Difference between revisions of "2022 AMC 10B Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | Consider functions <math>f</math> that satisfy < | + | 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== | + | ==Solution 2 (Lipschitz Condition)== |
Denote <math>f(900)-f(600) = a</math>. | Denote <math>f(900)-f(600) = a</math>. | ||
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\] | \] | ||
</cmath> | </cmath> | ||
− | |||
Thus, | Thus, | ||
<cmath> | <cmath> | ||
Line 45: | 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. | ||
Line 58: | Line 81: | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | |||
We have already construct instances in which the second inequality above is augmented to an equality. | We have already construct instances in which the second inequality above is augmented to an equality. | ||
Line 77: | 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>\boxed{\textbf{(B)}\ 50}</math>. |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
~Viliciri (LaTeX edits) | ~Viliciri (LaTeX edits) | ||
+ | |||
+ | ==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(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>. | ||
+ | |||
+ | ==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== | ==Video Solution== | ||
Line 90: | Line 123: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~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 | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2022|ab=B|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:53, 4 November 2023
Contents
Problem
Consider functions that satisfy
for all real numbers
and
. Of all such functions that also satisfy the equation
, what is the greatest possible value of
Solution 1 (Absolute Values and Inequalities)
We have
from which we eliminate answer choices
and
Note that
Let
Together, it follows that
We rewrite
as
~MRENTHUSIASM
Solution 2 (Lipschitz Condition)
Denote .
Because
,
.
Following from the Lipschitz condition given in this problem, and
and
Thus,
Thus,
is maximized at
,
,
, with the maximal value 100.
By symmetry, following from an analogous argument, we can show that is minimized at
,
,
, with the minimal value
.
Following from the Lipschitz condition,
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:
Therefore, the maximum value of
is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Viliciri (LaTeX edits)
Solution 3 (Slopes)
Divide both sides by to get
. This means that when we take any two points on
, the absolute value of the slope between the two points is at most
.
Let , and since we want to find the maximum value of
, we can take the most extreme case and draw a line with slope
down from
to
and a line with slope
up from
to
. Then
and
, so
, and this is attainable because the slope of the line connecting
and
still has absolute value less than
.
Therefore, .
Solution 4 (Translation)
Consider . Then
satisfies all the conditions and
. We would want
and
as distant from each other as possible. So assign
and
, the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for
as the answer.
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Algebraic Manipulation
~ pi_is_3.14
Video Solution by Interstigation
~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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.