Difference between revisions of "2021 Fall AMC 12A Problems/Problem 17"
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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 == | |
− | == | + | 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? |
− | + | <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: | We need to solve the following system of inequalities: | ||
<cmath> | <cmath> | ||
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~Steven Chen (www.professorchenedu.com) | ~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= | + | ==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.
Contents
Problem
How many ordered pairs of positive integers exist where both
and
do not have distinct, real solutions?
Solution 1 (Casework)
A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:
- Since
does not have real solutions, we have
- Since
does not have real solutions, we have
Squaring the first inequality, we get Multiplying the second inequality by
we get
Combining these results, we get
We apply casework to the value of
- If
then
from which
- If
then
from which
- If
then
from which
- If
then
from which
Together, there are ordered pairs
namely
and
~MRENTHUSIASM
Solution 2 (Graphing)
Similar to Solution 1, use the discriminant to get and
. These can be rearranged to
and
. Now, we can roughly graph these two inequalities, letting one of them be the
axis and the other be
.
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:
We are looking for lattice points (since
and
are positive integers), of which we can count
.
~aop2014
Solution 3 (Oversimplified but Risky)
A quadratic equation has one real solution if and only if
Similarly, it has imaginary solutions if and only if
We proceed as following:
We want both to be
value or imaginary and
to be
value or imaginary.
is one such case since
is
Also,
are always imaginary for both
and
We also have
along with
since the latter has one solution, while the first one is imaginary. Therefore, we have
total ordered pairs of integers, which is
~Arcticturn
Solution 4
We need to solve the following system of inequalities:
Feasible solutions are in the region formed between two parabolas and
.
Define and
.
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
For ,
and
.
Hence, the feasible
are 1, 2.
For ,
and
.
Hence, the feasible
are 1, 2.
For ,
and
.
Hence, the feasible
is 3.
For ,
and
.
Hence, the feasible
is 4.
For ,
. Hence, there is no feasible
.
Putting all cases together, the correct answer is .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.