Difference between revisions of "2021 AMC 10A Problems/Problem 15"
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==Solution 1 (Intuition):== | ==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 <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{C}</math>. ~IceWolf10 | + | 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 <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 | ||
==Solution 2 (Algebra):== | ==Solution 2 (Algebra):== | ||
− | Setting <math>y = Ax^2+B = Cx^2+D</math>, we find that <math>Ax^2-Cx^2 = x^2(A-C) = D-B</math>, so <math>x^2 = \frac {D-B}{A-C} \ge 0</math> by the trivial inequality. This implies that <math>D-B</math> and <math>A-C</math> must both be positive or negative. If two distinct values are chosen for <math>(A, C)</math> and <math>(B, D)</math> respectively, there are <math>2</math> ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by <math>2</math> at the end, however, since the <math>2</math> curves aren't considered distinct. Calculating, we get <cmath>\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = \boxed{\textbf{(C) }90}.</cmath> ~ ike.chen | + | Setting <math>y = Ax^2+B = Cx^2+D</math>, we find that <math>Ax^2-Cx^2 = x^2(A-C) = D-B</math>, so <math>x^2 = \frac {D-B}{A-C} \ge 0</math> by the trivial inequality. This implies that <math>D-B</math> and <math>A-C</math> must both be positive or negative. If two distinct values are chosen for <math>(A, C)</math> and <math>(B, D)</math> respectively, there are <math>2</math> ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by <math>2</math> at the end, however, since the <math>2</math> curves aren't considered distinct. Calculating, we get <cmath>\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = \boxed{\textbf{(C) }90}.</cmath> ~ike.chen |
==Video Solution (Quick & Simple)== | ==Video Solution (Quick & Simple)== |
Revision as of 17:53, 8 July 2021
Contents
[hide]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
Video Solution (Quick & Simple)
~ Education, the Study of Everything
Video Solution (Use of Combonatorics and Algebra)
https://www.youtube.com/watch?v=SRjtftj0tSE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=7&t=1s
~ North America Math Contest Go Go Go
Video Solution (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.