2005 AMC 12A Problems/Problem 24
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
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
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 |
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