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Difference between revisions of "2022 AMC 10B Problems/Problem 24"

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Let <math>f(300) = f(900) = c</math>, and since we want to maximize <math>|f(800) - f(400)|</math>, we can take the most extreme case and draw a line with slope <math>\frac{-1}{2}</math> down from <math>f(300)</math> to <math>f(400)</math> and a line with slope <math>\frac{-1}{2}</math> up from <math>f(900)</math> to <math>f(800)</math>. Then <math>f(400) = c - 50</math> and <math>f(800) = c + 50</math>, so <math>|f(800) - f(400)| = |c + 50 - (c - 50)| = 100</math>, and this is possible because the slope of the line connecting <math>f(400)</math> and <math>f(800)</math> is still less than <math>\frac{1}{2}</math>.
 
Let <math>f(300) = f(900) = c</math>, and since we want to maximize <math>|f(800) - f(400)|</math>, we can take the most extreme case and draw a line with slope <math>\frac{-1}{2}</math> down from <math>f(300)</math> to <math>f(400)</math> and a line with slope <math>\frac{-1}{2}</math> up from <math>f(900)</math> to <math>f(800)</math>. Then <math>f(400) = c - 50</math> and <math>f(800) = c + 50</math>, so <math>|f(800) - f(400)| = |c + 50 - (c - 50)| = 100</math>, and this is possible because the slope of the line connecting <math>f(400)</math> and <math>f(800)</math> is still less than <math>\frac{1}{2}</math>.
  
−
Therefore, <math>|f(f(800) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) \rightarrow \boxed{\textbf{(B) 50}}</math>
+
Therefore, <math>|f(f(800) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) \rightarrow \boxed{\textbf{(B) 50}}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 12:07, 19 November 2022

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 1

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)

~Viliciri (LaTeX edits)

Solution 2 (Cheap)

Divide both sides by $|x - y|$ to get $\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}$. This means that when you take any two points on $f$, the absolute value of the slope between the two points is at most $\frac{1}{2}$.

Let $f(300) = f(900) = c$, and since we want to maximize $|f(800) - f(400)|$, we can take the most extreme case and draw a line with slope $\frac{-1}{2}$ down from $f(300)$ to $f(400)$ and a line with slope $\frac{-1}{2}$ up from $f(900)$ to $f(800)$. Then $f(400) = c - 50$ and $f(800) = c + 50$, so $|f(800) - f(400)| = |c + 50 - (c - 50)| = 100$, and this is possible because the slope of the line connecting $f(400)$ and $f(800)$ is still less than $\frac{1}{2}$.

Therefore, $|f(f(800) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) \rightarrow \boxed{\textbf{(B) 50}}$.

Video Solution

https://youtu.be/2Li0IYOQCFQ

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

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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