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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]]}}
 +
 
 +
== Problem ==
 +
 
 
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?
 
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) } 12 \qquad \textbf{(E) } 16 \qquad</math>
 
<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad</math>
  
−
==Solution 1==
+
== Solution 1 (Casework) ==
 +
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;">
 +
  <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+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>
 +
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>
 +
 
 +
* If <math>b=4,</math> then <math>256\leq 16c^2\leq 256,</math> from which <math>c=4.</math>
 +
 
 +
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:
 +
<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)
 +
{
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return 0.25x^2;
 +
}
 +
real g(real x)
 +
{
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return 2*sqrt(x);
 +
}
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dot((1,1));
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dot((2,1));
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dot((1,2));
 +
dot((2,2));
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dot((3,3));
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dot((4,4));
 +
draw(graph(f,0,sqrt(20)));
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draw(graph(g,0,5));
 +
</asy>
 +
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>.
  
−
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>.
+
~aop2014
  
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== Solution 2 ==
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== Solution 3 (Graphing) ==
 
We need to solve the following system of inequalities:
 
We need to solve the following system of inequalities:
 
<cmath>
 
<cmath>
Line 20: Line 70:
 
\]
 
\]
 
</cmath>
 
</cmath>
−
 
 
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>.
  
Line 26: Line 75:
 
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>.
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Hence, the feasible <math>c</math> are 1, 2.
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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>.
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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.
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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>.
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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.
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Hence, the feasible <math>c</math> is <math>3</math>.
  
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For <math>b = 4</math>, <math>f \left( b \right) = 4</math> and <math>g \left( b \right) = 4</math>.
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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.
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Hence, the feasible <math>c</math> is <math>4</math>.
  
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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}</math>.
  
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
  
 +
==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:
 +
 +
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
 +
 +
==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
 +
 +
==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>(2,1)</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==
 +
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
  
−
{{AMC12 box|year=2021 Fall|ab=A|num-a=18|num-b=16}}
+
==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}}

Latest revision as of 16:26, 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.

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 $\sqrt{B^2-4AC}=0.$ 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 (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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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