2021 Fall AMC 12A Problems/Problem 17
- 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.