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

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Solution 3==
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<math>p(x)-p(2)</math> is a cubic with two integral real roots, therefore it has three real roots, which are all integers. There are exactly two distinct roots, so either <math>p(x)=p(2)+(x-2)^2(x-m)</math> or <math>p(x)=p(2)+(x-2)(x-m)^2</math> with <math>m\neq 2</math>. In the first case <math>p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)</math>, so <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1,3,4</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>410</math> polynomials. In the second case <math>p(x)=x^3-(1+2m)x^2+(4m+m^2)x-2m^2+p(2)</math>, and <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving 328 polynomials. The total is <math>\boxed{738}</math>.
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~EVIN-
  
 
==See also==
 
==See also==

Revision as of 16:23, 8 February 2023

Problem 9

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

Solution 1

Plugging $2$ into $P(x)$, we get $8+4a+2b+c = m^3+am^2+bm+c$. We can rewrite into $(2-m)(m^2+2m+4+a(2+m)+b)=0$, where $c$ can be any value in the range. Since $m\neq2, m^2+2m+4+a(2+m)+b$ must be $0$. The problem also asks for unique integers, meaning $m$ can only be one value for each polynomial, so the discriminant must be $0$. $m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0$, and $(2+a)^2-4(2a+b+4)=0$. Rewrite to be $a(a-4)=4(b+3)$. $a$ must be even for $4(b+3)$ to be an integer. $-6<=a<=10$ because $4(20+3) = 92$. However, plugging in $a=-6, b=12$ result in $m=2$. There are 8 pairs of $(a,b)$ and 41 integers for $c$, giving\[41\cdot8 = \boxed{328}\] ~chem1kall

Solution 2

Define q $\left( x \right) = p \left( x \right) - p \left( 2 \right)$. Hence, for $q \left( x \right)$, beyond having a root 2, it has a unique integer root that is not equal to 2.

We have $q(x)=p(x)-p(2)=(x-2)( (x^2+2x+4) + a(x+2) + b)$ Thus, the polynomial $x^2 + \left( 2 + a \right) x + 4 + 2a + b$ 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 $- 2 - a$. Because $a$ 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, $2 | a$. Denote $d = \frac{a-2}{2}$. Thus, (2) can be written as\[b = d^2 - 4 . \hspace{1cm} (3)\] Because $a \in \left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}$, $2 | a$, and $2 + a \neq -4$, we have $d \in \left\{ - 11, - 10, \cdots, 9 \right\} \backslash \left\{ 4 \right\}$.

Therefore, we have the following feasible solutions for $\left( b, d \right)$: $\left( -4 , 0 \right)$, $\left( -3 , \pm 1 \right)$, $\left( 0 , \pm 2 \right)$, $\left( 5, \pm 3 \right)$, $\left( 12 , 4 \right)$. Thus, the total number of $\left( b, d \right)$ is 8.

Because $c$ can take any value from $\left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}$, the number of feasible $c$ is 41.

Therefore, the number of $\left( a, b, c \right)$ is $8 \cdot 41 = \boxed{\textbf{(328) }}$.


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3

$p(x)-p(2)$ is a cubic with two integral real roots, therefore it has three real roots, which are all integers. There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$ with $m\neq 2$. In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, so $m$ can be $-6,-5,-4,-3,-2,-1,0,1,3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials. In the second case $p(x)=x^3-(1+2m)x^2+(4m+m^2)x-2m^2+p(2)$, and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving 328 polynomials. The total is $\boxed{738}$.

~EVIN-

See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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