2012 AMC 12B Problems/Problem 23

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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 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



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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