AMC 10 · 2022 · #24

Grade 8 algebra

lipschitz-conditionfunction-evaluationabsolute-valuebound-inequality-then-enumerateextremal-construction work-backwardsconvert-to-algebraeasier-related-problem ↑ Prerequisites: function-evaluationabsolute-value

📏 Long solution 💡 4 insights

Problem

Consider functions $f$ that satisfy $|f(x)-f(y)|\leq \frac{1}{2}|x-y|$ for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of
$f(f(800))-f(f(400))?$

Pick an answer.

(A)

(B)

(C)

100

(D)

150

(E)

200

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Toolkit + CCSS Solution

Understand

Restated: A function $f : \mathbb{R} \to \mathbb{R}$ has the property that $|f(x) - f(y)| \le \tfrac{1}{2}|x - y|$ for every pair of real numbers $x, y$, and also satisfies $f(300) = f(900)$. Among all such $f$, what is the greatest possible value of $f(f(800)) - f(f(400))$?

Givens: $|f(x) - f(y)| \le \tfrac{1}{2}|x - y|$ for all real $x, y$ (Lipschitz with constant $\tfrac{1}{2}$); $f(300) = f(900)$ — call this common value $c$; Answer choices: (A) $25$, (B) $50$, (C) $100$, (D) $150$, (E) $200$

Unknowns: Maximum possible value of $f(f(800)) - f(f(400))$

Understand

Plan

Primary tool: #11 Work Backwards

Secondary: #13 Convert to Algebra, #9 Solve an Easier Related Problem, #1 Draw a Diagram, #6 Guess and Check

Tool #11 (Work Backwards) — we want $f(f(800)) - f(f(400))$; apply the Lipschitz bound to peel off the outer $f$ first, reducing to bounding $|f(800) - f(400)|$. Then bound $|f(800) - f(400)|$ via the anchor $f(300) = f(900) = c$. Tool #9 (Easier) — solve the easier sub-problem: "given $f(300) = f(900) = c$, how far apart can $f(800)$ and $f(400)$ be?" Tool #13 (Algebra) — combine the two bounds to get $\le 50$. Tool #1 (Diagram) — sketch a piecewise linear $f$ (graph of slopes $\le \tfrac{1}{2}$ in absolute value) that hits the bound exactly, proving $50$ is achievable. Tool #6 (Guess & Check) — propose a specific $f$ and verify each slope $\le \tfrac{1}{2}$.

Execute — Answer: B

#11 Work Backwards 8.F.A.1 Step 1

Apply the Lipschitz bound to the outer $f$ first.
With $u = f(800), v = f(400)$ (real numbers!), $|f(u) - f(v)| \le \tfrac{1}{2}|u - v|$, that is $|f(f(800)) - f(f(400))| \le \tfrac{1}{2}|f(800) - f(400)|$.
So the whole problem reduces to bounding $|f(800) - f(400)|$.

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

💡 Grade 8 — the Lipschitz inequality applies to any two real inputs, including $f(800)$ and $f(400)$.

#9 Solve an Easier Related Problem 7.NS.A.3 Step 2

Bound $f(800)$ using the anchor $f(900) = c$: $|f(800) - c| = |f(800) - f(900)| \le \tfrac{1}{2}|800 - 900| = 50$.
Bound $f(400)$ using anchor $f(300) = c$: $|f(400) - c| \le \tfrac{1}{2}|400 - 300| = 50$.
So both $f(800), f(400) \in [c - 50, c + 50]$.

$$f(800), f(400) \in [c - 50, c + 50]$$

💡 Grade 7 — distance from the anchor $c$ is at most half the input distance to $300$ or $900$.

#13 Convert to Algebra 7.NS.A.3 Step 3

Difference: $f(800) - f(400) \le (c+50) - (c-50) = 100$ and $f(800) - f(400) \ge (c-50) - (c+50) = -100$, so $|f(800) - f(400)| \le 100$.
Substituting into step 1: $|f(f(800)) - f(f(400))| \le \tfrac{1}{2}(100) = 50$.
So the answer is at most $50$.

$$f(f(800)) - f(f(400)) \le 50$$

💡 Grade 7 — endpoints of intervals subtract to give the range of the difference.

#11 Work Backwards 8.F.B.4 Step 4

Now show $50$ is achievable.
We need every inequality above to become equality.
Pick $c = 0$ for simplicity, so $f(300) = f(900) = 0$.
We want $f(800) = 50, f(400) = -50$, and finally $f(50) - f(-50) = 50$ (since $f(800) = 50, f(400) = -50$ would mean $f(f(800)) - f(f(400)) = f(50) - f(-50)$).
To make $f(50) - f(-50) = 50$ at the Lipschitz limit, the slope between $-50$ and $50$ must be exactly $\tfrac{1}{2}$.

$$f(300) = f(900) = 0, \; f(800) = 50, \; f(400) = -50, \; f(50) - f(-50) = 50$$

💡 Grade 8 — work backwards from the equality conditions to design $f$'s required values.

#6 Guess and Check 8.F.B.4 Step 5

Construct piecewise linear $f$ with key knots $(-50, -25), (50, 25), (300, 0), (400, -50), (800, 50), (900, 0)$, extended outward as constants.
Compute the slope of every segment: $(-50, -25) \to (50, 25)$: slope $\tfrac{50}{100} = \tfrac{1}{2}$.
$(50, 25) \to (300, 0)$: slope $\tfrac{-25}{250} = -\tfrac{1}{10}$.
$(300, 0) \to (400, -50)$: slope $-\tfrac{1}{2}$.
$(400, -50) \to (800, 50)$: slope $\tfrac{100}{400} = \tfrac{1}{4}$.
$(800, 50) \to (900, 0)$: slope $-\tfrac{1}{2}$.
All slopes have $|\text{slope}| \le \tfrac{1}{2}$.

$$|\text{slope}| \le \tfrac{1}{2} \text{ on every segment}$$

💡 Grade 8 — for a piecewise linear function, Lipschitz constant $\le \tfrac{1}{2}$ is equivalent to every slope $\le \tfrac{1}{2}$ in absolute value.

#13 Convert to Algebra 5.NBT.B.7 Step 6

Verify the target.
$f(800) = 50, f(400) = -50$, so $f(f(800)) = f(50) = 25$ and $f(f(400)) = f(-50) = -25$.
Therefore $f(f(800)) - f(f(400)) = 25 - (-25) = 50$.
The upper bound $50$ is attained, so the maximum is exactly $50$.
Answer (B).

$$f(f(800)) - f(f(400)) = 25 - (-25) = 50 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 5 — simple subtraction to confirm the value.

[1] #11 8.F.A.1 Apply the Lipschitz bound to the outer $f$ first. With $u = f(800), v = f(400)$

[2] #9 7.NS.A.3 Bound $f(800)$ using the anchor $f(900) = c$: $|f(800) - c| = |f(800) - f(900)|

[3] #13 7.NS.A.3 Difference: $f(800) - f(400) \le (c+50) - (c-50) = 100$ and $f(800) - f(400) \ge

[4] #11 8.F.B.4 Now show $50$ is achievable. We need every inequality above to become equality.

[5] #6 8.F.B.4 Construct piecewise linear $f$ with key knots $(-50, -25), (50, 25), (300, 0), (

[6] #13 5.NBT.B.7 Verify the target. $f(800) = 50, f(400) = -50$, so $f(f(800)) = f(50) = 25$ and

Review

Reasonableness: Sanity. The Lipschitz constant $\tfrac{1}{2}$ halves every "distance budget" each time we apply the inequality. Applying it twice halves it twice: from the input gap $|800 - 400| = 400$, naïvely we'd get $\tfrac{1}{4}\cdot 400 = 100$ for $|f(f(800)) - f(f(400))|$. But the anchor $f(300) = f(900) = c$ pins both $f(800)$ and $f(400)$ to a band of total width $100$ (not $200$), so the inner difference is at most $100$, and one outer halving gives $50$. The construction shows $50$ is achievable, so (B). Choice (E) $200 = \tfrac{1}{2}\cdot 400$ ignores both the anchor and one application of Lipschitz; (C) $100$ ignores one Lipschitz application; (A) $25$ over-corrects.

Alternative: Tool #1 (Diagram). Plot $y = f(x)$. The constraint $|f(x)-f(y)| \le \tfrac{1}{2}|x-y|$ means the graph lives between lines of slope $\pm \tfrac{1}{2}$ from every point. From $f(300) = f(900) = c$, two double-cones of slope $\pm \tfrac{1}{2}$ converge from those points; the intersection at $x = 400$ has height $c \pm 50$, and likewise at $x = 800$. Reading the diagram, $f(800)$ and $f(400)$ each live in $[c-50, c+50]$, so their difference is in $[-100, 100]$. One more halving gives $|f(f(800)) - f(f(400))| \le 50$.

CCSS standards used (min grade 8)

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Final subtraction $25 - (-25) = 50$ to read off the maximum.)
7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers (Combining the two intervals $[c-50, c+50]$ for $f(800)$ and $f(400)$ to bound their difference by $100$.)
8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input (Applying the Lipschitz inequality to the outer $f$ with inputs $u = f(800), v = f(400)$.)
8.F.B.4 Construct a function to model a linear relationship between two quantities (Building the explicit piecewise linear $f$ with all slopes $\le \tfrac{1}{2}$ in absolute value that hits the upper bound.)

⭐ This AMC 10 problem only needs Grade 8 functions you already know — the Lipschitz rule "every step in $f$ is at most half the step in $x$" applies twice for the nested $f(f(\cdot))$. The anchor $f(300) = f(900) = c$ forces $f(800)$ and $f(400)$ each within $50$ of $c$, so their difference is at most $100$. One more halving gives the answer $\le 50$, and a piecewise linear graph with every slope $\le \tfrac{1}{2}$ in absolute value hits $50$ exactly.