Difference between revisions of "2007 AIME II Problems/Problem 14"
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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. | 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. | ||
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Case 1: <math>\deg f\geq1</math> | Case 1: <math>\deg f\geq1</math> | ||
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Subcase 1: <math>|r|>1</math> | Subcase 1: <math>|r|>1</math> | ||
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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. | 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. | ||
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Subcase 2: <math>|r|<1</math> | Subcase 2: <math>|r|<1</math> | ||
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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. | 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. | ||
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Subcase 3: <math>|r|=1.</math> | Subcase 3: <math>|r|=1.</math> | ||
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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> | 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> | ||
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Case 2: <math>\deg f=0.</math> | Case 2: <math>\deg f=0.</math> | ||
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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> | 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. | 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. |
Revision as of 14:54, 18 September 2021
Contents
Problem
Let be a polynomial with real coefficients such that and for all , Find
Solution
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 1
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.