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2022 AMC 10B Problems/Problem 24

Revision as of 13:54, 17 November 2022 by Stevenyiweichen (talk | contribs) (Created page with "==Problem== Consider functions <math>f</math> that satisfy <math>| f(x) - f(y) | \leq \frac{1}{2} |x-y|</math> for all real numbers <math>x</math> and <math>y</math>. Of all...")
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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

\[ f(f(800)) - f(f(400)) ? \]

Solution

Denote $f(900)-f(600) = a$. Because $f(300) = f(900)$, $f(300) - f(600) = a$.

Following from the Lipschitz condition given in this problem, $|a| \leq 150$ and \[ f(800) - f(600) \leq \min \left\{ a + 50 , 100 \right\} \] and \[ f(400) - f(600) \geq \max \left\{ a - 50 , -100 \right\} . \]

Thus, \begin{align*} f(800) - f(400) & \leq \min \left\{ a + 50 , 100 \right\} - \max \left\{ a - 50 , -100 \right\} \\ & = 100 + \min \left\{ a, 50 \right\} - \max \left\{ a , - 50 \right\} \\ & = 100 + \left\{ \begin{array}{ll} a + 50 & \mbox{ if } a \leq -50 \\ 0 & \mbox{ if } -50 < a < 50 \\ -a + 50 & \mbox{ if } a \geq 50 \end{array} \right. . \end{align*}

Thus, $f(800) - f(400)$ is maximized at $a = 0$, $f(800)-f(600) = 50$, $f(400)-f(600)=-50$, with the maximal value 100.

By symmetry, following from an analogous argument, we can show that $f(800) - f(400)$ is minimized at $a = 0$, $f(800)-f(600) = -50$, $f(400)-f(600)=50$, with the minimal value $-100$.

Following from the Lipschitz condition, \begin{align*} f(f(800)) - f(f(400)) & \leq \frac{1}{2} \left| f(800) - f(400) \right| \\ & \leq 50 . \end{align*}

We have already construct instances in which the second inequality above is augmented to an equality.

Now, we construct an instance in which the first inequality above is augmented to an equality.

Consider the following piecewise-linear function: \[ f(x) = \left\{ \begin{array}{ll} \frac{1}{2} \left( x - 300 \right) & \mbox{ if } x \leq 300 \\ -\frac{1}{2} \left( x - 300 \right) & \mbox{ if } 300 < x \leq 400 \\ \frac{1}{2} \left( x - 600 \right) & \mbox{ if } 400 < x \leq 800 \\ -\frac{1}{2} \left( x - 900 \right) & \mbox{ if } x > 800 \end{array} \right.. \]

Therefore, the maximum value of $f(f(800)) - f(f(400))$ is \boxed{\textbf{(B) 50}}.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/2Li0IYOQCFQ

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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