2021 AMC 10A Problem 15

P11-15~4 minPrint

Basic Counting Quadratics Map the objects to something easier to count Exploit symmetry to reduce work or pair up objects

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

Show solution without answering

Key insight

Curves meet only if the steeper parabola has the smaller constant. Swapping the two constants turns each good choice into a bad one, so half of all choices work.

Solution

Step 1: when do the curves meet?

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.
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.
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.
So the curves meet exactly when the bigger $x^2$ number goes with the smaller constant.

Step 2: count.

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 $6$ .
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.
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.
Total: $15 \times 6 \times 1 = 90$ .

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

Why this works

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.

Alternative approach

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: $180 \div 2 = 90$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(C)}\ 90}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Intermediate Counting & Probability, Chapter 4: Constructive Counting and 1-1 Correspondences (§4.4, §4.5, §4.6)