2021 AMC 10A Problem 15

Step 1: when do the curves meet?

1. Try an example. $y = 3x^2 + 1$ and $y = 2x^2 + 5$ meet when $3x^2 + 1 = 2x^2 + 5$, that is $x^2 = 4$, so $x = 2$ or $x = -2$. They meet.
2. Now swap the constants: $y = 3x^2 + 5$ and $y = 2x^2 + 1$. Setting them equal gives $x^2 = -4$, which is impossible. They never meet.
3. The picture explains it. Both curves are U-shapes opening upward from the $y$-axis. The one with the bigger number in front of $x^2$ is steeper. If the steeper one starts lower, it catches up and crosses the other. If it starts higher, it stays above forever.
4. So the curves meet exactly when the bigger $x^2$ number goes with the smaller constant.

Step 2: count.

1. Choose which $4$ of the $6$ numbers are used. Choosing $4$ to use is the same as choosing $2$ to leave out, and there are $15$ ways to pick $2$ of $6$.
2. Split those $4$ numbers into two $x^2$ numbers and two constants: pick which $2$ are the $x^2$ numbers, $6$ ways. For example, from $\{1, 2, 3, 4\}$ take $3$ and $4$ as the $x^2$ numbers and $1$ and $2$ as constants.
3. Now there is only one good way to match them: the bigger $x^2$ number with the smaller constant. In the example, $y = 4x^2 + 1$ and $y = 3x^2 + 2$. (The other matching, $4x^2 + 2$ with $3x^2 + 1$, never meets.) The order of the two curves doesn't matter, so this is one choice.
4. Total: $15 \times 6 \times 1 = 90$.

The answer is $\boxed{\textbf{(C)}\ 90}$.

Picturing the two U-shapes turns the algebra into a simple rule: the steeper curve must start lower. Counting by first choosing groups of numbers and then matching them avoids any worry about the order of the two curves, which is where most wrong answers come from.

Count all pairs of curves, then take half. Filling $A, B, C, D$ in order gives $6 \times 5 \times 4 \times 3 = 360$ ways, but each pair of curves is counted twice (once with each curve listed first), so there are $180$ pairs. Swapping just the two constants turns every pair that meets into a pair that doesn't, and back again. So exactly half the pairs meet: $180 \div 2 = 90$.

Setting the curves equal gives $(A - C)x^2 = D - B$, so $x^2 = \frac{D - B}{A - C}$. Since all four values are distinct, neither side is zero, and a real solution exists exactly when $D - B$ and $A - C$ have the same sign: the parabola with the larger coefficient must have the smaller constant term.

Count the total first. There are $6 \cdot 5 \cdot 4 \cdot 3 = 360$ ordered assignments, and swapping the two curves $(A,B) \leftrightarrow (C,D)$ pairs them up without changing anything, so there are $180$ distinct pairs of curves.

Now pair each configuration with the one obtained by swapping only the constants $B \leftrightarrow D$. This flips the sign of $D - B$ while leaving $A - C$ alone, so it turns every intersecting pair into a non-intersecting pair and vice versa. Exactly half of the $180$ configurations intersect:
$$\frac{180}{2} = 90.$$

The answer is $\boxed{\textbf{(C)}\ 90}$.