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

(→‎Solution 3 (Educated Guess))
Line 8: Line 8:
 
By definition, we have
 
By definition, we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
|f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\
+
|f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\
 
&\leq \frac12\left|\frac12|800-400|\right| \\
 
&\leq \frac12\left|\frac12|800-400|\right| \\
−
&= 100.
+
&= 100,
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Note that ...
+
from which we eliminate answer choices <math>\textbf{(D)}</math> and \textbf{(E)}.<math>
−
 
 
−
Let ...
 
−
 
 
−
We rewrite ... as ...
 
  
 +
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 </math>a=f(300)=f(900).<math> Together, we conclude 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
 
~MRENTHUSIASM
  
 
==Solution 2 (Lipschitz Condition)==
 
==Solution 2 (Lipschitz Condition)==
  
−
Denote <math>f(900)-f(600) = a</math>.
+
Denote </math>f(900)-f(600) = a<math>.
−
Because <math>f(300) = f(900)</math>, <math>f(300) - f(600) = a</math>.
+
Because </math>f(300) = f(900)<math>, </math>f(300) - f(600) = a<math>.
  
−
Following from the Lipschitz condition given in this problem, <math>|a| \leq 150</math> and
+
Following from the Lipschitz condition given in this problem, </math>|a| \leq 150<math> and
 
<cmath>
 
<cmath>
 
\[
 
\[
Line 54: Line 69:
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</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.
+
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.
  
−
By symmetry, following from an analogous argument, we can show that <math>f(800) - f(400)</math> is minimized at <math>a = 0</math>, <math>f(800)-f(600) = -50</math>, <math>f(400)-f(600)=50</math>, with the minimal value <math>-100</math>.
+
By symmetry, following from an analogous argument, we can show that </math>f(800) - f(400)<math> is minimized at </math>a = 0<math>, </math>f(800)-f(600) = -50<math>, </math>f(400)-f(600)=50<math>, with the minimal value </math>-100<math>.
  
 
Following from the Lipschitz condition,
 
Following from the Lipschitz condition,
Line 84: Line 99:
 
\]
 
\]
 
</cmath>
 
</cmath>
−
Therefore, the maximum value of <math>f(f(800)) - f(f(400))</math> is
+
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)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Line 93: Line 108:
 
==Solution 3 (Educated Guess)==
 
==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>.
+
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(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) = \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) = \boxed{\textbf{(B)}\ 50}$.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 06:10, 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)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*} from which we eliminate answer choices $\textbf{(D)}$ 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$. Because$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)$is maximized at$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$ (Error compiling LaTeX. Unknown error_msg)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, <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))$is$\boxed{\textbf{(B)}\ 50}$.

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

~Viliciri (LaTeX edits)

==Solution 3 (Educated Guess)==

Divide both sides by$ (Error compiling LaTeX. Unknown error_msg)|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$ (Error compiling LaTeX. Unknown error_msg)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,$ (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

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

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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