2021 Fall AMC 12A Problems/Problem 17

The following problem is from both the 2021 Fall AMC 10A #20 and 2021 Fall AMC 12A #17, so both problems redirect to this page.


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 (Casework)

A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that:

  1. Since $x^2+bx+c=0$ does not have real solutions, we have $b^2\leq 4c.$
  2. Since $x^2+cx+b=0$ does not have real solutions, we have $c^2\leq 4b.$

Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get \[b^4\leq 16c^2\leq 64b.\] We apply casework to the value of $b:$

  • If $b=1,$ then $1\leq 16c^2\leq 64,$ from which $c=1,2.$
  • If $b=2,$ then $16\leq 16c^2\leq 128,$ from which $c=1,2.$
  • If $b=3,$ then $81\leq 16c^2\leq 192,$ from which $c=3.$
  • If $b=4,$ then $256\leq 16c^2\leq 256,$ from which $c=4.$

Together, there are $\boxed{\textbf{(B) } 6}$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$


Solution 2 (Graphing)

Similar to Solution 1, use the discriminant to get $b^2\leq 4c$ and $c^2\leq 4b$. These can be rearranged to $c\geq \frac{1}{4}b^2$ and $b\geq \frac{1}{4}c^2$. Now, we can roughly graph these two inequalities, letting one of them be the $x$ axis and the other be $y$. The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs: [asy] unitsize(2); Label f;  f.p=fontsize(6);  xaxis("$x$",0,5,Ticks(f, 1.0));  yaxis("$y$",0,5,Ticks(f, 1.0));  real f(real x)  {  return 0.25x^2;  }  real g(real x)  {  return 2*sqrt(x);  }  dot((1,1)); dot((2,1)); dot((1,2)); dot((2,2)); dot((3,3)); dot((4,4)); draw(graph(f,0,sqrt(20))); draw(graph(g,0,5)); [/asy] We are looking for lattice points (since $b$ and $c$ are positive integers), of which we can count $\boxed{\textbf{(B) } 6}$.


Solution 3 (Graphing)

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

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

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

For $b = 4$, we have $f(b) = 4$ and $g(b) = 4$. Hence, the feasible $c$ is $4$.

For $b > 4$, we have $f(b) > g(b)$. Hence, there is no feasible $c$.

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

~Steven Chen (www.professorchenedu.com)

Solution 4 (Oversimplified but Risky)

A quadratic equation $Ax^2+Bx+C=0$ has one real solution if and only if $\sqrt{B^2-4AC}=0.$ Similarly, it has imaginary solutions if and only if $\sqrt{B^2-4AC}<0.$ We proceed as following:

We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case since $\sqrt {b^2-4ac}$ is $0.$ Also, $x^2+3x+3, x^2+2x+2, x^2+x+1$ are always imaginary for both $b$ and $c.$ We also have $x^2+x+2$ along with $x^2+2x+1$ since the latter has one solution, while the first one is imaginary. Therefore, we have $\boxed{\textbf{(B) } 6}$ total ordered pairs of integers.


Solution 5 (Quick and Easy)

We see that $b^2 \leq 4c$ and $c^2 \leq 4b.$ WLOG, assume that $b \geq c.$ Then we have that $b^2 \leq 4c \leq 4b$, so $b^2 \leq 4b$ and therefore $b \leq 4$, also meaning that $c \leq 4.$ This means that we only need to try 16 cases. Now we can get rid of the assumption that $b \geq c$, because we want ordered pairs. For $b = 1$ and $b = 2$, $c = 1$ and $c = 2$ work. When $b = 3$, $c$ can only be $3$, and when $b = 4$, only $c = 4$ works, for a total of $\boxed{\textbf{(B) } 6}$ ordered pairs of integers.


Solution 6 (Fastest)

We need both $b^2\leq 4c$ and $c^2\leq 4b$.

If $b=c$ then the above become $b^2\leq 4b\iff b\leq 4$, so we have four solutions $(k,k)$, where $k=1$,$2$,$3$,$4$.

If $b<c$ then we only need $c^2\leq 4b$ since it implies $b^2< 4c$. Now $c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2$, so $b=1$. We plug $b=1$, $c=2$ back into $c^2\leq 4b$ and it works. So there is another solution $(1,2)$.

By symmetry, if $b>c$ then $(b,c)=(2,1)$.

Therefore the total number of solutions is $\boxed{\textbf{(B) } 6}$.


Solution 7 (Shortest)

Since $b^{2} - 4c \le 0$ and $c^{2} - 4b \le 0$, adding the two together yields $b^{2} + c^{2} \le 4(c+b)$. Obviously, this is not true if either $b$ or $c$ get too large, and they are equal when $b = c = 4$, so the greatest pair is $(4,4)$ and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where $(b,c)$ are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are $(1,1)$, $(2,1)$, $(2,1)$, $(2,2)$, $(3,3)$, $(4,4)$ meaning there are $\boxed{\textbf{(B) } 6}$ pairs.

- youtube.com/indianmathguy

Video Solution by OmegaLearn


~ pi_is_3.14

Video Solution



Video Solution by Mathematical Dexterity


Video Solution by TheBeautyofMath



See Also

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
2021 Fall AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 10 Problems and Solutions

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