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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 17"

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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad</math>
 
<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad</math>
  
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==Solution==
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==Solution 1==
  
 
If a [[quadratic equation]] does not have two distinct real solutions, then its [[discriminant]] must be <math>\le0</math>. So, <math>b^2-4c\le0</math> and <math>c^2-4b\le0</math>. By inspection, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of positive integers that fulfill these criteria: <math>(1,1)</math>, <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, and <math>(4,4)</math>.
 
If a [[quadratic equation]] does not have two distinct real solutions, then its [[discriminant]] must be <math>\le0</math>. So, <math>b^2-4c\le0</math> and <math>c^2-4b\le0</math>. By inspection, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of positive integers that fulfill these criteria: <math>(1,1)</math>, <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, and <math>(4,4)</math>.
 +
 +
== Solution 2 ==
 +
We need to solve the following system of inequalities:
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{ll}
 +
b^2 - 4 c \leq 0 \\
 +
c^2 - 4 b \leq 0
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Feasible solutions are in the region formed between two parabolas <math>b^2 - 4 c = 0</math> and <math>c^2 - 4 b = 0</math>.
 +
 +
Define <math>f \left( b \right) = \frac{b^2}{4}</math> and <math>g \left( b \right) = 2 \sqrt{b}</math>.
 +
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
 +
 +
For <math>b = 1</math>, <math>f \left( b \right) = \frac{1}{4}</math> and <math>g \left( b \right) = 2</math>.
 +
Hence, the feasible <math>c</math> are 1, 2.
 +
 +
For <math>b = 2</math>, <math>f \left( b \right) = 1</math> and <math>g \left( b \right) = 2 \sqrt{2}</math>.
 +
Hence, the feasible <math>c</math> are 1, 2.
 +
 +
For <math>b = 3</math>, <math>f \left( b \right) = \frac{9}{4}</math> and <math>g \left( b \right) = 2 \sqrt{3}</math>.
 +
Hence, the feasible <math>c</math> is 3.
 +
 +
For <math>b = 4</math>, <math>f \left( b \right) = 4</math> and <math>g \left( b \right) = 4</math>.
 +
Hence, the feasible <math>c</math> is 4.
 +
 +
For <math>b > 4</math>, <math>f \left( b \right) > g \left( b \right)</math>. Hence, there is no feasible <math>c</math>.
 +
 +
Putting all cases together, the correct answer is <math>\boxed{\textbf{(B) }6}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
  
 
{{AMC12 box|year=2021 Fall|ab=A|num-a=18|num-b=16}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-a=18|num-b=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:26, 25 November 2021

Problem

For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$

Solution 1

If a quadratic equation does not have two distinct real solutions, then its discriminant must be $\le0$. So, $b^2-4c\le0$ and $c^2-4b\le0$. By inspection, there are $\boxed{\textbf{(B) } 6}$ ordered pairs of positive integers that fulfill these criteria: $(1,1)$, $(1,2)$, $(2,1)$, $(2,2)$, $(3,3)$, and $(4,4)$.

Solution 2

We need to solve the following system of inequalities: \[ \left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right.. \]

Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$.

Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \sqrt{b}$. Therefore, all feasible solutions are in the region formed between the graphs of these two functions.

For $b = 1$, $f \left( b \right) = \frac{1}{4}$ and $g \left( b \right) = 2$. Hence, the feasible $c$ are 1, 2.

For $b = 2$, $f \left( b \right) = 1$ and $g \left( b \right) = 2 \sqrt{2}$. Hence, the feasible $c$ are 1, 2.

For $b = 3$, $f \left( b \right) = \frac{9}{4}$ and $g \left( b \right) = 2 \sqrt{3}$. Hence, the feasible $c$ is 3.

For $b = 4$, $f \left( b \right) = 4$ and $g \left( b \right) = 4$. Hence, the feasible $c$ is 4.

For $b > 4$, $f \left( b \right) > g \left( b \right)$. Hence, there is no feasible $c$.

Putting all cases together, the correct answer is $\boxed{\textbf{(B) }6}$.

~Steven Chen (www.professorchenedu.com)


2021 Fall AMC 12A (Problems • Answer Key • Resources)
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Problem 16 		Followed by
Problem 18
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All AMC 12 Problems and Solutions

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