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

Revision as of 15:25, 5 February 2024

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.

1 Problem
2 Solution 1 (Casework)
3 Solution 2 (Graphing)
4 Solution 3 (Graphing)
5 Solution 4 (Oversimplified but Risky)
6 Solution 5 (Quick and Easy)
7 Solution 6 (Fastest)
8 Solution 7
9 Video Solution by OmegaLearn
10 Video Solution
11 Video Solution by Mathematical Dexterity
12 Video Solution by TheBeautyofMath
13 See Also

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 (Casework)

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

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

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 $(4,4).$

~MRENTHUSIASM

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 $\boxed{\textbf{(B) } 6}$ .

~aop2014

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.

~Arcticturn

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.

~littlefox_amc

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 $\boxed{\textbf{(B) } 6}$ .

~asops

Solution 7

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 $\boxed{\textbf{(B) } 6}$ solutions.

- youtube.com/indianmathguy

Video Solution by OmegaLearn

https://youtu.be/zfChnbMGLVQ?t=4254

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=ef-W3l94k00

~MathProblemSolvingSkills.com

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=EkaKfkQgFbI

Video Solution by TheBeautyofMath

https://youtu.be/RPnfZKv4DVA

~IceMatrix

@@ Line 105: / Line 105: @@
 ~littlefox_amc
-==Solution 6 (Fastest)==
+==Solution 6 (Fastest) ==
 We need both <math>b^2\leq 4c</math> and <math>c^2\leq 4b</math>.
@@ Line 118: / Line 118: @@
 ~asops
+==Solution 7==
+Since <math>b^{2} - 4c \le 0</math> and <math>c^{2} - 4b \le 0</math>, adding the two together yields <math>b^{2} + c^{2} \le 4(c+b)</math>. Obviously, this is not true if either <math>b</math> or <math>c</math> get too large, and they are equal when <math>b = c = 4</math>, so the greatest pair is <math>(4,4)</math> and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where <math>(b,c)</math> are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are <math>(1,1)</math>, <math>(2,1)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, <math>(4,4)</math> meaning there are <math>\boxed{\textbf{(B) } 6}</math> solutions.
+- youtube.com/indianmathguy
 == Video Solution by OmegaLearn ==

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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