Difference between revisions of "2024 AMC 12A Problems/Problem 25"
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And <math>y=\frac{(cx+d)(\frac{a}{c})+b-\frac{ad}{c}}{cx+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cx+d}</math> | And <math>y=\frac{(cx+d)(\frac{a}{c})+b-\frac{ad}{c}}{cx+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cx+d}</math> | ||
+ | |||
+ | So <math>\frac{a}{c}=-\frac{d}{c}</math>, or <math>a=-d</math> (<math>c\neq 0</math>), and substitude that into <math>\frac{b-\frac{ad}{c}}{cx-a}=\frac{b-\frac{ad}{c}}{cx+d}</math> gives us: | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=24|after=Last Problem}} | {{AMC12 box|year=2024|ab=A|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:48, 8 November 2024
Problem
A graph is about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers
, where
and
and
are not both
, is the graph of
symmetric about the line
?
Solution 1
Symmetric about the line implies that the inverse fuction
. Then we split the question into several cases to find the final answer.
Case 1:
Then and
.
Giving us
and
Therefore, we obtain 2 subcases: and
Case 2:
Then
And
So , or
(
), and substitude that into
gives us:
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 12 Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.