Difference between revisions of "2024 AMC 12A Problems/Problem 25"
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Case 2: <math>c\neq 0, bc-ad\neq 0, a=-d</math> | Case 2: <math>c\neq 0, bc-ad\neq 0, a=-d</math> | ||
− | <math>a=0</math>: <math>10\times 10=100</math> ways. | + | <math>a= 0</math>: <math>10\times 10=100</math> ways. |
− | <math>a=\pm 1</math>: <math>(11\times 10-2)\times 2=216</math> ways. | + | <math>a= \pm 1</math>: <math>(11\times 10-2)\times 2=216</math> ways. |
− | <math>a=\pm 2</math>: <math>(11\times 10-2)\times 2=216</math> ways. | + | <math>a= \pm 2</math>: <math>(11\times 10-2)\times 2=216</math> ways. |
− | <math>a=\pm 3</math>: <math>(11\times 10-2)\times 2=216</math> ways. | + | <math>a= \pm 3</math>: <math>(11\times 10-2)\times 2=216</math> ways. |
− | <math>a=\pm 4</math>: <math>(11\times 10-6)\times 2=208</math> ways. | + | <math>a= \pm 4</math>: <math>(11\times 10-6)\times 2=208</math> ways. |
− | <math>a=\pm 5</math>: <math>(11\times 10-2)\times 2=216</math> ways. | + | <math>a= \pm 5</math>: <math>(11\times 10-2)\times 2=216</math> ways. |
In total <math>100+208+216\times 4= 1172</math> ways. | In total <math>100+208+216\times 4= 1172</math> ways. |
Revision as of 23:35, 9 November 2024
Contents
[hide]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 substitute that into gives us:
(Otherwise , , and is not symmetric about )
Therefore we get three cases:
Case 1.1:
We have 10 choice of , 10 choice of and each choice of has one corresponding choice of . In total ways.
Case 1.2:
We have 10 choice for (), each choice of has 2 corresponding choice of , thus ways.
Case 2:
: ways.
: ways.
: ways.
: ways.
: ways.
: ways.
In total ways.
So the answer is
~ERiccc
Solution 2 (Rotation + Edge Cases)
First, observe that the only linear functions that are symmetric about are and , where is some constant.
We perform a counterclockwise rotation of the Cartesian plane. Let be sent to . Then and are the real and imaginary parts of respectively, which gives
so
.
The rotated function is symmetric about the y-axis, so the equation holds after replacing all instances of with (this is just switching the values of and which is a reflection over , but working in terms of allows more cancellations in the following calculations).
Writing and in terms of and , we have
Multiplying both equations by and subtracting the second equation from the first equation gives . Since are integers between and , this gives combinations. We need to subtract the edge cases that don't work, namely all undefined functions and linear functions except and . Consider the following cases:
Case 1: are all nonzero. Then the function is linear when is a multiple of , or .
If , or ; there are ways.
If , there are ways.
If , there are ways.
If , there are ways.
If , there are ways.
In total, this case has combinations.
Case 2: or
If then can take on values, and if , then can take on values, but is counted twice so this case has combinations.
Finally, we need to add the case where , which occurs when and . can be any integer from to except , so this case has combinations. Since occurs when and , this case has already been counted.
Thus, the answer is .
~babyhamster
Solution 3 (Asymptotes)
There are two cases: when and when .
If , then . This is the equation of a line, and the only lines symmetric about are those perpendicular to (i.e. those with slope ) and itself. To have a slope of , we need , and can be any of its possibilities from to . There are possibilities here. For the function to be , we need and . There are possibilities here. Thus, our total for Case 1 is possiblities.
When , we will first consider the case in which the graph is a hyperbola. Clearly, for this hyperbola to be symmetric about , the intersection of its horizontal and vertical asymptotes must be on . The location of the horizontal asymptote is , and the vertical asymptote occurs at . These asymptotes intersect on when , or, more simply, when .
If the asymptotes intersect on , then the hyperbola must be symmetric about . This is true because for any hyperbola with perpendicular asymptotes, we can rotate and translate the coordinate plane in a certain way such that that hyperbola has an equation of the form . Then, the hyperbola's asymptotes would intersect at the origin, and it would be symmetric about the coordinate axes (because it makes a distinction neither between and nor and ). The coordinate axes are the bisectors of the angles formed by the asymptotes, and the hyperbola is symmetric about them. Thus, because the angles formed by our hyperbola's asymptotes are bisected by , our hyperbola must be symmetric about .
Thus, with the conditions that and , there are possibilites for . However, not all of these ordered quadruples produce hyperbolas. If or , then the quadruples produce horizontal lines with a hole when the denominator equals . As seen in Case 1, these lines, with slope , cannot be symmetric about .
For the subcase where , there are possibilities for , which gives us wrongly counted quadruples.
For the subcase where , we wrongly counted cases where . Here, by cross-multiplication. The casework on the possible values of below counts the number of triples with which satisfy this condition.
If , , which yields possibilities.
If , , which yields possibilities.
If , , which yields possbilities. (recall that )
If , , which yields possibilities.
If , , which yields possibilities.
Adding the above values together for this subcase yields wrongly counted quadruples.
Subtracting the wrongly counted quadruples from our count for Case 2 yields .
Adding the possibilities for Case 1 and Case 2 yields our final answer of possible quadruples.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.