2021 AMC Fall 10A Problem 16

The floor function looks scary, so start by plugging in numbers and drawing the graph.

1. Whole numbers. For a whole number $x$, both floors just give back the number:
- $f(0) = |0| - |1| = -1$
- $f(1) = |1| - |0| = 1$
- $f(2) = |2| - |{-1}| = 2 - 1 = 1$
- $f(-1) = |{-1}| - |2| = 1 - 2 = -1$
- $f(3) = 3 - 2 = 1$ and $f(-2) = 2 - 3 = -1$.

The pattern: $f = 1$ at $1, 2, 3, \ldots$ and $f = -1$ at $0, -1, -2, \ldots$
2. Numbers between whole numbers. Try $x = 2.3$: $\lfloor 2.3 \rfloor = 2$ and $\lfloor 1 - 2.3 \rfloor = \lfloor -1.3 \rfloor = -2$. So $f(2.3) = 2 - 2 = 0$. Try $x = -0.6$: $\lfloor -0.6 \rfloor = -1$ and $\lfloor 1.6 \rfloor = 1$, so $f = 1 - 1 = 0$. In general, when $x$ is not a whole number, the two floors are opposites, so $f(x) = 0$.
3. The graph. It is the $x$-axis with the whole-number points moved: dots at height $1$ above $x = 1, 2, 3, \ldots$ and dots at height $-1$ below $x = 0, -1, -2, \ldots$
4. Find the center. Pair the dots up: $(1, 1)$ with $(0, -1)$, $(2, 1)$ with $(-1, -1)$, $(3, 1)$ with $(-2, -1)$, and so on. Each pair has midpoint $\left(\frac12, 0\right)$. Turning the graph halfway around that point swaps each pair, and the flat part along the $x$-axis turns into itself. So the graph is symmetric about $\left(\frac12, 0\right)$.
5. Rule out the rest.
- $y$-axis: $f(1) = 1$ but $f(-1) = -1$. Fails.
- Line $x = 1$: $f(0) = -1$ but $f(2) = 1$. Fails.
- Origin: a graph symmetric about a point on the $x$-axis must have $f = 0$ there, but $f(0) = -1$. Fails.
- Point $(1, 0)$: the same reason, since $f(1) = 1$. Fails.

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

The formula pairs $x$ with $1 - x$. Replacing $x$ by $1 - x$ swaps the two parts, and since they are subtracted, the value flips sign: $f(1 - x) = -f(x)$. The numbers $x$ and $1 - x$ are mirror images around $\frac12$, and a flipped sign means a half-turn around the point $\left(\frac12, 0\right)$. When a formula mixes $x$ and $c - x$, check what happens when you swap them.

Do the swap directly. Replace $x$ by $1 - x$:
$$f(1 - x) = |\lfloor 1 - x \rfloor| - |\lfloor x \rfloor| = -f(x).$$
So the point $(x, y)$ on the graph always comes with the point $(1 - x, -y)$. These two points have midpoint $\left(\frac12, 0\right)$, which is exactly symmetry about that point.

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