Question

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 $A=4,$ $B=1,$ $C=3,$ $D=2.$ )

$30$

$60$

$90$

$180$

$360$

Answer: C

Concepts:parabola combinations

Difficulty rating: 1540

Video solution:

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Written 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.

2021 AMC 10A Spring Problem 15

Answer: C

Difficulty rating: 1540

Video solution:

Written solution:

Problem 15 in Other Years