2021 Fall AMC 12A Problems/Problem 17

Revision as of 20:26, 25 November 2021 by Stevenyiweichen (talk | contribs)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png