2022 AMC 10 B Problem 24

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To help you master 2022 AMC 10 B Problem 24, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2022 AMC10 B Problem 24

Solution

Solution 1

Common-Value Bound and Attaining Construction

First bound the distance between the inner values $f(800)$ and $f(400)$ by comparing each with the common value $f(300)=f(900)$. Then apply the same Lipschitz condition to those inner values. Because the problem asks for a greatest possible value, we must also construct a function that reaches the bound.

A function satisfying $|f(x)-f(y)|\le L|x-y|$ is called Lipschitz with constant $L$. Here $L=\frac{1}{2}$, so changing an input by $d$ can change the output by at most $\frac{d}{2}$. Also, to prove a greatest possible value, we need both an upper bound and an example attaining it.

Step 1: Bound each inner output using the common value

Let the equal outputs at inputs 300 and 900 be denoted by $c$. The input 400 is 100 units from 300, while 800 is 100 units from 900. Applying the given condition to these two pairs places both relevant inner outputs within 50 of $c$.

$$c=f(300)=f(900)$$

$$|f(400)-c|=|f(400)-f(300)|\le \frac{1}{2}|400-300|=50$$

$$|f(800)-c|=|f(800)-f(900)|\le \frac{1}{2}|800-900|=50$$

Step 2: Bound the distance between the inner outputs

The triangle inequality allows us to travel from one inner output to the other through their common reference value $c$. Since each is at most 50 from $c$, their distance is at most 100.

$$|f(800)-f(400)|\le |f(800)-c|+|c-f(400)|$$

$$|f(800)-f(400)|\le 50+50=100$$

Step 3: Apply the Lipschitz condition to the outer function values

Now treat the two inner outputs as the inputs to the outer applications of $f$. The desired signed difference cannot exceed its absolute value, and the Lipschitz condition reduces the bound by another factor of one-half.

$$f(f(800))-f(f(400))\le |f(f(800))-f(f(400))|$$

$$|f(f(800))-f(f(400))|\le \frac{1}{2}|f(800)-f(400)|$$

$$f(f(800))-f(f(400))\le \frac{1}{2}\cdot 100=50$$

Step 4: Construct a function that attains the upper bound

To show that 50 is genuinely possible, define a continuous piecewise-linear function. Its segment slopes all have absolute value at most one-half, so the output change across each segment is at most one-half of the input change. Splitting any interval at the breakpoints and adding these bounds proves that the condition holds for every pair of real inputs.

$$f(x)=\{\begin{array}{ll}-25& x\le -50\\ \frac{x}{2}& -50\le x\le 50\\ 30-\frac{x}{10}& 50\le x\le 300\\ -\frac{x-300}{2}& 300\le x\le 400\\ -50+\frac{x-400}{4}& 400\le x\le 800\\ 50-\frac{x-800}{2}& 800\le x\le 900\\ 0& x\ge 900\end{array}$$

$$0,\frac{1}{2},-\frac{1}{10},-\frac{1}{2},\frac{1}{4},-\frac{1}{2},0$$

$$|\text{each slope}|\le \frac{1}{2}$$

Step 5: Verify that the construction reaches the bound

The constructed function satisfies the required equal-value condition. Its inner outputs are the two endpoints where the central segment has slope one-half, so the outer difference attains the upper bound.

$$f(300)=0=f(900)$$

$$f(400)=-50$$

$$f(800)=50$$

$$f(f(800))-f(f(400))=f(50)-f(-50)$$

$$f(50)-f(-50)=25-(-25)=50$$

Correct answer: B. $50$
Every legal function gives a value at most $50$, and the displayed function satisfies all conditions and produces exactly $50$. Therefore the greatest possible value is $\boxed{50}$, corresponding to choice $\boxed{\mathrm{B}}$.

Similar Problems in Real Exams

The following five AMC 10 B problems are closely related to AMC 10 B 2022 Question 24. These real AMC 10 B problems test the same key concept—Algebra. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2024 AMC 10 B Problem 5

2022 AMC 10 B Problem 1

2019 AMC 10 B Problem 9

2018 AMC 10 B Problem 25

2025 AMC 10 B Problem 1

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Lipschitz inequality:

$x$ and $y$ are any real inputs, and $L$ is a fixed nonnegative constant. In this problem, $L=\frac{1}{2}$. It bounds how far apart two outputs can be based on the distance between their inputs. It is used first on the original inputs and then again on the inner outputs.

• Triangle inequality through a reference value:

$a$, $b$, and $c$ are real numbers. Both $f(400)$ and $f(800)$ are close to the same value $c=f(300)=f(900)$, so their distance is at most the sum of those two distances.

• Piecewise-linear Lipschitz test:

The function must be continuous at the breakpoints, and every linear piece must have slope with absolute value at most $L$. It verifies that the constructed function is legal on every interval, including intervals crossing several breakpoints.

• Requirement for proving a greatest value:

This applies whenever a problem asks for a greatest or maximum possible value. The inequalities prove that no value above 50 is possible, while the constructed function proves that 50 is achievable.

Key Takeaway: Remember This Strategy for Next Time

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