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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 17"
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==Problem==
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{{duplicate|[[2021 Fall AMC 10A Problems/Problem 20|2021 Fall AMC 10A #20]] and [[2021 Fall AMC 12A Problems/Problem 17|2021 Fall AMC 12A #17]]}}
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For how many ordered pairs <math>(b,c)</math> of positive integers does neither <math>x^2+bx+c=0</math> nor <math>x^2+cx+b=0</math> have two distinct real solutions?
 
  
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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>
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== Problem ==
  
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==Solution 1==
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How many ordered pairs of positive integers <math>(b,c)</math> exist where both <math>x^2+bx+c=0</math> and <math>x^2+cx+b=0</math> do not have distinct, real solutions?
  
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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>.
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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad</math>
  
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== Solution 2 ==
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== Solution 1 (Casework) ==
 +
A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:
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<ol style="margin-left: 1.5em;">
 +
  <li>Since <math>x^2+bx+c=0</math> does not have real solutions, we have <math>b^2\leq 4c.</math></li><p>
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  <li>Since <math>x^2+cx+b=0</math> does not have real solutions, we have <math>c^2\leq 4b.</math></li><p>
 +
</ol>
 +
Squaring the first inequality, we get <math>b^4\leq 16c^2.</math> Multiplying the second inequality by <math>16,</math> we get <math>16c^2\leq 64b.</math> Combining these results, we get <cmath>b^4\leq 16c^2\leq 64b.</cmath>
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We apply casework to the value of <math>b:</math>
 +
 
 +
* If <math>b=1,</math> then <math>1\leq 16c^2\leq 64,</math> from which <math>c=1,2.</math>
 +
 
 +
* If <math>b=2,</math> then <math>16\leq 16c^2\leq 128,</math> from which <math>c=1,2.</math>
 +
 
 +
* If <math>b=3,</math> then <math>81\leq 16c^2\leq 192,</math> from which <math>c=3.</math>
 +
 
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* If <math>b=4,</math> then <math>256\leq 16c^2\leq 256,</math> from which <math>c=4.</math>
 +
 
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Together, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs <math>(b,c),</math> namely <math>(1,1),(1,2),(2,1),(2,2),(3,3),</math> and <math>(4,4).</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
== Solution 2 (Graphing) ==
 +
Similar to Solution 1, use the discriminant to get <math>b^2\leq 4c</math> and <math>c^2\leq 4b</math>. These can be rearranged to <math>c\geq \frac{1}{4}b^2</math> and <math>b\geq \frac{1}{4}c^2</math>. Now, we can roughly graph these two inequalities, letting one of them be the <math>x</math> axis and the other be <math>y</math>.
 +
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:
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<asy>
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unitsize(2);
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Label f;
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f.p=fontsize(6);
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xaxis("$x$",0,5,Ticks(f, 1.0));
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yaxis("$y$",0,5,Ticks(f, 1.0));
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real f(real x)
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{
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return 0.25x^2;
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}
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real g(real x)
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{
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return 2*sqrt(x);
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}
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dot((1,1));
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dot((2,1));
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dot((1,2));
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dot((2,2));
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dot((3,3));
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dot((4,4));
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draw(graph(f,0,sqrt(20)));
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draw(graph(g,0,5));
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</asy>
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We are looking for lattice points (since <math>b</math> and <math>c</math> are positive integers), of which we can count <math>\boxed{\textbf{(B) } 6}</math>.
 +
 
 +
~aop2014
 +
 
 +
==Solution 3 (Oversimplified but Risky)==
 +
A quadratic equation <math>Ax^2+Bx+C=0</math> has one real solution if and only if <math>\sqrt{B^2-4AC}=0.</math> Similarly, it has imaginary solutions if and only if <math>\sqrt{B^2-4AC}<0.</math> We proceed as following:
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 +
We want both <math>x^2+bx+c</math> to be <math>1</math> value or imaginary and <math>x^2+cx+b</math> to be <math>1</math> value or imaginary. <math>x^2+4x+4</math> is one such case since <math>\sqrt {b^2-4ac}</math> is <math>0.</math> Also, <math>x^2+3x+3, x^2+2x+2, x^2+x+1</math> are always imaginary for both <math>b</math> and <math>c.</math> We also have <math>x^2+x+2</math> along with <math>x^2+2x+1</math> since the latter has one solution, while the first one is imaginary. Therefore, we have <math>6</math> total ordered pairs of integers, which is <math>\boxed{\textbf{(B) } 6}.</math>
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 +
~Arcticturn
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 +
== Solution 4 ==
 
We need to solve the following system of inequalities:
 
We need to solve the following system of inequalities:
 
<cmath>
 
<cmath>
Line 44: Line 101:
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
  
 +
==Video Solution by Mathematical Dexterity==
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https://www.youtube.com/watch?v=EkaKfkQgFbI
  
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{{AMC12 box|year=2021 Fall|ab=A|num-a=18|num-b=16}}
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==See Also==
 +
{{AMC12 box|year=2021 Fall|ab=A|num-b=16|num-a=18}}
 +
{{AMC10 box|year=2021 Fall|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:27, 26 November 2021

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.

Problem

How many ordered pairs of positive integers $(b,c)$ exist where both $x^2+bx+c=0$ and $x^2+cx+b=0$ do not have distinct, real solutions?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad$

Solution 1 (Casework)

A quadratic equation does not have 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 $(4,4).$

~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 (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 $6$ total ordered pairs of integers, which is $\boxed{\textbf{(B) } 6}.$

~Arcticturn

Solution 4

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)

Video Solution by Mathematical Dexterity

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

See Also

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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