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Difference between revisions of "2012 AMC 12B Problems/Problem 23"
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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>
 
<math>\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 </math>
  
−
== Solution ==
 
−
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>.
 
  
−
Second, assume that </math>z_0\in \mathbb{C} \backslash \mathbb{R}<math>.
 
  
 
==Solution (doubtful) ==
 
==Solution (doubtful) ==
  
−
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>.  
  
−
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.
+
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.
  
−
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.  
+
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.  
  
−
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>.  
+
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
Line 27: Line 24:
 
<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)$. 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 ==
 +
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> 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>.
 +
 
 +
Second, assume that <math>z_0\in \mathbb{C} \backslash \mathbb{R}</math>, so <math>z_0=x_0+iy_0</math> for some real <math>x_0, y_0</math>, <math>|x_0|<1</math>. By conjugate roots theorem we have that <math>P(z_0)=P(z_0^{*})=0</math>, therefore <math>(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0 + 1)</math> is a factor of <math>P(z)</math>, and we may assume that
 +
 
 +
<cmath>P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)</cmath>
 +
 
 +
for some real <math>p</math>. Expanding this polynomial and comparing the coefficients, we have the following equations:
 +
 
 +
<cmath>p-8x_0 = a</cmath>
 +
<cmath>d+4-2px_0 = b</cmath>
 +
<cmath>p-2dx_0 = c</cmath>
 +
 
 +
From the first and the third we may deduce that <math>2x_0 = \frac{a-c}{d-4}</math> and that <math>p=\frac{da-4c}{d-4}</math>, if <math>d\neq 4</math> (we will consider <math>d=4</math> by the end). Let <math>k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}</math>. From the second equation, we know that <math>k=d+4-b</math> is a non-negative integer. Consider the following cases:
 +
 
 +
Case 1: <math>a=c</math>. Then <math>k=0</math>, <math>b=d+4</math>, so <math>a=b=c=4</math>, <math>d=0</math>. <math>x_0=0</math>. In this case, <math>P(z)</math> has a root <math>z_0=i</math>. However, this case is already covered before (<math>t=4</math>).
 +
Case 2: <math>a>c\geq 0</math>. Then since <math>k\geq 0</math>, we have <math>da-4c\geq 0</math>. However, <math>da \leq 4c</math>, therefore <math>da-4c=0</math>. This is true only when <math>d=c</math>. Also, we get <math>k=0</math> again. In this case, <math>b=d+4</math>, so <math>a=b=4</math>, <math>c=d=0</math>, <math>x_0=-1/2</math>. <math>P(z)</math> has a root <math>z_0=e^{i2\pi/3}</math>. <math>P(1)=12</math>.
 +
Last case: <math>d=4</math>. We have <math>a=b=c=d=4</math> and that <math>P(z)</math> has a root <math>z_0=e^{i2\pi/5}</math>. <math>P(1)=20</math>.
 +
 
 +
Therefore the desired sum is <math>60+12+20=92 ...\framebox{B}</math>.
 +
 
 +
 
 +
The case of <math>d=4</math>. We get <math>a=b=c=d=4</math>. In this case, <math>z_0=e^{i2\pi/5}</math> satisfies the conditions.

Revision as of 03:03, 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 (doubtful)

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

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 $z_0\in \mathbb{C} \backslash \mathbb{R}$, 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 $(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0 + 1)$ 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 a non-negative integer. Consider the following cases:

Case 1: $a=c$. Then $k=0$, $b=d+4$, so $a=b=c=4$, $d=0$. $x_0=0$. In this case, $P(z)$ has a root $z_0=i$. However, this case is already covered before ($t=4$). 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 $60+12+20=92 ...\framebox{B}$.


The case of $d=4$. We get $a=b=c=d=4$. In this case, $z_0=e^{i2\pi/5}$ satisfies the conditions.

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