2021 AMC 10A Problems/Problem 15
Contents
Problem
Values for and are to be selected from without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves and intersect? (The order in which the curves are listed does not matter; for example, the choices is considered the same as the choices )
Solution 1 (Intuition):
Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then and . Therefore the number of ways to choose the four integers is , and the answer is .
~IceWolf10
Solution 2 (Algebra):
Setting , we find that , so by the trivial inequality. This implies that and must both be positive or negative. If two distinct values are chosen for and respectively, there are ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by at the end, however, since the curves aren't considered distinct. Calculating, we get ~ike.chen
Solution 3 (Symmetry):
Like in Solution 2, we find . Notice that, since , this expression can never equal , and since , there won't be a divide-by-. This means that every choice results in either a positive or a negative value.
For every choice of that results in a positive value, we can flip and to obtain a corresponding negative value. This is a bijection (we could flip and again to obtain the original choice (injectivity) and we could flip and from any negative choice to obtain the corresponding positive choice (surjectivity)), so half of the choices are positive (where the curves intersect) and half are negative (where they don't).
This means that of the total choices (dividing by because the order of the curves does not matter), half of them, or , lead to intersecting curves.
Video Solution (Quick & Simple)
~ Education, the Study of Everything
Video Solution (Use of Combinatorics and Algebra)
https://www.youtube.com/watch?v=SRjtftj0tSE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=7&t=1s
~ North America Math Contest Go Go Go
Video Solution by OmegaLearn(Using Vieta's Formulas and clever combinatorics)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/t-EEP2V4nAE?t=1376
~IceMatrix
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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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