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

Revision as of 22:32, 11 November 2018

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 (dubious)

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 $60+12+20=92 ...\framebox{B}$ .

@@ Line 26: / Line 26: @@
 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 ==
+== 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>.

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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Difference between revisions of "2012 AMC 12B Problems/Problem 23"

Revision as of 22:32, 11 November 2018

Contents

Problem 23

Solution (dubious)

Solution 2

See Also