ||
+ | |||
+ | This means that <math>|2r^3+r|\geq2|r^3|-|r|=|r|(|2r^2|-1)<|r|.</math> It follows that <math>|2r^3+r|<|r|</math> for all <math>r.</math> This implies that <math>Q(r)</math> can't be cyclic. Thus, it is impossible for <math>|r|>1</math> to be true. | ||
+ | |||
+ | Subcase 3: <math>|r|=1.</math> | ||
+ | |||
+ | Since <math>|r|</math> is not greater than or less than 1, <math>|r|=1.</math> This means that all the roots of the polynomial have a magnitude of <math>1.</math> More specifically, <math>|2r^3+r|</math> has a magnitude of one. Since this would mean an equality condition from the triangle inequality, <math>2r^3</math> and <math>r</math> are collinear with the origin in the complex plane. In other words, <math>\frac{2r^3}{r}=\pm c\Leftrightarrow cr=2r^3\Leftrightarrow 2r^2=c\Leftrightarrow r=\pm\sqrt{\pm\frac{c}{2}},</math> for some real constant <math>c.</math> Now, from <math>|r|=1,</math> we find that <math>\left|\pm\sqrt{\pm\frac{c}{2}}\right|=1\Leftrightarrow \sqrt{\pm\frac{c}{2}}=1\Leftrightarrow \pm\frac{c}{2}=1\Leftrightarrow c=\pm2.</math> Putting this back into the equation, we find that <math>r=1,-1,i,-i.</math> Now, this means that <math>2r^3+r=3,-3,i,-i.</math> <math>3</math> and <math>-3</math> obviously doesn't have a magnitude of <math>1.</math> Thus, <math>i,-i</math> are the only possible roots of the polynomial. Since roots come in conjugate pairs, <math>f(x)=[(x-i)(x+i)]^n=(x^2+1)^n,</math> works for all constants <math>n\neq0.</math> | ||
+ | |||
+ | Case 2: <math>\deg f=0.</math> | ||
+ | |||
+ | This means that <math>f(x)=c,</math> for some constant <math>c.</math> In other words, <math>c^2=c.</math> We can easily find that this means that <math>c=0,1.</math> | ||
+ | Combining all the cases, we conclude that <math>f(x)=(x^2+1)^n,0,1</math> are the only polynomials that satisfy this equation. | ||
+ | Now, we can test! <math>f(x)=0,1</math> obviously don't satisfy <math>f(2)+f(3)=125.</math> Thus, <math>f(x)=(x^2+1)^n.</math> Substituting, we find that <math>5^n+10^n=125\Leftrightarrow n=2.</math> We conclude that <math>f(5)=(5^2+1)^2=26^2=\boxed{676}.</math> | ||
+ | |||
+ | ~ pinkpig | ||
== See also == | == See also == | ||
Line 9: | Line 55: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:44, 25 January 2022
Problem
Let be a polynomial with real coefficients such that
and for all
,
Find
Official Solution (MAA)
If the leading term of is
, then the leading term of
, and the leading term of
. Hence
, and
. Because
, the product of all the roots
of
is
. If
, then
. Assume that there exists a root
with
. Then there must be such a root
with
. Then
But then
would have infinitely many roots, given by
, for
.
Therefore
for all of the roots of the polynomial. Thus
, and
.
Solving these equations simultaneously for
yields
,
, and so
. Because the polynomial has real coefficients, the polynomial must have the form
for some integer
. The condition
implies
, giving
.
Solution 1
Let be a root of
. Then we have
; since
is a root, we have
; therefore
is also a root. Thus, if
is real and non-zero,
, so
has infinitely many roots. Since
is a polynomial (thus of finite degree) and
is nonzero,
has no real roots.
Note that is not constant. We then find two complex roots:
. We find that
, and that
. This means that
. Thus,
are roots of the polynomial, and so
will be a factor of the polynomial. (Note: This requires the assumption that
. Clearly,
, because that would imply the existence of a real root.)
The polynomial is thus in the form of . Substituting into the given expression, we have
Thus either is 0 for any
, or
satisfies the same constraints as
. Continuing, by infinite descent,
for some
.
Since for some
, we have
; so
.
Comment:
The answer is clearly correct, but the proof has a gap, i.e. there is no reason that . Since
has no real roots, the degree must be even. Consider
. Then since
is non-zero,
. Now the function
applied repeatedly from some real starting value of x becomes arbitrarily large, and the limit of
as
approaches infinity is 1, so
=1 for all x, or
. Then
for some polynomial
, and
. Now suppose h has degree m. It is clearly monic. Assume that the next highest non-zero coefficient in h is k. Then, subtracting
from both sides of the equation yields a polynomial equality with degree
on the left and degree
on the right, a contradiction. So
, and
.
Solution 2
Let be a root of
This means that
In other words,
is a root of
too. Since
can't have infinitely many roots,
is cyclic, where
Now, we will do casework.
Case 1:
Subcase 1:
This means that It follows that
for all
This implies that
can't be cyclic. Thus, it is impossible for
to be true.
Subcase 2:
This means that It follows that
for all
This implies that
can't be cyclic. Thus, it is impossible for
to be true.
Subcase 3:
Since is not greater than or less than 1,
This means that all the roots of the polynomial have a magnitude of
More specifically,
has a magnitude of one. Since this would mean an equality condition from the triangle inequality,
and
are collinear with the origin in the complex plane. In other words,
for some real constant
Now, from
we find that
Putting this back into the equation, we find that
Now, this means that
and
obviously doesn't have a magnitude of
Thus,
are the only possible roots of the polynomial. Since roots come in conjugate pairs,
works for all constants
Case 2:
This means that for some constant
In other words,
We can easily find that this means that
Combining all the cases, we conclude that
are the only polynomials that satisfy this equation.
Now, we can test!
obviously don't satisfy
Thus,
Substituting, we find that
We conclude that
~ pinkpig
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.