2021 AMC Fall 10A Problem 16

The two terms of $f$ are built from $x$ and $1 - x$, and the substitution $x \mapsto 1 - x$ exchanges these two expressions. So
$$f(1 - x) = |\lfloor 1 - x \rfloor| - |\lfloor 1 - (1 - x) \rfloor| = |\lfloor 1 - x \rfloor| - |\lfloor x \rfloor| = -f(x).$$

A graph is symmetric about the point $(a, 0)$ exactly when $f(2a - x) = -f(x)$ for all $x$. Here $2a = 1$, so $a = \tfrac12$: the graph is symmetric about $\left(\tfrac12, 0\right)$.

To rule out the other choices, compute a few values: $f(-1) = -1$, $f(0) = -1$, $f(1) = 1$, $f(2) = 1$. The $y$-axis fails since $f(-1) \ne f(1)$; the origin fails since $f(0) \ne 0$; the line $x = 1$ fails since $f(0) \ne f(2)$; and the point $(1, 0)$ fails since point symmetry about $(a, 0)$ forces $f(a) = 0$, but $f(1) = 1$.

The answer is $\boxed{\textbf{(D)}\ \text{the point }\left(\frac12, 0\right)}$.

When a formula pairs $x$ with $c - x$, test the substitution $x \to c - x$: if the two pieces swap and the operation between them is subtraction, you get $f(c - x) = -f(x)$, point symmetry about $\left(\tfrac c2, 0\right)$; if the operation is addition, you get $f(c - x) = f(x)$, line symmetry about $x = \tfrac c2$. The floor and absolute value are decoration; the structure of the expression decides.

Draw the graph. For non-integer $x$, $\lfloor 1 - x \rfloor = -\lfloor x \rfloor$, so $f(x) = |\lfloor x \rfloor| - |{-\lfloor x \rfloor}| = 0$. For an integer $n$, $f(n) = |n| - |1 - n|$, which is $1$ for $n \ge 1$ and $-1$ for $n \le 0$. The graph is the $x$-axis with isolated dots at $(n, 1)$ for $n \ge 1$ and $(n, -1)$ for $n \le 0$; a half-turn about $\left(\tfrac12, 0\right)$ sends $(n, 1)$ to $(1 - n, -1)$, matching the picture.

• Testing only a few integer inputs and settling on the origin or the y-axis; f(0) = -1 alone rules out both.
• Confusing the symmetry conditions: point symmetry needs $f(2a - x) = -f(x)$, line symmetry needs $f(2a - x) = f(x)$.