Difference between revisions of "2022 AMC 10B Problems/Problem 24"
Pratimakarpe (talk | contribs) (→Solution 4 (Translation)) |
MRENTHUSIASM (talk | contribs) (→Solution 1 (Absolute Values and Inequalities)) |
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Note that | Note that | ||
− | <cmath>\begin{ | + | <cmath>\begin{alignat*}{8} |
− | |f(800)-f(300)|&\leq 250, \\ | + | |f(800)-f(300)|&\leq \frac12|800-300|&&= 250, \\ |
− | |f(800)-f(900)|&\leq 50, \\ | + | |f(800)-f(900)|&\leq \frac12|800-900|&&= 50, \\ |
− | |f(400)-f(300)|&\leq 50, \\ | + | |f(400)-f(300)|&\leq \frac12|400-300|&&= 50, \\ |
− | |f(400)-f(900)|&\leq 250. \\ | + | |f(400)-f(900)|&\leq \frac12|400-900|&&= 250. \\ |
− | \end{ | + | \end{alignat*}</cmath> |
− | Let <math>a=f(300)=f(900).</math> Together, | + | Let <math>a=f(300)=f(900).</math> Together, it follows that |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
|f(800)-a|&\leq 50, \\ | |f(800)-a|&\leq 50, \\ |
Revision as of 11:12, 3 January 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 (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, .
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
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.