Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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== 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;"> | <div style="text-align:center;"> | ||
<math> | <math> | ||
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− | However, we have included <math>Q(x)</math> which are not quadratics. Namely, | + | However, we have included <math>Q(x)</math> which are not quadratics: lines. Namely, |
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
<math> | <math> | ||
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</math> | </math> | ||
</div> | </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> | + | 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>. |
== See also == | == See also == |
Revision as of 22:57, 26 November 2018
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 .
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.