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

Revision as of 06:00, 4 December 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 $f(f(800))-f(f(400))?$ $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$

Solution 1 (Absolute Values and Inequalities)

By definition, we have $\begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100. \end{align*}$ Note that ...

Let ...

We rewrite ... as ...

~MRENTHUSIASM

Solution 2 (Lipschitz Condition)

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 3 (Educated Guess)

Divide both sides by $|x - y|$ to get $\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}$ . This means that when we 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 find the maximum value of $|f(800) - f(400)|$ , 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 attainable because the slope of the line connecting $f(400)$ and $f(800)$ still has absolute value 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)

Video Solution by OmegaLearn Using Algebraic Manipulation

https://youtu.be/-yzpw6b3_KA

~ pi_is_3.14

@@ Line 1: / Line 1: @@
 ==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 such functions that also satisfy the equation <math>f(300) = f(900)</math>, what is the greatest possible value of
+Consider functions <math>f</math> that satisfy <cmath>|f(x)-f(y)|\leq \frac{1}{2}|x-y|</cmath> for all real numbers <math>x</math> and <math>y</math>. Of all such functions that also satisfy the equation <math>f(300) = f(900)</math>, what is the greatest possible value of
 <cmath>f(f(800))-f(f(400))?</cmath>
 <math>\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200</math>
-==Solution 1==
+==Solution 1 (Absolute Values and Inequalities)==
+By definition, we have
+<cmath>\begin{align*}
+|f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\
+&\leq \frac12\left|\frac12|800-400|\right| \\
+&= 100.
+\end{align*}</cmath>
+Note that ...
+Let ...
+We rewrite ... as ...
+~MRENTHUSIASM
+==Solution 2 (Lipschitz Condition)==
 Denote <math>f(900)-f(600) = a</math>.
@@ Line 22: / Line 37: @@
 \]
 </cmath>
 Thus,
 <cmath>
@@ Line 40: / Line 54: @@
 \end{align*}
 </cmath>
 Thus, <math>f(800) - f(400)</math> is maximized at <math>a = 0</math>, <math>f(800)-f(600) = 50</math>, <math>f(400)-f(600)=-50</math>, with the maximal value 100.
@@ Line 53: / Line 66: @@
 \end{align*}
 </cmath>
 We have already construct instances in which the second inequality above is augmented to an equality.
@@ Line 72: / Line 84: @@
 \]
 </cmath>
 Therefore, the maximum value of <math>f(f(800)) - f(f(400))</math> is
-<math>\boxed{\textbf{(B) 50}}</math>.
+<math>\boxed{\textbf{(B)}\ 50}</math>.
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
@@ Line 80: / Line 91: @@
 ~Viliciri (LaTeX edits)
-==Solution 2 (If you're low on time)==
+==Solution 3 (Educated Guess)==
 Divide both sides by <math>|x - y|</math> to get <math>\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}</math>. This means that when we take any two points on <math>f</math>, the absolute value of the slope between the two points is at most <math>\frac{1}{2}</math>.
@@ Line 86: / Line 97: @@
 Let <math>f(300) = f(900) = c</math>, and since we want to find the maximum value of <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 attainable because the slope of the line connecting <math>f(400)</math> and <math>f(800)</math> still has absolute value 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==

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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