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Difference between revisions of "2024 AMC 12A Problems/Problem 25"

(Created page with "==Problem== A graph is <math>\textit{symmetric}</math> about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers <math>(a,...")
 
(→‎Solution 1)
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<math>\textbf{(A) }1282\qquad\textbf{(B) }1292\qquad\textbf{(C) }1310\qquad\textbf{(D) }1320\qquad\textbf{(E) }1330</math>
 
<math>\textbf{(A) }1282\qquad\textbf{(B) }1292\qquad\textbf{(C) }1310\qquad\textbf{(D) }1320\qquad\textbf{(E) }1330</math>
 +
==Solution 1==
 +
Symmetric about the line <math>y=x</math> implies that the inverse fuction <math>y^{-1}=y</math>. Then we split the question into several cases to find the final answer.
 +
 +
 +
 +
Case 1: <math>c=0</math>
 +
 +
Then <math>y=\frac{a}{d}x+\frac{b}{d}</math> and <math>y^{-1}=\frac{d}{a}x-\frac{b}{a}</math>.
 +
Giving us <math>\frac{a}{d}=\frac{d}{a}</math> and <math>\frac{b}{d}=-\frac{b}{a}</math>
 +
 +
Therefore, we obtain 2 subcases: <math>b\neq 0, a+d=0</math> and <math>b=0, a^2=d^2</math>
 +
 +
 +
 +
Case 2: <math>c\neq 0</math>
 +
 +
Then <math>y^{-1}=\frac{b-dx}{cx-a}=\frac{(cx-a)(-\frac{d}{c})+b-\frac{ad}{c}}{cx-a}=-\frac{d}{c}+\frac{b-\frac{ad}{c}}{cx-a}</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:
 +
 +
<math>bc-ad\neq 0</math> (Otherwise <math>y=\frac{a}{c}</math>, <math>y^{-1}=-\frac{d}{c}=\frac{a}{c}</math>, and is not symmetric about <math>y=x</math>)
 +
 +
 +
 +
Therefore we get three cases:
 +
 +
Case 1.1: <math>c= 0, b\neq 0, d\neq 0, a+d=0</math>
 +
 +
We have 10 choice of <math>b</math>, 10 choice of <math>d</math> and each choice of <math>d</math> has one corresponding choice of <math>a</math>. In total <math>10\times 10=100</math> ways.
 +
 +
 +
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Case 1.2: <math>c= 0, b = 0, d\neq 0, a^2=d^2</math>
 +
 +
We have 10 choice for <math>d</math> (<math>d\neq 0</math>), each choice of <math>d</math> has 2 corresponding choice of <math>a</math>, thus <math>10\times 2=20</math> ways.
 +
 +
 +
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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=-1,1</math>: <math>(11\times 10-2)\times 2=216</math> ways.
 +
 +
<math>a=-2,2</math>: <math>(11\times 10-2)\times 2=216</math> ways.
 +
 +
<math>a=-3,3</math>: <math>(11\times 10-2)\times 2=216</math> ways.
 +
 +
<math>a=-4,4</math>: <math>(11\times 10-6)\times 2=208</math> ways.
 +
 +
<math>a=-5,5</math>: <math>(11\times 10-2)\times 2=216</math> ways.
 +
 +
In total <math>100+208+216\times 4= 1172</math> ways.
 +
 +
 +
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So the answer is <math>100+20+1172= \boxed{\textbf{B) }1292}</math>
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 +
~ERiccc
  
 
==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 21:13, 8 November 2024

Problem

A graph is $\textit{symmetric}$ about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers $(a,b,c,d)$, where $|a|,|b|,|c|,|d|\le5$ and $c$ and $d$ are not both $0$, is the graph of \[y=\frac{ax+b}{cx+d}\]symmetric about the line $y=x$?

$\textbf{(A) }1282\qquad\textbf{(B) }1292\qquad\textbf{(C) }1310\qquad\textbf{(D) }1320\qquad\textbf{(E) }1330$

Solution 1

Symmetric about the line $y=x$ implies that the inverse fuction $y^{-1}=y$. Then we split the question into several cases to find the final answer.


Case 1: $c=0$

Then $y=\frac{a}{d}x+\frac{b}{d}$ and $y^{-1}=\frac{d}{a}x-\frac{b}{a}$. Giving us $\frac{a}{d}=\frac{d}{a}$ and $\frac{b}{d}=-\frac{b}{a}$

Therefore, we obtain 2 subcases: $b\neq 0, a+d=0$ and $b=0, a^2=d^2$


Case 2: $c\neq 0$

Then $y^{-1}=\frac{b-dx}{cx-a}=\frac{(cx-a)(-\frac{d}{c})+b-\frac{ad}{c}}{cx-a}=-\frac{d}{c}+\frac{b-\frac{ad}{c}}{cx-a}$

And $y=\frac{(cx+d)(\frac{a}{c})+b-\frac{ad}{c}}{cx+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cx+d}$

So $\frac{a}{c}=-\frac{d}{c}$, or $a=-d$ ($c\neq 0$), and substitude that into $\frac{b-\frac{ad}{c}}{cx-a}=\frac{b-\frac{ad}{c}}{cx+d}$ gives us:

$bc-ad\neq 0$ (Otherwise $y=\frac{a}{c}$, $y^{-1}=-\frac{d}{c}=\frac{a}{c}$, and is not symmetric about $y=x$)


Therefore we get three cases:

Case 1.1: $c= 0, b\neq 0, d\neq 0, a+d=0$

We have 10 choice of $b$, 10 choice of $d$ and each choice of $d$ has one corresponding choice of $a$. In total $10\times 10=100$ ways.


Case 1.2: $c= 0, b = 0, d\neq 0, a^2=d^2$

We have 10 choice for $d$ ($d\neq 0$), each choice of $d$ has 2 corresponding choice of $a$, thus $10\times 2=20$ ways.


Case 2: $c\neq 0, bc-ad\neq 0, a=-d$

$a=0$: $10\times 10=100$ ways.

$a=-1,1$: $(11\times 10-2)\times 2=216$ ways.

$a=-2,2$: $(11\times 10-2)\times 2=216$ ways.

$a=-3,3$: $(11\times 10-2)\times 2=216$ ways.

$a=-4,4$: $(11\times 10-6)\times 2=208$ ways.

$a=-5,5$: $(11\times 10-2)\times 2=216$ ways.

In total $100+208+216\times 4= 1172$ ways.


So the answer is $100+20+1172= \boxed{\textbf{B) }1292}$

~ERiccc

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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