Difference between revisions of "2022 AMC 10B Problems/Problem 24"
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&= 100, | &= 100, | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | from which we eliminate answer choices <math>\textbf{(D)}</math> and \textbf{(E)}.<math> | + | from which we eliminate answer choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}.</math> |
Note that | Note that | ||
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|f(400)-f(900)|&\leq 250. \\ | |f(400)-f(900)|&\leq 250. \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Let < | + | Let <math>a=f(300)=f(900).</math> Together, we conclude that |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
|f(800)-a|&\leq 50, \\ | |f(800)-a|&\leq 50, \\ | ||
|f(400)-a|&\leq 50. \\ | |f(400)-a|&\leq 50. \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | We rewrite < | + | We rewrite <math>(\bigstar)</math> as |
<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)| \\ | ||
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==Solution 2 (Lipschitz Condition)== | ==Solution 2 (Lipschitz Condition)== | ||
− | Denote < | + | Denote <math>f(900)-f(600) = a</math>. |
− | Because < | + | Because <math>f(300) = f(900)</math>, <math>f(300) - f(600) = a</math>. |
− | Following from the Lipschitz condition given in this problem, < | + | Following from the Lipschitz condition given in this problem, <math>|a| \leq 150</math> and |
<cmath> | <cmath> | ||
\[ | \[ | ||
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\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | Thus, < | + | Thus, <math>f(800) - f(400)</math> is maximized at <math>a = 0</math>, <math>f(800)-f(600) = 50</math>, <math>f(400)-f(600)=-50</math>, with the maximal value 100. |
− | By symmetry, following from an analogous argument, we can show that < | + | By symmetry, following from an analogous argument, we can show that <math>f(800) - f(400)</math> is minimized at <math>a = 0</math>, <math>f(800)-f(600) = -50</math>, <math>f(400)-f(600)=50</math>, with the minimal value <math>-100</math>. |
Following from the Lipschitz condition, | Following from the Lipschitz condition, | ||
Line 99: | Line 99: | ||
\] | \] | ||
</cmath> | </cmath> | ||
− | Therefore, the maximum value of < | + | Therefore, the maximum value of <math>f(f(800)) - f(f(400))</math> is |
− | < | + | <math>\boxed{\textbf{(B)}\ 50}</math>. |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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==Solution 3 (Educated Guess)== | ==Solution 3 (Educated Guess)== | ||
− | Divide both sides by < | + | 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 < | + | 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, < | + | 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>. |
==Video Solution== | ==Video Solution== |
Revision as of 07:10, 4 December 2022
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)
By definition, we have
from which we eliminate answer choices
and
Note that
Let
Together, we conclude 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 (Educated Guess)
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, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.