Difference between revisions of "2023 AIME I Problems/Problem 9"
(→Solution 1) |
|||
Line 12: | Line 12: | ||
===Solution 2=== | ===Solution 2=== | ||
a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328? | a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328? | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>. | ||
+ | Hence, for <math>q \left( x \right)</math>, beyond having a root 2, it has a unique integer root that is not equal to 2. | ||
+ | |||
+ | We have | ||
+ | \begin{align*} | ||
+ | q \left( x \right) & = p \left( x \right) - p \left( 2 \right) \ | ||
+ | & = \left( x - 2 \right) | ||
+ | \left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) . | ||
+ | \end{align*} | ||
+ | |||
+ | Thus, the polynomial <math>x^2 + \left( 2 + a \right) x + 4 + 2a + b</math> has a unique integer root and it is not equal to 2. | ||
+ | |||
+ | Following from Vieta' formula, the sum of two roots of this polynomial is <math>- 2 - a</math>. | ||
+ | Because <math>a</math> is an integer, if a root is an integer, the other root is also an integer. | ||
+ | Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0. | ||
+ | Thus, | ||
+ | \[ | ||
+ | \left( 2 + a \right)^2 = 4 \left( 4 + 2a + b \right) . \hspace{1cm} (1) | ||
+ | \] | ||
+ | |||
+ | In addition, because two identical roots are not 2, we have | ||
+ | \[ | ||
+ | 2 + a \neq - 4 . | ||
+ | \] | ||
+ | |||
+ | Equation (1) can be reorganized as | ||
+ | \[ | ||
+ | 4 b = \left( a - 2 \right)^2 - 16 . \hspace{1cm} (2) | ||
+ | \] | ||
+ | |||
+ | Thus, <math>2 | a</math>. Denote <math>d = \frac{a-2}{2}</math>. | ||
+ | Thus, (2) can be written as | ||
+ | \[ | ||
+ | b = d^2 - 4 . \hspace{1cm} (3) | ||
+ | \] | ||
+ | |||
+ | Because <math>a \in \left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}</math>, <math>2 | a</math>, and <math>2 + a \neq -4</math>, we have <math>d \in \left\{ - 11, - 10, \cdots, 9 \right\} \backslash \left\{ 4 \right\}</math>. | ||
+ | |||
+ | Therefore, we have the following feasible solutions for <math>\left( b, d \right)</math>: <math>\left( -4 , 0 \right)</math>, <math>\left( -3 , \pm 1 \right)</math>, <math>\left( 0 , \pm 2 \right)</math>, <math>\left( 5, \pm 3 \right)</math>, <math>\left( 12 , 4 \right)</math>. | ||
+ | Thus, the total number of <math>\left( b, d \right)</math> is 8. | ||
+ | |||
+ | Because <math>c</math> can take any value from <math>\left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}</math>, the number of feasible <math>c</math> is 41. | ||
+ | |||
+ | Therefore, the number of <math>\left( a, b, c \right)</math> is <math>8 \cdot 41 = \boxed{\textbf{(328) }}</math>. | ||
+ | |||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 14:22, 8 February 2023
Contents
[hide]Problem (Unofficial, please update when official one comes out):
is a polynomial with integer coefficients in the range
, inclusive. There is exactly one integer
such that
. How many possible values are there for the ordered triple
?
Solution
Solution 1
Plugging into
, we get
. We can rewrite into
, where
can be any value in the range. Since
must be
. The problem also asks for unique integers, meaning
can only be one value for each polynomial, so the discriminant must be
.
, and
. Rewrite to be
.
must be even for
to be an integer.
because
. There are 9 pairs of
and 41 integers for
, giving
~chem1kall
Solution 2
a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328?
Solution
Define .
Hence, for
, beyond having a root 2, it has a unique integer root that is not equal to 2.
We have
Thus, the polynomial has a unique integer root and it is not equal to 2.
Following from Vieta' formula, the sum of two roots of this polynomial is .
Because
is an integer, if a root is an integer, the other root is also an integer.
Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0.
Thus,
\[
\left( 2 + a \right)^2 = 4 \left( 4 + 2a + b \right) . \hspace{1cm} (1)
\]
In addition, because two identical roots are not 2, we have \[ 2 + a \neq - 4 . \]
Equation (1) can be reorganized as \[ 4 b = \left( a - 2 \right)^2 - 16 . \hspace{1cm} (2) \]
Thus, . Denote
.
Thus, (2) can be written as
\[
b = d^2 - 4 . \hspace{1cm} (3)
\]
Because ,
, and
, we have
.
Therefore, we have the following feasible solutions for :
,
,
,
,
.
Thus, the total number of
is 8.
Because can take any value from
, the number of feasible
is 41.
Therefore, the number of is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)