2021 AMC 10A Problems/Problem 15

Problem

Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ 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 $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$ )

$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

Solution

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 $C>A$ and $B>D$ . Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$ , and the answer is $\boxed{C}$ . ~IceWolf10

Video Solution (Using Vieta's Formulas and clever combinatorics)

https://youtu.be/l85Qah1vGgc

~ pi_is_3.14

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2021 AMC 10A Problems/Problem 15

Contents

Problem

Solution

Video Solution (Using Vieta's Formulas and clever combinatorics)

See also