Difference between revisions of "2021 Fall AMC 10A Problems/Problem 20"
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~Arcticturn | ~Arcticturn | ||
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| + | == Solution 3 == | ||
| + | Similarly 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 x axis and the other be y. The graph of solutions should look like this: | ||
| + | [asy] | ||
| + | Label f; | ||
| + | f.p=fontsize(6); | ||
| + | xaxis(0,5,Ticks(f, 1.0)); | ||
| + | yaxis(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,0,5)); | ||
| + | draw(graph(g,0,5)); | ||
| + | [/asy] | ||
| + | We are looking for lattice points(since b and c are positive integers), of which we can count <math>\boxed{\textbf{(B) } 6}</math>. | ||
| + | |||
| + | ~aop2014 | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:29, 23 November 2021
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 ![\[b^4\leq 16c^2\leq 64b.\]](http://latex.artofproblemsolving.com/c/f/8/cf8fa6bf7c791041cdad74405e8789551ecc94c1.png)
We apply casework to the value of 
- If
then 
from which 
then 
from which 
- If
then 
from which
then 
from which - If
then 
from which 
then 
from which 
- If
then 
from which 
then 
from which 
Together, there are 
ordered pairs 
namely 
and 
~MRENTHUSIASM
Solution 2 (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 3
Similarly 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 x axis and the other be y. The graph of solutions should look like this:
[asy]
Label f;
f.p=fontsize(6);
xaxis(0,5,Ticks(f, 1.0));
yaxis(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,0,5));
draw(graph(g,0,5));
[/asy]
We are looking for lattice points(since b and c are positive integers), of which we can count .
~aop2014
See Also
| 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. 
