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

Revision as of 01:27, 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:

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 (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

@@ Line 1: / Line 1: @@
-==Problem==
+{{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]]}}
-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>
+== Problem ==
-==Solution 1==
+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?
-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>.
+<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad</math>
-== Solution 2 ==
+== Solution 1 (Casework) ==
+A quadratic equation does not have 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)
+{
+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 <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:
+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>
+~Arcticturn
+== Solution 4 ==
 We need to solve the following system of inequalities:
 <cmath>
@@ Line 44: / Line 101: @@
 ~Steven Chen (www.professorchenedu.com)
+==Video Solution by Mathematical Dexterity==
+https://www.youtube.com/watch?v=EkaKfkQgFbI
-{{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}}

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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