Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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<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> | <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 | 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) | + | <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 | ||
<math>\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32</math> | <math>\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32</math> |
Revision as of 19:49, 19 March 2023
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(incorrect)
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
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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