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

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== Problem 23 ==
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== Problem ==
 
 
 
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?
 
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?
  
 
<math>\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 </math>
 
<math>\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 </math>
  
 
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== Solutions ==
 
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=== Solution 1 ===
==Solution (dubious) ==
 
 
 
 
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>.  
 
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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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.
 
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.
  
== Solution 2 ==
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=== Solution 2 ===
 
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</math>. 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> gives <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>.
 
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</math>. 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> gives <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>.
  
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Therefore the desired sum is <math>60+12+20=92 ...\framebox{B}</math>.
 
Therefore the desired sum is <math>60+12+20=92 ...\framebox{B}</math>.
  
== Solution 3 ==
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=== Solution 3 ===
First, notice that <math>z=1</math> cannot be a root of the polynomial because <math>a,b,c,d \geq 0</math>. Multiplying the polynomial by <math>(z-1)</math> yields <math>P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d</math>, so for <math>z \neq 1</math> to be a root of <math>P(z)</math>, <math>4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d</math>. Now we consider the root <math>z_0</math> with <math>|z_0|=1</math>. <math>|4z_0^5| = 4</math>, so the right hand side must have absolute value 4. By the triangle inequality, <math>|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|</math> <math>\geq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d </math> <math>= (4-a)+(a-b)+(b-c)+(c-d)+d=4</math>, with equality if and only if each of <math>4z_0^5</math>, <math>(4-a)z_0^4</math>, <math>(a-b)z_0^3</math>, <math>(b-c)z_0^2</math>, <math>(c-d)z_0</math>, and <math>d</math> is either zero or in the same direction as all the others when looked at as vectors in the complex plane.  
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First, notice that <math>z=1</math> cannot be a root of the polynomial because <math>a,b,c,d \geq 0</math>. Multiplying the polynomial by <math>(z-1)</math> yields <math>P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d</math>, so for <math>z \neq 1</math> to be a root of <math>P(z)</math>, <math>4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d</math>. Now we consider the root <math>z_0</math> with <math>|z_0|=1</math>. <math>|4z_0^5| = 4</math>, so the right hand side must have absolute value 4. By the triangle inequality, <math>|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|</math> <math>\leq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d </math> <math>= (4-a)+(a-b)+(b-c)+(c-d)+d=4</math>, with equality if and only if each of <math>4z_0^5</math>, <math>(4-a)z_0^4</math>, <math>(a-b)z_0^3</math>, <math>(b-c)z_0^2</math>, <math>(c-d)z_0</math>, and <math>d</math> is either zero or in the same direction as all the others when looked at as vectors in the complex plane.  
  
 
We can now divide into two cases: <math>d \neq 0</math> and <math>d=0</math>. If <math>d \neq 0</math>, then <math>4z_0^5</math> must be real by the previous argument, so <math>z_0</math> is a fifth root of unity. Also, <math>(4-a)z_0^4</math>, <math>(a-b)z_0^3</math>, <math>(b-c)z_0^2</math>, and <math>(c-d)z_0</math> must all be zero because if <math>z_0</math> is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, <math>a=4</math>, <math>b=a</math>, <math>c=b</math>, and <math>d=c</math>, leading to the solution <math>(a,b,c,d)=(4,4,4,4)</math>. Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.
 
We can now divide into two cases: <math>d \neq 0</math> and <math>d=0</math>. If <math>d \neq 0</math>, then <math>4z_0^5</math> must be real by the previous argument, so <math>z_0</math> is a fifth root of unity. Also, <math>(4-a)z_0^4</math>, <math>(a-b)z_0^3</math>, <math>(b-c)z_0^2</math>, and <math>(c-d)z_0</math> must all be zero because if <math>z_0</math> is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, <math>a=4</math>, <math>b=a</math>, <math>c=b</math>, and <math>d=c</math>, leading to the solution <math>(a,b,c,d)=(4,4,4,4)</math>. Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.
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Then, the complete list of solutions is <math>(a,b,c,d)=(4,4,4,4),(4,4,4,0),(4,3,3,0),(4,2,2,0),(4,1,1,0),(4,0,0,0),(4,4,0,0)</math>, leading to a sum of <math>\framebox{B}=92</math>.
 
Then, the complete list of solutions is <math>(a,b,c,d)=(4,4,4,4),(4,4,4,0),(4,3,3,0),(4,2,2,0),(4,1,1,0),(4,0,0,0),(4,4,0,0)</math>, leading to a sum of <math>\framebox{B}=92</math>.
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== Video Solution by Richard Rusczyk ==
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https://artofproblemsolving.com/videos/amc/2012amc12b/278
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~dolphin7
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:14, 19 January 2021

Problem

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?

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

Solutions

Solution 1

Since $z_0$ 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: [1]). 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 $z=i$. 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 $z=-i$. 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 $z^2+z+1$ 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 \[4x^4+4x^3+4x^2+4x\] \[4x^4+4x^3+3x^2+3x\] \[4x^4+4x^3+2x^2+2x\] \[4x^4+4x^3+x^2+x\] \[4x^4+4x^3\] \[4x^4+4x^3+4x^2\] However, by inspection $4x^4+4x^3+4x^2+4x+4$ 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.

Solution 2

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 $4+b=4+c$ gives $b=c$, 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 $z_0\in \mathbb{C} \backslash \mathbb{R}$, so $z_0=x_0+iy_0$ for some real $x_0, y_0$, $|x_0|<1$. By conjugate roots theorem we have that $P(z_0)=P(z_0^{*})=0$, therefore $(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0*z + 1)$ is a factor of $P(z)$, and we may assume that

\[P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)\]

for some real $p$. Expanding this polynomial and comparing the coefficients, we have the following equations:

\[p-8x_0 = a\] \[d+4-2px_0 = b\] \[p-2dx_0 = c\]

From the first and the third we may deduce that $2x_0 = \frac{a-c}{d-4}$ and that $p=\frac{da-4c}{d-4}$, if $d\neq 4$ (we will consider $d=4$ by the end). Let $k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}$. From the second equation, we know that $k=d+4-b$ is non-negative.

Consider the following cases:

Case 1: $a=c$. Then $k=0$, $b=d+4$, so $a=b=c=4$, $d=0$. However, this has already been found (i.e. the form of $(4,t,t,0)$).

Case 2: $a>c\geq 0$. Then since $k\geq 0$, we have $da-4c\geq 0$. However, $da \leq 4c$, therefore $da-4c=0$. This is true only when $d=c$. Also, we get $k=0$ again. In this case, $b=d+4$, so $a=b=4$, $c=d=0$, $x_0=-1/2$. $P(z)$ has a root $z_0=e^{i2\pi/3}$. $P(1)=12$.

Last case: $d=4$. We have $a=b=c=d=4$ and that $P(z)$ has a root $z_0=e^{i2\pi/5}$. $P(1)=20$.

Therefore the desired sum is $60+12+20=92 ...\framebox{B}$.

Solution 3

First, notice that $z=1$ cannot be a root of the polynomial because $a,b,c,d \geq 0$. Multiplying the polynomial by $(z-1)$ yields $P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d$, so for $z \neq 1$ to be a root of $P(z)$, $4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d$. Now we consider the root $z_0$ with $|z_0|=1$. $|4z_0^5| = 4$, so the right hand side must have absolute value 4. By the triangle inequality, $|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|$ $\leq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d$ $= (4-a)+(a-b)+(b-c)+(c-d)+d=4$, with equality if and only if each of $4z_0^5$, $(4-a)z_0^4$, $(a-b)z_0^3$, $(b-c)z_0^2$, $(c-d)z_0$, and $d$ is either zero or in the same direction as all the others when looked at as vectors in the complex plane.

We can now divide into two cases: $d \neq 0$ and $d=0$. If $d \neq 0$, then $4z_0^5$ must be real by the previous argument, so $z_0$ is a fifth root of unity. Also, $(4-a)z_0^4$, $(a-b)z_0^3$, $(b-c)z_0^2$, and $(c-d)z_0$ must all be zero because if $z_0$ is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, $a=4$, $b=a$, $c=b$, and $d=c$, leading to the solution $(a,b,c,d)=(4,4,4,4)$. Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.

If $d=0$, then each of $4z_0^5$, $(4-a)z_0^4$, $(a-b)z_0^3$, $(b-c)z_0^2$, and $cz_0$ must either be zero or in the same direction as all the others, so each of $4z_0^4$, $(4-a)z_0^3$, $(a-b)z_0^2$, $(b-c)z_0$, and $c$ must either be zero or in the same direction as all the others. We can divide this into two cases: $c \neq 0$ and $c=0$. If $c \neq 0$, then $4z_0^4$ must be real. Then, $z_0$ is a fourth root of unity. If $z_0$ is not a second root of unity, $(4-a)z_0^3$, $(a-b)z_0^2$, and $(b-c)z_0$ must be zero, implying that $a=4$, $b=a=4$, and $c=b=4$, leading to the solution $(a,b,c,d)=(4,4,4,0)$. If $z_0$ is also a second root of unity, $(4-a)z_0^3$ and $(b-c)z_0$ must be zero but $(a-b)z_0^2$ can be anything. This implies $a=4$ and $b=c$ with no other restrictions, leading to the new solutions $(a,b,c,d) = (4,3,3,0), (4,2,2,0), (4,1,1,0), (4,0,0,0)$.

If $c=0$, then we can similarly show that each of $4z_0^3$, $(4-a)z_0^2$, $(a-b)z_0$, and $b$ must be zero or in the same direction as all the others. If $b \neq 0$, then $z_0$ must be a third root of unity, so $(4-a)z_0^2$ and $(a-b)z_0$ must be zero, implying $a=b=4$, leading to the new solution $(a,b,c,d)=(4,4,0,0)$.

If $b=0$, then we can similarly show that each of $4z_0^2$, $(4-a)z_0$, and $a$ must be zero or in the same direction as the others. For $|z_0|=1$, $a = 4$, but we have already counted the solution $(a,b,c,d)=(4,0,0,0)$.

Then, the complete list of solutions is $(a,b,c,d)=(4,4,4,4),(4,4,4,0),(4,3,3,0),(4,2,2,0),(4,1,1,0),(4,0,0,0),(4,4,0,0)$, leading to a sum of $\framebox{B}=92$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12b/278

~dolphin7

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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