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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 17"
m (→‎Solution 4: Improved formatting.)
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~Arcticturn
 
~Arcticturn
  
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== Solution 4 ==
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== Solution 4 (Graphing) ==
 
We need to solve the following system of inequalities:
 
We need to solve the following system of inequalities:
 
<cmath>
 
<cmath>
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\]
 
\]
 
</cmath>
 
</cmath>
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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>.
 
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>.
  
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Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
 
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>.
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For <math>b = 1</math>, we have <math>f(b) = \frac{1}{4}</math> and <math>g(b) = 2</math>.
−
Hence, the feasible <math>c</math> are 1, 2.
+
Hence, the feasible <math>c</math> are <math>1, 2</math>.
  
−
For <math>b = 2</math>, <math>f \left( b \right) = 1</math> and <math>g \left( b \right) = 2 \sqrt{2}</math>.
+
For <math>b = 2</math>, we have <math>f(b) = 1</math> and <math>g(b) = 2 \sqrt{2}</math>.
−
Hence, the feasible <math>c</math> are 1, 2.
+
Hence, the feasible <math>c</math> are <math>1, 2</math>.
  
−
For <math>b = 3</math>, <math>f \left( b \right) = \frac{9}{4}</math> and <math>g \left( b \right) = 2 \sqrt{3}</math>.
+
For <math>b = 3</math>, we have <math>f(b) = \frac{9}{4}</math> and <math>g(b) = 2 \sqrt{3}</math>.
−
Hence, the feasible <math>c</math> is 3.
+
Hence, the feasible <math>c</math> is <math>3</math>.
  
−
For <math>b = 4</math>, <math>f \left( b \right) = 4</math> and <math>g \left( b \right) = 4</math>.
+
For <math>b = 4</math>, we have <math>f(b) = 4</math> and <math>g(b) = 4</math>.
−
Hence, the feasible <math>c</math> is 4.
+
Hence, the feasible <math>c</math> is 4<math>.
  
−
For <math>b > 4</math>, <math>f \left( b \right) > g \left( b \right)</math>. Hence, there is no feasible <math>c</math>.
+
For </math>b > 4<math>, we have </math>f(b) > g(b)<math>. Hence, there is no feasible </math>c<math>.
  
−
Putting all cases together, the correct answer is <math>\boxed{\textbf{(B) }6}</math>.
+
Putting all cases together, the correct answer is </math>\boxed{\textbf{(B) }6}$.
  
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)

Revision as of 16:53, 28 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 (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$ (Error compiling LaTeX. Unknown error_msg)b > 4$, we have$f(b) > g(b)$. Hence, there is no feasible$c$.

Putting all cases together, the correct answer is$ (Error compiling LaTeX. Unknown error_msg)\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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