2005 AMC 12A Problems/Problem 24
Contents
Problem
Let . For how many polynomials
does there exist a polynomial
of degree
such that
?
Solution 1
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 polynomials which are linear, rather than quadratic. Namely,
Clearly, we could not have included any other constant functions. For any linear function, we have because
is the
-coordinate of the midpoint of
and
. So we have not included any other linear functions. Therefore, the desired answer is
.
Solution 2
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
.) However, we have two functions which are constant, which are
and
. So we subtract
to get a final answer of
.
~Williamgolly
Solution 3 (fastest)
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 (i.e. that give
), and
that are linear (i.e. that map
or
). Hence there are
solutions, so
.
~Maximilian113
Non-Solution 4 (guessing, only if desperate!)
As must have degree
, let its roots be
,
, and
, giving
If we now guess that is a quadratic polynomial, it follows that e.g.
must be equal to a product of
factors from the expression
. As this expression contains
factors in total, the number of ways of choosing
factors from it, where the order matters, would be
.
While none of the answer choices are , we can now eliminate
as
. Looking for answers that are similar, we further observe that
,
, or
would have been derived from a
in the actual solution, to which
was possibly added or subtracted. We can therefore guess, if sufficiently desperate, that the answer should be either
or
.
On the other hand, if you're desperate for time, it's probably a better idea to go back and review Problems to
, rather than guessing on Problem
!
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.