2022 AMC 10B Problems/Problem 24
Contents
Problem
Consider functions 
that satisfy ![\[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\]](http://latex.artofproblemsolving.com/8/0/7/807767b613ba502ffb8480dcb6ec61bc4f0f8fd2.png)
for all real numbers 
and 
. Of all such functions that also satisfy the equation 
, what is the greatest possible value of
![\[f(f(800))-f(f(400))?\]](http://latex.artofproblemsolving.com/d/c/0/dc03e872a46ec49f679240b312b43547ea012bd9.png)
Solution 1 (Absolute Values and Inequalities)
By definition, we have
from which we eliminate answer choices 
and \textbf{(E)}.$Note that
<cmath>\begin{align*}
|f(800)-f(300)|&\leq 250, \\
|f(800)-f(900)|&\leq 50, \\
|f(400)-f(300)|&\leq 50, \\
|f(400)-f(900)|&\leq 250. \\
\end{align*}</cmath>
Let$ (Error compiling LaTeX. Unknown error_msg)a=f(300)=f(900).$Together, we conclude that
<cmath>\begin{align*}
|f(800)-a|&\leq 50, \\
|f(400)-a|&\leq 50. \\
\end{align*}</cmath>
We rewrite$ (Error compiling LaTeX. Unknown error_msg)(\bigstar)$as
<cmath>\begin{align*}
|f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\
&= \frac12|(f(800)-a)-(f(400)-a)| \\
&\leq \frac12|50-(-50)| \\
&=\boxed{\textbf{(B)}\ 50}.
\end{align*}</cmath>
~MRENTHUSIASM
==Solution 2 (Lipschitz Condition)==
Denote$ (Error compiling LaTeX. Unknown error_msg)f(900)-f(600) = a
f(300) = f(900)
f(300) - f(600) = a$.
Following from the Lipschitz condition given in this problem,$ (Error compiling LaTeX. Unknown error_msg)|a| \leq 150$and
<cmath>
\[
f(800) - f(600) \leq \min \left\{ a + 50 , 100 \right\}
\]
</cmath>
and
<cmath>
\[
f(400) - f(600) \geq \max \left\{ a - 50 , -100 \right\} .
\]
</cmath>
Thus,
<cmath>
\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*}
</cmath>
Thus,$ (Error compiling LaTeX. Unknown error_msg)f(800) - f(400)
a = 0f(800)-f(600) = 50f(400)-f(600)=-50$, with the maximal value 100.
By symmetry, following from an analogous argument, we can show that$ (Error compiling LaTeX. Unknown error_msg)f(800) - f(400)
a = 0f(800)-f(600) = -50f(400)-f(600)=50
-100$.
Following from the Lipschitz condition, <cmath> \begin{align*} f(f(800)) - f(f(400)) & \leq \frac{1}{2} \left| f(800) - f(400) \right| \\ & \leq 50 . \end{align*} </cmath> 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:
<cmath>
\[
f(x) =
\left\{
\begin{array}{ll}
\frac{1}{2} \left( x - 300 \right) & \mbox{ if } x \leq 300 \\
-\frac{1}{2} \left( x - 300 \right) & \mbox{ if } 300 < x \leq 400 \\
\frac{1}{2} \left( x - 600 \right) & \mbox{ if } 400 < x \leq 800 \\
-\frac{1}{2} \left( x - 900 \right) & \mbox{ if } x > 800
\end{array}
\right..
\]
</cmath>
Therefore, the maximum value of$ (Error compiling LaTeX. Unknown error_msg)f(f(800)) - f(f(400))
\boxed{\textbf{(B)}\ 50}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Viliciri (LaTeX edits)
==Solution 3 (Educated Guess)==
Divide both sides by$ (Error compiling LaTeX. Unknown error_msg)|x - y|
\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}
f
\frac{1}{2}$.
Let$ (Error compiling LaTeX. Unknown error_msg)f(300) = f(900) = c
|f(800) - f(400)|
\frac{-1}{2}
f(300)
f(400)
\frac{-1}{2}
f(900)f(800)
f(400) = c - 50
f(800) = c + 50
|f(800) - f(400)| = |c + 50 - (c - 50)| = 100
f(400)f(800)
\frac{1}{2}$.
Therefore,$ (Error compiling LaTeX. Unknown error_msg)|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}$.
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Algebraic Manipulation
~ pi_is_3.14
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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