Difference between revisions of "2023 AIME I Problems/Problem 9"

(Solution)
(Solution)
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<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
 
<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
  
==Solution==
+
Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
 
+
are integers in <math>{-20, -10, -18, . . . , 18, 19, 20}</math>, such that there is a unique integer
Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>.
+
<math>m \neq 2</math> with <math>p(m) = p(2)</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
 
<cmath>
 
\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*}
 
</cmath>
 
 
 
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,
 
<cmath>
 
\[
 
\left( 2 + a \right)^2 = 4 \left( 4 + 2a + b \right) . \hspace{1cm} (1)
 
\]
 
</cmath>
 
 
 
In addition, because two identical roots are not 2, we have
 
<cmath>
 
\[
 
2 + a \neq - 4 .
 
\]
 
</cmath>
 
 
 
Equation (1) can be reorganized as
 
<cmath>
 
\[
 
4 b = \left( a - 2 \right)^2 - 16 .  \hspace{1cm} (2)
 
\]
 
</cmath>
 
 
 
Thus, <math>2 | a</math>. Denote <math>d = \frac{a-2}{2}</math>.
 
Thus, (2) can be written as
 
<cmath>
 
\[
 
b = d^2 - 4 .  \hspace{1cm} (3)
 
\]
 
</cmath>
 
 
 
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)
 
  
 
==Solution==
 
==Solution==

Revision as of 15:50, 8 February 2023

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are integers in ${−20, −10, −18, . . . , 18, 19, 20}$ (Error compiling LaTeX. Unknown error_msg), such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$.

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are integers in ${-20, -10, -18, . . . , 18, 19, 20}$, such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$.

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are integers in ${-20, -10, -18, . . . , 18, 19, 20}$, such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$.

Solution

It can be easily noticed that $c$ is independent of the condition $P(m) = P(2)$, and can thus safely take all $41$ possible values between $-20$ and $20$.

There are two possible ways for $m\ne2$ to be the only integer satisfying $P(m) = P(2)$: $P$ has a double root at $2$ or a double root at $m$.

Case 1: $P$ has a double root at $2$:

In this case, $\frac{dP}{dx}(2) = 0$, or $12 + 4a + b = 0$. Thus $a$ ranges from $-8$ to $2$. One of these values, $(a,b) = (-6,-12)$ corresponds to a triple root at $2$, which means $m=2$. Thus there are $10$ possible values of $(a,b)$. (It can be verified that $m$ is an integer).

Case 2: $P$ has a double root at $m$:

See the above solution. This yields $8$ possible combinations of $a$ and $b$.

Thus, in total we have $41*18 = \boxed{738}$ combinations of $(a,b,c)$.


-Alex_Z