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

Line 5: | Line 5: | ||

<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 (doubtful) == | ==Solution (doubtful) == | ||

− | Since < | + | 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 < | + | 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 < | + | 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 < | + | 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 < | + | 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 04:03, 6 December 2012

## Problem 23

Consider all polynomials of a complex variable, , where and are integers, , and the polynomial has a zero with What is the sum of all values over all the polynomials with these properties?

## Solution (doubtful)

Since is a root of , and has integer coefficients, must be algebraic. Since is algebraic and lies on the unit circle, must be a root of unity (Comment: this is not true. See this link: [1]). Since has degree 4, it seems reasonable (and we will assume this only temporarily) that must be a 2nd, 3rd, or 4th root of unity. These are among the set . Since complex roots of polynomials come in conjugate pairs, we have that has one (or more) of the following factors: , , , or . If then ; a contradiction since are non-negative. On the other hand, suppose . Then . This implies while correspondingly. After listing cases, the only such valid are , , , , and .

Now suppose . Then whereupon and . But then and . This gives only the cases equals , which we have already counted in a previous case.

Suppose . Then so that and . This only gives rise to equal which we have previously counted.

Finally suppose divides . Using polynomial division ((or that to make the same deductions) we ultimately obtain that . This can only happen if is .

Hence we've the polynomials However, by inspection has roots on the unit circle, because 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 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 in an th root of unity where , and is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If is prime, then \textit{every} th root of unity except 1 must satisfy our polynomial, but since and the degree of our polynomial is 4, this is impossible. Suppose is composite. If it has a prime factor greater than 5 then again every th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose is divisible only by 2,3,or 5. Since by hypothesis is not a 2nd or 3rd root of unity, 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 . But has exactly all 5th roots of unity excluding 1, and . Thus this must divide which implies . This completes the proof.

## Solution

First, assume that , so or . does not work because . Assume that . Then , we have , so . Also, has to be true since . Now deduces , therefore the only possible choices for are . In these cases, . The sum of over these cases is .

Second, assume that , so for some real , . By conjugate roots theorem we have that , therefore is a factor of , and we may assume that

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

From the first and the third we may deduce that and that , if (we will consider by the end). Let . From the second equation, we know that is a non-negative integer. Consider the following cases:

Case 1: . Then , , so , . . In this case, has a root . However, this case is already covered before (). Case 2: . Then since , we have . However, , therefore . This is true only when . Also, we get again. In this case, , so , , . has a root . . Last case: . We have and that has a root . .

Therefore the desired sum is .

The case of . We get . In this case, satisfies the conditions.