Difference between revisions of "2012 AMC 12B Problems/Problem 23"

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==Solution 1==
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== Problem 23 ==
  
Since <math>z_0</math> is a root of <math>P</math>, and <math>P</math> has integer coefficients, <math>z_0</math> must be algebraic. Since <math>z_0</math> is algebraic and lies on the unit circle, <math>z_0</math> must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since <math>P</math> has degree 4, it seems reasonable (and we will assume this only temporarily) that <math>z_0</math> must be a 2nd, 3rd, or 4th root of unity. These are among the set <math>\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}</math>. Since complex roots of polynomials come in conjugate pairs, we have that <math>P</math> has one (or more) of the following factors: <math>z+1</math>, <math>z-1</math>, <math>z^2+1</math>, or <math>z^2+z+1</math>. If <math>z=1</math> then <math>a+b+c+d+4=0</math>; a contradiction since <math>a,b,c,d</math> are non-negative. On the other hand, suppose <math>z=-1</math>. Then <math>(a+c)-(b+d)=4</math>. This implies <math>a+b=8,7,6,5,4</math> while <math>b+d=4,3,2,1,0</math> correspondingly. After listing cases, the only such valid <math>a,b,c,d</math> are <math>4,4,4,0</math>, <math>4,3,3,0</math>, <math>4,2,2,0</math>, <math>4,1,1,0</math>, and <math>4,0,0,0</math>.
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Consider all polynomials of a complex variable, <math>P(z)=4z^4+az^3+bz^2+cz+d</math>, where <math>a,b,c,</math> and <math>d</math> are integers, <math>0\le d\le c\le b\le a\le 4</math>, and the polynomial has a zero <math>z_0</math> with <math>|z_0|=1.</math> What is the sum of all values <math>P(1)</math> over all the polynomials with these properties?
  
Now suppose <math>z=i</math>. Then <math>4=(a-c)i+(b-d)</math> whereupon <math>a=c</math> and <math>b-d=4</math>. But then <math>a=b=c</math> and <math>d=a-4</math>. This gives only the cases <math>a,b,c,d</math> equals <math>4,4,4,0</math>, which we have already counted in a previous case.
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<math>\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 </math>
  
Suppose <math>z=-i</math>. Then <math>4=i(c-a)+(b-d)</math> so that <math>a=c</math> and <math>b=4+d</math>. This only gives rise to <math>a,b,c,d</math> equal <math>4,4,4,0</math> which we have previously counted.  
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== Solution ==
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First, assume that <math>z_0\in \mathbb{R}</math>, so <math>z_0=1</math> or <math>-1</math>. <math>1</math> does not work because <math>P(1)\geq 4</math>. Assume that <math>z_0=-1</math>. Then <math>0=P(-1)=4-a+b-c+d</math>,  we have <math>4+b+d=a+c\leq 4+b</math>, so <math>d=0. Also, </math>a=4<math> has to be true since </math>4+b=a+c \leq a+b<math>. Now </math>4+b=4+c<math> deduces </math>b=c<math>, therefore the only possible choices for </math>(a,b,c,d)<math> are </math>(4,t,t,0)<math>. In these cases, </math>P(-1)=4-4+t-t+0=0<math>. The sum of </math>P(1)<math> over these cases is </math>\sum_{t=0}^{4} (4+4+t+t) = 40+20=60<math>.
  
Finally suppose <math>z^2+z+1</math> divides <math>P</math>. Using polynomial division ((or that <math>z^3=1</math> to make the same deductions) we ultimately obtain that <math>b=4+c</math>. This can only happen if <math>a,b,c,d</math> is <math>4,4,0,0</math>.  
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Second, assume that </math>z_0\in \mathbb{C} \backslash \mathbb{R}<math>.
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==Solution (doubtful) ==
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Since </math>z_0<math> is a root of </math>P<math>, and </math>P<math> has integer coefficients, </math>z_0<math> must be algebraic. Since </math>z_0<math> is algebraic and lies on the unit circle, </math>z_0<math> must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since </math>P<math> has degree 4, it seems reasonable (and we will assume this only temporarily) that </math>z_0<math> must be a 2nd, 3rd, or 4th root of unity. These are among the set </math>\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}<math>. Since complex roots of polynomials come in conjugate pairs, we have that </math>P<math> has one (or more) of the following factors: </math>z+1<math>, </math>z-1<math>, </math>z^2+1<math>, or </math>z^2+z+1<math>. If </math>z=1<math> then </math>a+b+c+d+4=0<math>; a contradiction since </math>a,b,c,d<math> are non-negative. On the other hand, suppose </math>z=-1<math>. Then </math>(a+c)-(b+d)=4<math>. This implies </math>a+b=8,7,6,5,4<math> while </math>b+d=4,3,2,1,0<math> correspondingly. After listing cases, the only such valid </math>a,b,c,d<math> are </math>4,4,4,0<math>, </math>4,3,3,0<math>, </math>4,2,2,0<math>, </math>4,1,1,0<math>, and </math>4,0,0,0<math>.
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Now suppose </math>z=i<math>. Then </math>4=(a-c)i+(b-d)<math> whereupon </math>a=c<math> and </math>b-d=4<math>. But then </math>a=b=c<math> and </math>d=a-4<math>. This gives only the cases </math>a,b,c,d<math> equals </math>4,4,4,0<math>, which we have already counted in a previous case.
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Suppose </math>z=-i<math>. Then </math>4=i(c-a)+(b-d)<math> so that </math>a=c<math> and </math>b=4+d<math>. This only gives rise to </math>a,b,c,d<math> equal </math>4,4,4,0<math> which we have previously counted.
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Finally suppose </math>z^2+z+1<math> divides </math>P<math>. Using polynomial division ((or that </math>z^3=1<math> to make the same deductions) we ultimately obtain that </math>b=4+c<math>. This can only happen if </math>a,b,c,d<math> is </math>4,4,0,0<math>.  
  
 
Hence we've the polynomials
 
Hence we've the polynomials
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<cmath>4x^4+4x^3</cmath>
 
<cmath>4x^4+4x^3</cmath>
 
<cmath>4x^4+4x^3+4x^2</cmath>
 
<cmath>4x^4+4x^3+4x^2</cmath>
However, by inspection <math>4x^4+4x^3+4x^2+4x+4</math> has roots on the unit circle, because <math>x^4+x^3+x^2+x+1=(x^5-1)/(x-1)</math> which brings the sum to 92 (choice B).  Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that <math>z_0</math> is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that <math>z_0</math> in an <math>n</math>th root of unity where <math>n>5</math>, and <math>z_0</math> is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If <math>n</math> is prime, then \textit{every} <math>n</math>th root of unity except 1 must satisfy our polynomial, but since <math>n>5</math> and the degree of our polynomial is 4, this is impossible. Suppose <math>n</math> is composite. If it has a prime factor <math>p</math> greater than 5 then again every <math>p</math>th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose <math>n</math> is divisible only by 2,3,or 5. Since by hypothesis <math>z_0</math> is not a 2nd or 3rd root of unity, <math>z_0</math> must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy <math>P(z_0)=0</math>. But <math>(x^5-1)/(x-1)</math> has exactly all 5th roots of unity excluding 1, and <math>(x^5-1)/(x-1)=x^4+x^3+x^2+x+1</math>. Thus this must divide <math>P</math> which implies <math>P(x)=4(x^4+x^3+x^2+x+1)</math>. This completes the proof.
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However, by inspection </math>4x^4+4x^3+4x^2+4x+4<math> has roots on the unit circle, because </math>x^4+x^3+x^2+x+1=(x^5-1)/(x-1)<math> which brings the sum to 92 (choice B).  Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that </math>z_0<math> is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that </math>z_0<math> in an </math>n<math>th root of unity where </math>n>5<math>, and </math>z_0<math> is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If </math>n<math> is prime, then \textit{every} </math>n<math>th root of unity except 1 must satisfy our polynomial, but since </math>n>5<math> and the degree of our polynomial is 4, this is impossible. Suppose </math>n<math> is composite. If it has a prime factor </math>p<math> greater than 5 then again every </math>p<math>th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose </math>n<math> is divisible only by 2,3,or 5. Since by hypothesis </math>z_0<math> is not a 2nd or 3rd root of unity, </math>z_0<math> must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy </math>P(z_0)=0<math>. But </math>(x^5-1)/(x-1)<math> has exactly all 5th roots of unity excluding 1, and </math>(x^5-1)/(x-1)=x^4+x^3+x^2+x+1<math>. Thus this must divide </math>P<math> which implies </math>P(x)=4(x^4+x^3+x^2+x+1)$. This completes the proof.

Revision as of 02:31, 6 December 2012

Problem 23

Consider all polynomials of a complex variable, $P(z)=4z^4+az^3+bz^2+cz+d$, where $a,b,c,$ and $d$ are integers, $0\le d\le c\le b\le a\le 4$, and the polynomial has a zero $z_0$ with $|z_0|=1.$ What is the sum of all values $P(1)$ over all the polynomials with these properties?

$\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120$

Solution

First, assume that $z_0\in \mathbb{R}$, so $z_0=1$ or $-1$. $1$ does not work because $P(1)\geq 4$. Assume that $z_0=-1$. Then $0=P(-1)=4-a+b-c+d$, we have $4+b+d=a+c\leq 4+b$, so $d=0. Also,$a=4$has to be true since$4+b=a+c \leq a+b$. Now$4+b=4+c$deduces$b=c$, therefore the only possible choices for$(a,b,c,d)$are$(4,t,t,0)$. In these cases,$P(-1)=4-4+t-t+0=0$. The sum of$P(1)$over these cases is$\sum_{t=0}^{4} (4+4+t+t) = 40+20=60$.

Second, assume that$ (Error compiling LaTeX. ! Missing $ inserted.)z_0\in \mathbb{C} \backslash \mathbb{R}$.

==Solution (doubtful) ==

Since$ (Error compiling LaTeX. ! Missing $ inserted.)z_0$is a root of$P$, and$P$has integer coefficients,$z_0$must be algebraic. Since$z_0$is algebraic and lies on the unit circle,$z_0$must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since$P$has degree 4, it seems reasonable (and we will assume this only temporarily) that$z_0$must be a 2nd, 3rd, or 4th root of unity. These are among the set$\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}$. Since complex roots of polynomials come in conjugate pairs, we have that$P$has one (or more) of the following factors:$z+1$,$z-1$,$z^2+1$, or$z^2+z+1$. If$z=1$then$a+b+c+d+4=0$; a contradiction since$a,b,c,d$are non-negative. On the other hand, suppose$z=-1$. Then$(a+c)-(b+d)=4$. This implies$a+b=8,7,6,5,4$while$b+d=4,3,2,1,0$correspondingly. After listing cases, the only such valid$a,b,c,d$are$4,4,4,0$,$4,3,3,0$,$4,2,2,0$,$4,1,1,0$, and$4,0,0,0$.

Now suppose$ (Error compiling LaTeX. ! Missing $ inserted.)z=i$. Then$4=(a-c)i+(b-d)$whereupon$a=c$and$b-d=4$. But then$a=b=c$and$d=a-4$. This gives only the cases$a,b,c,d$equals$4,4,4,0$, which we have already counted in a previous case.

Suppose$ (Error compiling LaTeX. ! Missing $ inserted.)z=-i$. Then$4=i(c-a)+(b-d)$so that$a=c$and$b=4+d$. This only gives rise to$a,b,c,d$equal$4,4,4,0$which we have previously counted.

Finally suppose$ (Error compiling LaTeX. ! Missing $ inserted.)z^2+z+1$divides$P$. Using polynomial division ((or that$z^3=1$to make the same deductions) we ultimately obtain that$b=4+c$. This can only happen if$a,b,c,d$is$4,4,0,0$.

Hence we've the polynomials <cmath>4x^4+4x^3+4x^2+4x</cmath> <cmath>4x^4+4x^3+3x^2+3x</cmath> <cmath>4x^4+4x^3+2x^2+2x</cmath> <cmath>4x^4+4x^3+x^2+x</cmath> <cmath>4x^4+4x^3</cmath> <cmath>4x^4+4x^3+4x^2</cmath> However, by inspection$ (Error compiling LaTeX. ! Missing $ inserted.)4x^4+4x^3+4x^2+4x+4$has roots on the unit circle, because$x^4+x^3+x^2+x+1=(x^5-1)/(x-1)$which brings the sum to 92 (choice B).  Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that$z_0$is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that$z_0$in an$n$th root of unity where$n>5$, and$z_0$is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If$n$is prime, then \textit{every}$n$th root of unity except 1 must satisfy our polynomial, but since$n>5$and the degree of our polynomial is 4, this is impossible. Suppose$n$is composite. If it has a prime factor$p$greater than 5 then again every$p$th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose$n$is divisible only by 2,3,or 5. Since by hypothesis$z_0$is not a 2nd or 3rd root of unity,$z_0$must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy$P(z_0)=0$. But$(x^5-1)/(x-1)$has exactly all 5th roots of unity excluding 1, and$(x^5-1)/(x-1)=x^4+x^3+x^2+x+1$. Thus this must divide$P$which implies$P(x)=4(x^4+x^3+x^2+x+1)$. This completes the proof.

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