Difference between revisions of "2016 AIME I Problems/Problem 11"
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<cmath>(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)</cmath> | <cmath>(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)</cmath> | ||
<cmath>Q(x+1) = Q(x)</cmath> | <cmath>Q(x+1) = Q(x)</cmath> | ||
− | From here, we know that <math>Q(x) = C</math> for a constant <math>C</math>, so <math>P(x) = Cx(x-1)(x+1)</math>. We know that <math>\left(P(2)\right)^2 = P(3)</math>. Plugging those into our definition of <math>P(x)</math>: <math>(C \cdot 2 \cdot (2-1) \cdot (2+1))^2 = C \cdot 3 \cdot (3-1) \cdot (3+1) \Rightarrow (6C)^2 = 24C \Rightarrow 36C^2 - 24C = 0 \Rightarrow C = / | + | From here, we know that <math>Q(x) = C</math> for a constant <math>C</math>, so <math>P(x) = Cx(x-1)(x+1)</math>. We know that <math>\left(P(2)\right)^2 = P(3)</math>. Plugging those into our definition of <math>P(x)</math>: <math>(C \cdot 2 \cdot (2-1) \cdot (2+1))^2 = C \cdot 3 \cdot (3-1) \cdot (3+1) \Rightarrow (6C)^2 = 24C \Rightarrow 36C^2 - 24C = 0 \Rightarrow C = 0</math> or <math>\frac{2}{3}</math>. So we know that <math>P(x) = \frac{2}{3}x(x-1)(x+1)</math>. So <math>P(\frac{7}{2}) = \frac{2}{3} \cdot \frac{7}{2} \cdot (\frac{7}{2} - 1) \cdot (\frac{7}{2} + 1) = \frac{105}{4}</math>. Thus, the answer is <math>105 + 4 = \boxed{109}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 09:20, 5 March 2016
Problem
Let be a nonzero polynomial such that
for every real
, and
. Then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Plug in to get
. Plug in
to get
. Now plug in
to get that
. So
for some polynomial
. Using the initial equation, once again,
From here, we know that
for a constant
, so
. We know that
. Plugging those into our definition of
:
or
. So we know that
. So
. Thus, the answer is
.
Solution 2
So from the equation we see that divides
and
divides
so we can conclude that
and
divide
. This means that
and
are roots of
. Plug in
and we see that
so
is also a root.
Suppose we had another root that is not those . Notice that the equation above indicates that if
is a root then
and
is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.
That means . We can use
to get
. Plugging in
is now trivial and we see that it is
so our answer is
Solution 3
Although this may not be the most mathematically rigorous answer, we see that . Using a bit of logic, we can make a guess that
has a factor of
, telling us
has a factor of
. Similarly, we guess that
has a factor of
, which means
has a factor of
. Now, since
and
have so many factors that are off by one, we may surmise that when you plug
into
, the factors "shift over," i.e.
, which goes to
. This is useful because these, when divided, result in
. If
, then we get
and
,
. This gives us
and
, and at this point we realize that there has to be some constant
multiplied in front of the factors, which won't affect our fraction
but will give us the correct values of
and
. Thus
, and we utilize
to find
. Evaluating
is then easy, and we see it equals
, so the answer is
Solution 4
As above, we find that . Now for integers
, we know that
Applying this repeatedly, we find that
Therefore, as
, we find
for all positive integers
. This cubic polynomial matches the values
for infinitely many numbers, hence the two polynomials are identically equal. In particular,
, and the answer is
.
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.