2021 AMC 10A Problem 15

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.

)

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Solution

Setting the equations equal to each other, we get

Ax^2 + B = Cx^2 + D

x^2(A - C) = D - B

x^2 = \dfrac{D - B}{A - C} \geq 0

since squares are non-negative. This means

D - B

and

A - C

must both have the same sign. If we choose two distinct values for

(A, C)

and

(B, D),

there are

2

ways to arrange them such that the numerator and denominator both have the same sign. We have to divide by

2,

however, since the two curves are not considered distinct. Therefore, the total number of tuples is

\dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90.

Thus, C is the correct answer.