2021 AMC Fall 10A Problem 14

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 $\boxed{\textbf{(D)}\ 5}$.

Absolute-value equations describe piecewise-linear shapes; sketch them and the problem becomes counting intersections edge by edge. Symmetry halves the work, and a discriminant check settles each edge quickly. The delicate part is the inner diamond, where the parabola passes exactly through three vertices without crossing any edge; always test whether a contact point is a crossing or just a touch.

• Assuming the parabola crosses the inner diamond's edges in extra points, or miscounting the tangency-type contacts at the vertices (0,3), (3,0), (-3,0).
• Accepting the solution of $x^2 + 3x - 24 = 0$ with the negative root for the right-hand edge without checking the domain, or forgetting to mirror to $x < 0$.