Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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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>. | 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 | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | 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, Q(x)-3</math> to see that there are <math>3!+3 \cdot 2 \cdot \binom{3}{2}</math> = 24 without restrictions (the count was made by pairing off the linear factors of <math>P(x)</math> with <math>Q(x)-1, Q(x)-2, Q(x)-3</math> and notice 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> The answer is <math>22</math>. | ||
+ | |||
+ | ~Williamgolly | ||
== See also == | == See also == |
Revision as of 00:01, 3 March 2021
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 to see that there are = 24 without restrictions (the count was made by pairing off the linear factors of with and notice that the degree of is 2). However, we have two functions which are constant, which are and The answer is .
~Williamgolly
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.