Difference between revisions of "2021 AMC 10A Problems/Problem 15"
(new solution) |
Erringbubble (talk | contribs) m (added WLOG) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 5: | Line 5: | ||
==Solution 1 (Intuition):== | ==Solution 1 (Intuition):== | ||
− | Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. | + | Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. WLOG the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>, and the answer is <math>\boxed{\textbf{(C) }90}</math>. |
~IceWolf10 | ~IceWolf10 | ||
Line 20: | Line 20: | ||
~[[User:emerald_block|emerald_block]] | ~[[User:emerald_block|emerald_block]] | ||
+ | |||
+ | ==Solution 4 (Simple)== | ||
+ | Notice that one of the parabolas must be the wider one and the other one must be the thinner one. There are only two options: the wider one is above the thinner one, or the thinner one is above the wider one. Only the first options works. Therefore, out of the <math>\frac{1}{2}\cdot 6\cdot 5\cdot 4\cdot 3=180</math> ways to pick <math>A, B, C,</math> and <math>D</math> without regard to order, only <math>\frac{1}{2}\cdot 180=90</math> of these work. | ||
+ | |||
+ | ~[[Mathkiddie]] | ||
==Video Solution (Quick & Simple)== | ==Video Solution (Quick & Simple)== | ||
Line 31: | Line 36: | ||
~ North America Math Contest Go Go Go | ~ North America Math Contest Go Go Go | ||
− | == Video Solution (Using Vieta's Formulas and clever combinatorics) == | + | == Video Solution by OmegaLearn (Using Vieta's Formulas and clever combinatorics) == |
https://youtu.be/l85Qah1vGgc | https://youtu.be/l85Qah1vGgc | ||
Latest revision as of 12:55, 31 October 2024
Contents
[hide]- 1 Problem
- 2 Solution 1 (Intuition):
- 3 Solution 2 (Algebra):
- 4 Solution 3 (Symmetry):
- 5 Solution 4 (Simple)
- 6 Video Solution (Quick & Simple)
- 7 Video Solution (Use of Combinatorics and Algebra)
- 8 Video Solution by OmegaLearn (Using Vieta's Formulas and clever combinatorics)
- 9 Video Solution by TheBeautyofMath
- 10 See also
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. WLOG 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.
Solution 4 (Simple)
Notice that one of the parabolas must be the wider one and the other one must be the thinner one. There are only two options: the wider one is above the thinner one, or the thinner one is above the wider one. Only the first options works. Therefore, out of the ways to pick and without regard to order, only of these work.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.