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

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== Problem ==
 
== Problem ==
  
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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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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?
  
−
<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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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad</math>
  
 
== Solution 1 (Casework) ==
 
== Solution 1 (Casework) ==
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A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:
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A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that:
 
<ol style="margin-left: 1.5em;">
 
<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>
 
   <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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==Solution 4 (Oversimplified but Risky)==
 
==Solution 4 (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:
+
A quadratic equation <math>Ax^2+Bx+C=0</math> has one real solution if and only if <math>B^2-4AC=0.</math> Similarly, it has imaginary solutions if and only if <math>B^2-4AC<0.</math> We proceed as following:
  
 
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>\boxed{\textbf{(B) } 6}</math> total ordered pairs of integers.
 
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>\boxed{\textbf{(B) } 6}</math> total ordered pairs of integers.
  
 
~Arcticturn
 
~Arcticturn
 +
 +
==Solution 5 (Quick and Easy)==
 +
We see that <math>b^2 \leq 4c</math> and <math>c^2 \leq 4b.</math> WLOG, assume that <math>b \geq c.</math> Then we have that <math>b^2 \leq 4c \leq 4b</math>, so <math>b^2 \leq 4b</math> and therefore <math>b \leq 4</math>, also meaning that <math>c \leq 4.</math> This means that we only need to try 16 cases. Now we can get rid of the assumption that <math>b \geq c</math>, because we want ordered pairs. For <math>b = 1</math> and <math>b = 2</math>, <math>c = 1</math> and <math>c = 2</math> work. When <math>b = 3</math>, <math>c</math> can only be <math>3</math>, and when <math>b = 4</math>, only <math>c = 4</math> works, for a total of <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of integers.
 +
 +
~littlefox_amc
 +
 +
==Solution 6 (Fastest) ==
 +
We need both <math>b^2\leq 4c</math> and <math>c^2\leq 4b</math>.
 +
 +
If <math>b=c</math> then the above become <math>b^2\leq 4b\iff b\leq 4</math>, so we have four solutions <math>(k,k)</math>, where <math>k=1</math>,<math>2</math>,<math>3</math>,<math>4</math>.
 +
 +
If <math>b<c</math> then we only need <math>c^2\leq 4b</math> since it implies <math>b^2< 4c</math>. Now
 +
<math>c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2</math>, so <math>b=1</math>. We plug <math>b=1</math>, <math>c=2</math> back into <math>c^2\leq 4b</math> and it works. So there is another solution <math>(1,2)</math>.
 +
 +
By symmetry, if <math>b>c</math> then <math>(b,c)=(2,1)</math>.
 +
 +
Therefore the total number of solutions is <math>\boxed{\textbf{(B) } 6}</math>.
 +
 +
~asops
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 +
==Solution 7 (Shortest) ==
 +
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>(1,2)</math>, <math>(2,2)</math>, <math>(3,3)</math>, <math>(4,4)</math> meaning there are <math>\boxed{\textbf{(B) } 6}</math> pairs.
 +
 +
- youtube.com/indianmathguy
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/zfChnbMGLVQ?t=4254
 +
 +
~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
−
https://youtube/ef-W3l94k00
+
https://www.youtube.com/watch?v=ef-W3l94k00
 +
 
 +
~MathProblemSolvingSkills.com
  
 
==Video Solution by Mathematical Dexterity==
 
==Video Solution by Mathematical Dexterity==

Latest revision as of 14:44, 20 October 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.

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:

  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 (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 $B^2-4AC=0.$ Similarly, it has imaginary solutions if and only if $B^2-4AC<0.$ 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 (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 $(1,1)$, $(2,1)$, $(1,2)$, $(2,2)$, $(3,3)$, $(4,4)$ meaning there are $\boxed{\textbf{(B) } 6}$ pairs.

- 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

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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