2021 AMC Fall 10A Problem 14

P11-15~6 minPrint

Absolute Value & Inequalities Coordinate Geometry Quadratics Split into exhaustive cases and handle each Exploit symmetry to reduce work or pair up objects

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? $\begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*}$

Show solution without answering

Key insight

The second equation is two diamonds, |x| + |y| = 3 or 5. The parabola touches the small one at 3 corners and crosses the big one twice.

Solution

1. The first equation. Solve for $y$ : $y = 3 - \frac{x^2}{3}$ . This is a parabola opening downward. Some points: $x = 0$ gives $y = 3$ , and $x = \pm 3$ gives $y = 0$ . So it passes through $(0, 3)$ , $(3, 0)$ , $(-3, 0)$ .
2. The second equation. A square equals $1$ when the thing being squared is $1$ or $-1$ . So $|x| + |y| - 4 = \pm 1$ , which means
$|x| + |y| = 3 \quad \text{or} \quad |x| + |y| = 5.$
The graph of $|x| + |y| = 3$ is a diamond (a tilted square) with corners $(3,0)$ , $(0,3)$ , $(-3,0)$ , $(0,-3)$ . The graph of $|x| + |y| = 5$ is a bigger diamond with corners at distance $5$ .
3. Use symmetry. Both graphs look the same on the left and right of the $y$ -axis. So count points with $x \ge 0$ , and then mirror them.
4. Small diamond, right half. Its top edge is $y = 3 - x$ for $x$ from $0$ to $3$ . Compare with the parabola at a middle point, $x = 1.5$ : the parabola has $y = 3 - 0.75 = 2.25$ and the edge has $y = 1.5$ . In fact the parabola minus the edge is $x - \frac{x^2}{3} = x\left(1 - \frac{x}{3}\right)$ , which is positive between $0$ and $3$ and zero only at the ends. So they meet only at the corners $(0, 3)$ and $(3, 0)$ .
The bottom edge has $y \le 0$ , while the parabola has $y \ge 0$ for $x$ from $0$ to $3$ . They meet only at $(3, 0)$ again.
With the mirror image $(-3, 0)$ , the small diamond gives $3$ points: $(0,3)$ , $(3,0)$ , $(-3,0)$ .
5. Big diamond, top-right edge $y = 5 - x$ . Set $3 - \frac{x^2}{3} = 5 - x$ and multiply by $3$ : $x^2 - 3x + 6 = 0$ . Its discriminant is $9 - 24 < 0$ , so there is no solution. (The parabola's top is at height $3$ , below this edge.)
6. Big diamond, bottom-right edge $y = x - 5$ , for $x$ from $0$ to $5$ . Set $3 - \frac{x^2}{3} = x - 5$ and multiply by $3$ : $x^2 + 3x - 24 = 0$ . The positive root is $x = \frac{-3 + \sqrt{105}}{2} \approx 3.6$ , which is between $0$ and $5$ . So there is $1$ point here, and $1$ more as its mirror image. The big diamond gives $2$ points.
7. Total. $3 + 2 = 5$ .

The answer is $\boxed{\textbf{(D)}\ 5}$ .

Why this works

Equations with absolute values usually draw shapes made of straight pieces. Draw them, then count crossings one edge at a time. Symmetry cuts the work in half. The careful part is the small diamond: the parabola goes exactly through three of its corners without crossing any edge, so each corner counts once.

Concise solution

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

The first equation is the downward parabola $y = 3 - \dfrac{x^2}{3}$ , through $(0, 3)$ and $(\pm 3, 0)$ . The second says $|x| + |y| = 3$ or $|x| + |y| = 5$ : two diamonds with vertices on the axes at distance $3$ and $5$ . Everything is symmetric about the $y$ -axis, so work with $x \ge 0$ and mirror.

Diamond $|x| + |y| = 3$ . Its upper-right edge is $y = 3 - x$ for $0 \le x \le 3$ . On this interval $3 - \tfrac{x^2}{3} - (3 - x) = x\left(1 - \tfrac{x}{3}\right) \ge 0$ , with equality only at $x = 0$ and $x = 3$ . So the parabola rides above the edge, touching it only at the vertices $(0, 3)$ and $(3, 0)$ . The lower-right edge $y = x - 3$ has $y \le 0$ while the parabola has $y \ge 0$ on $[0, 3]$ ; they meet only at $(3, 0)$ . Mirroring gives $(-3, 0)$ . Total from this diamond: $3$ points.

Diamond $|x| + |y| = 5$ . Upper-right edge $y = 5 - x$ : setting $3 - \tfrac{x^2}{3} = 5 - x$ gives $x^2 - 3x + 6 = 0$ , discriminant $9 - 24 < 0$ , no solutions. Lower-right edge $y = x - 5$ for $0 \le x \le 5$ : $3 - \tfrac{x^2}{3} = x - 5$ gives $x^2 + 3x - 24 = 0$ , so $x = \dfrac{-3 + \sqrt{105}}{2} \approx 3.6$ , which lies in $[0, 5]$ with $y \approx -1.4 \le 0$ . One valid point, and its mirror image. Total: $2$ points.

Altogether $3 + 2 = 5$ ordered pairs.

The answer is $\boxed{\textbf{(D)}\ 5}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)