Difference between revisions of "2012 AMC 12B Problems/Problem 23"
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− | == Problem | + | == 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? | 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> | <math>\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 </math> | ||
− | == Solution == | + | == Solution 1 == |
− | + | 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. | |
− | + | 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 < | ||
Hence we've the polynomials | Hence we've the polynomials | ||
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<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 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>. | ||
+ | |||
+ | 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*z + 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 non-negative. | ||
+ | |||
+ | 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>. However, this has already been found (i.e. the form of <math>(4,t,t,0)</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>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | First, notice that <math>z=1</math> cannot be a root of the polynomial because <math>a,b,c,d \geq 0</math>. Multiplying the polynomial by <math>(z-1)</math> yields <math>P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d</math>, so for <math>z \neq 1</math> to be a root of <math>P(z)</math>, <math>4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d</math>. Now we consider the root <math>z_0</math> with <math>|z_0|=1</math>. <math>|4z_0^5| = 4</math>, so the right hand side must have absolute value 4. By the triangle inequality, <math>|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|</math> <math>\leq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d </math> <math>= (4-a)+(a-b)+(b-c)+(c-d)+d=4</math>, with equality if and only if each of <math>4z_0^5</math>, <math>(4-a)z_0^4</math>, <math>(a-b)z_0^3</math>, <math>(b-c)z_0^2</math>, <math>(c-d)z_0</math>, and <math>d</math> 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: <math>d \neq 0</math> and <math>d=0</math>. If <math>d \neq 0</math>, then <math>4z_0^5</math> must be real by the previous argument, so <math>z_0</math> is a fifth root of unity. Also, <math>(4-a)z_0^4</math>, <math>(a-b)z_0^3</math>, <math>(b-c)z_0^2</math>, and <math>(c-d)z_0</math> must all be zero because if <math>z_0</math> is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, <math>a=4</math>, <math>b=a</math>, <math>c=b</math>, and <math>d=c</math>, leading to the solution <math>(a,b,c,d)=(4,4,4,4)</math>. 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 <math>d=0</math>, then each of <math>4z_0^5</math>, <math>(4-a)z_0^4</math>, <math>(a-b)z_0^3</math>, <math>(b-c)z_0^2</math>, and <math>cz_0</math> must either be zero or in the same direction as all the others, so each of <math>4z_0^4</math>, <math>(4-a)z_0^3</math>, <math>(a-b)z_0^2</math>, <math>(b-c)z_0</math>, and <math>c</math> must either be zero or in the same direction as all the others. We can divide this into two cases: <math>c \neq 0</math> and <math>c=0</math>. If <math>c \neq 0</math>, then <math>4z_0^4</math> must be real. Then, <math>z_0</math> is a fourth root of unity. If <math>z_0</math> is not a second root of unity, <math>(4-a)z_0^3</math>, <math>(a-b)z_0^2</math>, and <math>(b-c)z_0</math> must be zero, implying that <math>a=4</math>, <math>b=a=4</math>, and <math>c=b=4</math>, leading to the solution <math>(a,b,c,d)=(4,4,4,0)</math>. If <math>z_0</math> is also a second root of unity, <math>(4-a)z_0^3</math> and <math>(b-c)z_0</math> must be zero but <math>(a-b)z_0^2</math> can be anything. This implies <math>a=4</math> and <math>b=c</math> with no other restrictions, leading to the new solutions <math>(a,b,c,d) = (4,3,3,0), (4,2,2,0), (4,1,1,0), (4,0,0,0)</math>. | ||
+ | |||
+ | If <math>c=0</math>, then we can similarly show that each of <math>4z_0^3</math>, <math>(4-a)z_0^2</math>, <math>(a-b)z_0</math>, and <math>b</math> must be zero or in the same direction as all the others. If <math>b \neq 0</math>, then <math>z_0</math> must be a third root of unity, so <math>(4-a)z_0^2</math> and <math>(a-b)z_0</math> must be zero, implying <math>a=b=4</math>, leading to the new solution <math>(a,b,c,d)=(4,4,0,0)</math>. | ||
+ | |||
+ | If <math>b=0</math>, then we can similarly show that each of <math>4z_0^2</math>, <math>(4-a)z_0</math>, and <math>a</math> must be zero or in the same direction as the others. For <math>|z_0|=1</math>, <math>a = 4</math>, but we have already counted the solution <math>(a,b,c,d)=(4,0,0,0)</math>. | ||
+ | |||
+ | 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 == | ||
+ | |||
+ | {{AMC12 box|year=2012|ab=B|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Latest revision as of 17:25, 6 September 2021
Contents
Problem
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 1
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 2
First, assume that , so
or
.
does not work because
. Assume that
. Then
, we have
, so
. Also,
has to be true since
. Now
gives
, 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 non-negative.
Consider the following cases:
Case 1: . Then
,
, so
,
. However, this has already been found (i.e. the form of
).
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 .
Solution 3
First, notice that cannot be a root of the polynomial because
. Multiplying the polynomial by
yields
, so for
to be a root of
,
. Now we consider the root
with
.
, so the right hand side must have absolute value 4. By the triangle inequality,
, with equality if and only if each of
,
,
,
,
, and
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: and
. If
, then
must be real by the previous argument, so
is a fifth root of unity. Also,
,
,
, and
must all be zero because if
is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore,
,
,
, and
, leading to the solution
. 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 , then each of
,
,
,
, and
must either be zero or in the same direction as all the others, so each of
,
,
,
, and
must either be zero or in the same direction as all the others. We can divide this into two cases:
and
. If
, then
must be real. Then,
is a fourth root of unity. If
is not a second root of unity,
,
, and
must be zero, implying that
,
, and
, leading to the solution
. If
is also a second root of unity,
and
must be zero but
can be anything. This implies
and
with no other restrictions, leading to the new solutions
.
If , then we can similarly show that each of
,
,
, and
must be zero or in the same direction as all the others. If
, then
must be a third root of unity, so
and
must be zero, implying
, leading to the new solution
.
If , then we can similarly show that each of
,
, and
must be zero or in the same direction as the others. For
,
, but we have already counted the solution
.
Then, the complete list of solutions is , leading to a sum of
.
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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