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

Latest revision as of 18:53, 4 November 2023

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)

We have $\begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*}$ from which we eliminate answer choices $\textbf{(D)}$ and $\textbf{(E)}.$

Note that $\begin{alignat*}{8} |f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\ |f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\ |f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\ |f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\ \end{alignat*}$ Let $a=f(300)=f(900).$ Together, it follows that $\begin{align*} |f(800)-a|&\leq 50, \\ |f(400)-a|&\leq 50. \\ \end{align*}$ We rewrite $(\bigstar)$ as $\begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &= \frac12|(f(800)-a)-(f(400)-a)| \\ &\leq \frac12|50-(-50)| \\ &=\boxed{\textbf{(B)}\ 50}. \end{align*}$ ~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 (Slopes)

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 $f(800)$ to $f(900)$ . 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, $|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}$ .

Solution 4 (Translation)

Consider $g(x) = f(x)-f(300)$ . Then $g(x)$ satisfies all the conditions and $g(300) = g(900) = 0$ . We would want $g(400)$ and $g(800)$ as distant from each other as possible. So assign $g(400) = -50$ and $g(800) = 50$ , the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for $|g(g(800)) - g(g(400))| = |g(50) - g(-50 )| \leq \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}$ as the answer.

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

Video Solution by Interstigation

https://youtu.be/n0ZTR6edOBs

~Interstigation

@@ 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
+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
-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>
-<cmath>
+==Solution 1 (Absolute Values and Inequalities)==
-\[
+We have
-f(f(800)) - f(f(400)) ?
+<cmath>\begin{align*}
-\]
+|f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\
-</cmath>
+&\leq \frac12\left|\frac12|800-400|\right| \\
+&= 100,
+\end{align*}</cmath>
+from which we eliminate answer choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}.</math>
+Note that
+<cmath>\begin{alignat*}{8}
+|f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\
+|f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\
+|f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\
+|f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\
+\end{alignat*}</cmath>
+Let <math>a=f(300)=f(900).</math> Together, it follows that
+<cmath>\begin{align*}
+|f(800)-a|&\leq 50, \\
+|f(400)-a|&\leq 50. \\
+\end{align*}</cmath>
+We rewrite <math>(\bigstar)</math> 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 1==
+==Solution 2 (Lipschitz Condition)==
 Denote <math>f(900)-f(600) = a</math>.
@@ Line 27: / Line 52: @@
 \]
 </cmath>
 Thus,
 <cmath>
@@ Line 45: / Line 69: @@
 \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 58: / Line 81: @@
 \end{align*}
 </cmath>
 We have already construct instances in which the second inequality above is augmented to an equality.
@@ Line 77: / Line 99: @@
 \]
 </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 85: / Line 106: @@
 ~Viliciri (LaTeX edits)
-==Solution 2 (Good if you're low on time)==
+==Solution 3 (Slopes)==
 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>.
-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>.
+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(800)</math> to <math>f(900)</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) = \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>.
+==Solution 4 (Translation)==
+Consider <math> g(x) = f(x)-f(300) </math>. Then <math> g(x)</math> satisfies all the conditions and <math> g(300) = g(900) = 0 </math>. We would want <math> g(400)</math> and <math> g(800)</math> as distant from each other as possible. So assign <math> g(400) = -50</math> and <math> g(800) = 50</math>, the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for <math>|g(g(800)) - g(g(400))| = |g(50) - g(-50  )| \leq \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}</math> as the answer.
 ==Video Solution==
@@ Line 104: / Line 129: @@
 ~ pi_is_3.14
+==Video Solution by Interstigation==
+https://youtu.be/n0ZTR6edOBs
+~Interstigation
 ==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
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