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

Latest revision as of 16:25, 6 September 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?

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

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

@@ Line 1: / Line 1: @@
-== Problem 23 ==
+== 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?
 <math>\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 </math>
+== 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>.
@@ Line 63: / Line 59: @@
 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>.
+== Video Solution by Richard Rusczyk ==
+https://artofproblemsolving.com/videos/amc/2012amc12b/278
+~dolphin7
 == See Also ==

2012 AMC 12B (Problems • Answer Key • Resources)
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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