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). Note that we must divide by at the end, however, since the curves aren't considered distinct. Calculating, we get ![\[\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = 90 = \boxed{C}.\]](http://latex.artofproblemsolving.com/8/b/7/8b79c250f72fe1c07fb8d49dc49981aa8c1e96de.png)
~ ike.chen
Video Solution (Using Vieta's Formulas and clever combinatorics)
~ pi_is_3.14
See also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
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| All AMC 10 Problems and Solutions | ||
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