Difference between revisions of "2022 AMC 10B Problems/Problem 24"
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Let <math>f(300) = f(900) = c</math>, and since we want to maximize <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 possible because the slope of the line connecting <math>f(400)</math> and <math>f(800)</math> is still less than <math>\frac{1}{2}</math>. | Let <math>f(300) = f(900) = c</math>, and since we want to maximize <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 possible because the slope of the line connecting <math>f(400)</math> and <math>f(800)</math> is still 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== | ==Video Solution== |
Revision as of 13:07, 19 November 2022
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
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 2 (Cheap)
Divide both sides by to get . This means that when you take any two points on , the absolute value of the slope between the two points is at most .
Let , and since we want to maximize , 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 possible because the slope of the line connecting and is still less than .
Therefore, .
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.