2022 AMC 10B Problem 24

Upper bound. Applying the condition to the outer $f$,
$$|f(f(800)) - f(f(400))| \le \tfrac12 |f(800) - f(400)|.$$
Now bound $|f(800) - f(400)|$ using the equal values at $300$ and $900$:
$$|f(800) - f(400)| \le |f(800) - f(900)| + |f(900) - f(300)| + |f(300) - f(400)| \le 50 + 0 + 50 = 100.$$
Hence the target is at most $50$.

Construction. Equality needs $f(400) = f(300) - 50$, $f(800) = f(900) + 50$, and slope exactly $+\tfrac12$ between the two inner values. Take $f(300) = f(900) = 0$, so $f(400) = -50$ and $f(800) = 50$, and make $f$ have slope $\tfrac12$ on $[-50, 50]$. One piecewise-linear choice, with every slope of absolute value at most $\tfrac12$:
$$f(x) = \begin{cases} \tfrac{x}{2} & -50 \le x \le 50,\\ 25 - \tfrac{x - 50}{10} & 50 \le x \le 300,\\ -\tfrac{x - 300}{2} & 300 \le x \le 400,\\ \tfrac{x - 600}{4} & 400 \le x \le 800,\\ -\tfrac{x - 900}{2} & 800 \le x \le 900, \end{cases}$$
extended by constants outside $[-50, 900]$. It is continuous (check the joints: $25, 0, -50, 50, 0$) and satisfies the condition. Then $f(f(800)) - f(f(400)) = f(50) - f(-50) = 25 - (-25) = 50$.

The answer is $\boxed{\textbf{(B)}\ 50}$.

A contraction with factor $\tfrac12$ shrinks every gap by half, and composing it shrinks again. The hidden ingredient is $f(300) = f(900)$, which lets you travel from $400$ to $800$ through a shortcut of total length $100$ instead of $400$. Maximum-value problems then demand a construction: piecewise-linear functions with slopes $\pm\tfrac12$ and $0$ are the standard way to make every inequality tight.

• Bounding |f(800) − f(400)| by ½·400 = 200 directly (ignoring f(300) = f(900)) and answering 100.
• Stopping at the bound $50$ without checking it is attainable, or trying a symmetric $f$ like $\tfrac12|x - 600|$, which forces $f(800) = f(400)$ and gives $0$.