Difference between revisions of "2021 Fall AMC 12A Problems/Problem 17"
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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> | + | 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. | ||
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==Solution 7 (Shortest) == | ==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> | + | 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 | - youtube.com/indianmathguy |
Latest revision as of 17:58, 5 August 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.
Contents
- 1 Problem
- 2 Solution 1 (Casework)
- 3 Solution 2 (Graphing)
- 4 Solution 3 (Graphing)
- 5 Solution 4 (Oversimplified but Risky)
- 6 Solution 5 (Quick and Easy)
- 7 Solution 6 (Fastest)
- 8 Solution 7 (Shortest)
- 9 Video Solution by OmegaLearn
- 10 Video Solution
- 11 Video Solution by Mathematical Dexterity
- 12 Video Solution by TheBeautyofMath
- 13 See Also
Problem
For how many ordered pairs of positive integers does neither
nor
have two distinct real solutions?
Solution 1 (Casework)
A quadratic equation does not have two distinct 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 (Graphing)
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 , we have
and
.
Hence, the feasible
are
.
For , we have
and
.
Hence, the feasible
are
.
For , we have
and
.
Hence, the feasible
is
.
For , we have
and
.
Hence, the feasible
is
.
For , we have
. Hence, there is no feasible
.
Putting all cases together, the correct answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4 (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.
~Arcticturn
Solution 5 (Quick and Easy)
We see that and
WLOG, assume that
Then we have that
, so
and therefore
, also meaning that
This means that we only need to try 16 cases. Now we can get rid of the assumption that
, because we want ordered pairs. For
and
,
and
work. When
,
can only be
, and when
, only
works, for a total of
ordered pairs of integers.
~littlefox_amc
Solution 6 (Fastest)
We need both and
.
If then the above become
, so we have four solutions
, where
,
,
,
.
If then we only need
since it implies
. Now
, so
. We plug
,
back into
and it works. So there is another solution
.
By symmetry, if then
.
Therefore the total number of solutions is .
~asops
Solution 7 (Shortest)
Since and
, adding the two together yields
. Obviously, this is not true if either
or
get too large, and they are equal when
, so the greatest pair is
and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where
are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are
,
,
,
,
,
meaning there are
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
~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.