Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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== Problem == | == Problem == | ||
− | Let <math>P(x)=(x-1)(x-2)(x-3)</math>. For how many | + | Let <math>P(x)=(x-1)(x-2)(x-3)</math>. For how many [[polynomial]]s <math>Q(x)</math> does there exist a polynomial <math>R(x)</math> of degree 3 such that <math>P(Q(x))=P(x) \cdot R(x)</math>? |
+ | |||
+ | |||
+ | <math>\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32</math> | ||
== Solution == | == Solution == | ||
− | {{ | + | We can write the problem as |
+ | <div style="text-align:center;"> | ||
+ | <math>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x)</math>. | ||
+ | </div> | ||
+ | |||
+ | |||
+ | Since <math>\deg P(x) = 3</math> and <math>\deg R(x) = 3</math>, <math>\deg P(x)\cdot R(x) = 6</math>. Thus, <math>\deg P(Q(x)) = 6</math>, so <math>\deg Q(x) = 2</math>. | ||
+ | <div style="text-align:center;"> | ||
+ | <math> | ||
+ | P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\cdot R(1)=0, | ||
+ | </math><br /><math> | ||
+ | P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\cdot R(2)=0, | ||
+ | </math><br /><math> | ||
+ | P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\cdot R(3)=0. | ||
+ | </math> | ||
+ | </div> | ||
+ | Hence, we conclude <math>Q(1)</math>, <math>Q(2)</math>, and <math>Q(3)</math> must each be <math>1</math>, <math>2</math>, or <math>3</math>. Since a [[quadratic equation|quadratic]] is uniquely determined by three points, there can be <math>3*3*3 = 27</math> different quadratics <math>Q(x)</math> after each of the values of <math>Q(1)</math>, <math>Q(2)</math>, and <math>Q(3)</math> are chosen. | ||
+ | |||
+ | |||
+ | However, we have included <math>Q(x)</math> which are not quadratics: lines. Namely, | ||
+ | <div style="text-align:center;"> | ||
+ | <math> | ||
+ | Q(1)=Q(2)=Q(3)=1 \Rightarrow Q(x)=1, | ||
+ | </math><br /><math> | ||
+ | Q(1)=Q(2)=Q(3)=2 \Rightarrow Q(x)=2, | ||
+ | </math><br /><math> | ||
+ | Q(1)=Q(2)=Q(3)=3 \Rightarrow Q(x)=3, | ||
+ | </math><br /><math> | ||
+ | Q(1)=1, Q(2)=2, Q(3)=3 \Rightarrow Q(x)=x, | ||
+ | </math><br /><math> | ||
+ | Q(1)=3, Q(2)=2, Q(3)=1 \Rightarrow Q(x)=4-x. | ||
+ | </math> | ||
+ | </div> | ||
+ | Clearly, we could not have included any other constant functions. For any linear function, we have <math>2\cdot Q(2) = Q(1) + Q(3)</math> because <math>Q(2)</math> is y-value of the midpoint of <math>(1, Q(1))</math> and <math>(3, Q(3))</math>. So we have not included any other linear functions. Therefore, the desired answer is <math>27 - 5 = \boxed{\textbf{(B) }22}</math>. | ||
+ | |||
+ | ==Quicker Solution== | ||
+ | We see that | ||
+ | <cmath>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x).</cmath> | ||
+ | Therefore, <math>P(x) | P(Q(x))</math>. Since <math>\deg Q = 2,</math> we must have <math>x-1, x-2, x-3</math> divide <math>P(Q(x))</math>. So, we pair them off with one of <math>Q(x)-1, Q(x)-2,</math> and <math>Q(x)-3</math> to see that there are <math>3!+3 \cdot 2 \cdot \binom{3}{2} = 24</math> without restrictions. (Note that this count was made by pairing off linear factors of <math>P(x)</math> with <math>Q(x)-1, Q(x)-2,</math> and <math>Q(x)-3</math>, and also note that the degree of <math>Q</math> is 2.) However, we have two functions which are constant, which are <math>Q(x) = x</math> and <math>Q(x) = 4-x.</math> So, we subtract <math>2</math> to get a final answer of <math>\boxed{22} \implies \boxed{B}</math>. | ||
+ | |||
+ | ~Williamgolly | ||
+ | |||
+ | ==Quickest Solution== | ||
+ | Obviously, <math>\deg Q = 2</math> and the RHS has roots <math>1, 2, 3.</math> Thus, the only requirement is that <math>Q</math> is a quadratic that maps each of the numbers <math>1, 2, 3</math> to one of <math>{1, 2, 3}.</math> It suffices to count the number of such mappings, as this uniquely determines <math>Q</math>, through polynomial interpolation. Thus there are <math>3^3</math> possibilities. But we have three functions that are constant (<math>Q(1) =Q(2) = Q(3)</math>), and <math>2</math> that are linear (<math>\{1, 2, 3\} \rightarrow \{3, 2, 1\}</math> and <math>\{1, 2, 3\} \rightarrow \{1, 2, 3\}</math>). Hence there are <math>3^3-3-2 = \boxed{22}</math> solutions, so <math>\boxed{B}</math>. | ||
+ | |||
+ | ~Maximilian113 | ||
+ | |||
+ | ==Guessing Solution(desperate),== | ||
+ | |||
+ | <cmath>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x).</cmath> | ||
+ | |||
+ | rewrite it as | ||
+ | <cmath>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)(x-r1)(x-r2)(x-r3).</cmath> | ||
+ | |||
+ | say Q(x)= 2nd degree polymonial | ||
+ | |||
+ | that means (Q(x)-1) must equal to 2 factors of (R(x) times P(x)) | ||
+ | |||
+ | we have 6 factors <cmath>(x-1)(x-2)(x-3)(x-r1)(x-r2)(x-r3).</cmath> | ||
+ | |||
+ | We need 2 factors,so it must be | ||
+ | 6 choices, choose 2 or | ||
+ | |||
+ | 6!/4!=30 | ||
+ | none of choices are 30, so lets use the answers | ||
+ | <math>\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32</math> | ||
+ | |||
+ | it cannot be E because it is above 30. Now we look for answers that are similar | ||
+ | |||
+ | so we see 22,27,32 which shows that we added or subtracted 5 from 27 in the actual solution. | ||
+ | since it can't be greater than 30, the answer should be 22 or 27. Which means choose B or D if you are desperate. | ||
+ | |||
+ | |||
+ | only use if you are desperate | ||
+ | Note: If you're desperate, please go back and review 1-20 instead of guessing on a 24 | ||
+ | |||
== See also == | == See also == | ||
− | + | {{AMC12 box|year=2005|ab=A|num-b=23|num-a=25}} | |
− | + | ||
− | + | [[Category:Intermediate Algebra Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 23:42, 29 September 2024
Contents
Problem
Let . For how many polynomials does there exist a polynomial of degree 3 such that ?
Solution
We can write the problem as
.
Since and , . Thus, , so .
Hence, we conclude , , and must each be , , or . Since a quadratic is uniquely determined by three points, there can be different quadratics after each of the values of , , and are chosen.
However, we have included which are not quadratics: lines. Namely,
Clearly, we could not have included any other constant functions. For any linear function, we have because is y-value of the midpoint of and . So we have not included any other linear functions. Therefore, the desired answer is .
Quicker Solution
We see that Therefore, . Since we must have divide . So, we pair them off with one of and to see that there are without restrictions. (Note that this count was made by pairing off linear factors of with and , and also note that the degree of is 2.) However, we have two functions which are constant, which are and So, we subtract to get a final answer of .
~Williamgolly
Quickest Solution
Obviously, and the RHS has roots Thus, the only requirement is that is a quadratic that maps each of the numbers to one of It suffices to count the number of such mappings, as this uniquely determines , through polynomial interpolation. Thus there are possibilities. But we have three functions that are constant (), and that are linear ( and ). Hence there are solutions, so .
~Maximilian113
Guessing Solution(desperate),
rewrite it as
say Q(x)= 2nd degree polymonial
that means (Q(x)-1) must equal to 2 factors of (R(x) times P(x))
we have 6 factors
We need 2 factors,so it must be 6 choices, choose 2 or
6!/4!=30 none of choices are 30, so lets use the answers
it cannot be E because it is above 30. Now we look for answers that are similar
so we see 22,27,32 which shows that we added or subtracted 5 from 27 in the actual solution. since it can't be greater than 30, the answer should be 22 or 27. Which means choose B or D if you are desperate.
only use if you are desperate
Note: If you're desperate, please go back and review 1-20 instead of guessing on a 24
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.