2021 AMC 10A Problem 24

Take square roots: the first graph is the pair of parallel lines $x + ay = 2a$ and $x + ay = -2a$; the second is the pair $ax - y = a$ and $ax - y = -a$.

The first pair has normal vector $(1, a)$ and the second has normal vector $(a, -1)$. Their dot product is $a - a = 0$, so the two families are perpendicular, and the quadrilateral they bound is a rectangle.

The side lengths of the rectangle are the distances between the parallel lines in each pair, using $\frac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$:
$$\frac{2a - (-2a)}{\sqrt{1 + a^2}} = \frac{4a}{\sqrt{a^2+1}},\qquad \frac{a - (-a)}{\sqrt{a^2 + 1}} = \frac{2a}{\sqrt{a^2+1}}.$$

The area is their product:
$$\frac{4a}{\sqrt{a^2+1}} \cdot \frac{2a}{\sqrt{a^2+1}} = \frac{8a^2}{a^2+1}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{8a^2}{a^2+1}}$.

A squared linear expression equal to a constant is two parallel lines, and the coefficients tell you the direction. Checking perpendicularity first turns a vertex-hunting problem into "width times height," where the width and height are the distances between parallel lines. The answer must be valid for every $a$, which is also a hint to test a convenient value.

Plug in $a = 2$: the lines $x + 2y = \pm4$ and $2x - y = \pm2$ bound a rectangle with sides $\frac{8}{\sqrt5}$ and $\frac{4}{\sqrt5}$, area $\frac{32}{5}$. Among the choices only (D) gives $\frac{32}{5}$ at $a = 2$ (choice (C) gives $\frac{16}{3}$ and (E) gives $\frac{16}{5}$). Note that $a = 1$ does not separate (C), (D) and (E), which all give $4$.

• Assuming the figure is a square because the two families are perpendicular; the gaps 4a/sqrt(a^2+1) and 2a/sqrt(a^2+1) differ.
• Testing only $a = 1$, where three choices coincide, and guessing among them.