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

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(→‎Solution 1)
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~ERiccc
 
~ERiccc
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==Solution 2 (Rotation + Edge Cases)==
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Observe that the only linear functions that are symmetric about <math>y = x</math> are <math>y = x</math> and <math>y = -x</math>.
 +
 +
We perform a <math>45^\circ</math> counterclockwise rotation of the Cartesian plane. Let <math>(x, y)</math> be sent to <math>(u, v)</math>. Then <math>u</math> and <math>v</math> are the real and imaginary parts of <math>(x + yi)(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)</math> respectively, which gives
 +
 +
<cmath>u = \frac{x - y}{\sqrt{2}}</cmath>
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<cmath>v = \frac{x + y}{\sqrt{2}}</cmath>
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 +
so
 +
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<cmath>x = \frac{v + u}{\sqrt{2}}</cmath>
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<cmath>y = \frac{v - u}{\sqrt{2}}</cmath>.
 +
 +
The rotated function is symmetric about the y-axis, so the equation holds after replacing all instances of <math>u</math> with <math>-u</math> (this is just switching the values of <math>x</math> and <math>y</math> which is a reflection over <math>y = x</math>, but working in terms of <math>(u, v)</math> allows more cancellations in the following calculations).
 +
 +
Writing <math>x</math> and <math>y</math> in terms of <math>u</math> and <math>v</math>, we have
 +
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<cmath>\frac{v - u}{\sqrt{2}} = \frac{a(v + u) + b\sqrt{2}}{c(v + u) + d\sqrt{2}}</cmath>
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<cmath>\frac{v + u}{\sqrt{2}} = \frac{a(v - u) + b\sqrt{2}}{c(v - u) + d\sqrt{2}}</cmath>
 +
 +
Multiplying both equations by <math>\sqrt{2}</math> and subtracting the second equation from the first equation gives <math>d = -a</math>. Since <math>a, b, c, d</math> are integers between <math>-5</math> and <math>5</math>, this gives <math>11^3 = 1331</math> combinations. We need to subtract the edge cases that don't work, namely all linear functions except <math>y = x</math> and <math>y = -x</math>. Consider the following cases:
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 +
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Case 1: <math>a, b, c, d</math> are all nonzero. Then the function is linear when <math>ax + b</math> is a multiple of <math>cx + d</math>, or <math>\frac{a}{b} = \frac{c}{-a}</math>.
 +
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If <math>a = \pm 1</math>, <math>(b,c) = (1, -1)</math> or <math>(-1, 1)</math>; there are <math>2*2 = 4</math> ways.
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If <math>a = \pm 2</math>, there are <math>12</math> ways.
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If <math>a = \pm 3</math>, there are <math>4</math> ways.
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If <math>a = \pm 4</math>, there are <math>4</math> ways.
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If <math>a = \pm 5</math>, there are <math>4</math> ways.
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In total, this case has <math>28</math> combinations.
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Case 2: <math>a = b = d = 0</math> or <math>a = c = d = 0</math>
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If <math>a = b = d = 0</math> then <math>c</math> can take on <math>11</math> values, and if <math>a = c = d = 0</math>, then <math>b</math> can take on <math>11</math> values, but <math>a = b = c = d = 0</math> is counted twice so this case has <math>11 + 11 - 1 = 21</math> combinations.
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Finally, we need to add the case where <math>y = x</math>, which occurs when <math>a = d</math> and <math>b = c = 0</math>. <math>a</math> can be any integer from <math>-5</math> to <math>5</math> except <math>0</math>, so this case has <math>10</math> combinations. Since <math>y = -x</math> occurs when <math>a = -d</math> and <math>b = c = 0</math>, this case is already counted.
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The answer is <math>1331 - 28 - 21 + 10 = \boxed{\textbf{B) }1292}</math>.
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~babyhamster
  
 
==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 20:04, 9 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

Solution 2 (Rotation + Edge Cases)

Observe that the only linear functions that are symmetric about $y = x$ are $y = x$ and $y = -x$.

We perform a $45^\circ$ counterclockwise rotation of the Cartesian plane. Let $(x, y)$ be sent to $(u, v)$. Then $u$ and $v$ are the real and imaginary parts of $(x + yi)(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)$ respectively, which gives

\[u = \frac{x - y}{\sqrt{2}}\] \[v = \frac{x + y}{\sqrt{2}}\]

so

\[x = \frac{v + u}{\sqrt{2}}\] \[y = \frac{v - u}{\sqrt{2}}\].

The rotated function is symmetric about the y-axis, so the equation holds after replacing all instances of $u$ with $-u$ (this is just switching the values of $x$ and $y$ which is a reflection over $y = x$, but working in terms of $(u, v)$ allows more cancellations in the following calculations).

Writing $x$ and $y$ in terms of $u$ and $v$, we have

\[\frac{v - u}{\sqrt{2}} = \frac{a(v + u) + b\sqrt{2}}{c(v + u) + d\sqrt{2}}\] \[\frac{v + u}{\sqrt{2}} = \frac{a(v - u) + b\sqrt{2}}{c(v - u) + d\sqrt{2}}\]

Multiplying both equations by $\sqrt{2}$ and subtracting the second equation from the first equation gives $d = -a$. Since $a, b, c, d$ are integers between $-5$ and $5$, this gives $11^3 = 1331$ combinations. We need to subtract the edge cases that don't work, namely all linear functions except $y = x$ and $y = -x$. Consider the following cases:


Case 1: $a, b, c, d$ are all nonzero. Then the function is linear when $ax + b$ is a multiple of $cx + d$, or $\frac{a}{b} = \frac{c}{-a}$.

If $a = \pm 1$, $(b,c) = (1, -1)$ or $(-1, 1)$; there are $2*2 = 4$ ways.

If $a = \pm 2$, there are $12$ ways.

If $a = \pm 3$, there are $4$ ways.

If $a = \pm 4$, there are $4$ ways.

If $a = \pm 5$, there are $4$ ways.

In total, this case has $28$ combinations.


Case 2: $a = b = d = 0$ or $a = c = d = 0$

If $a = b = d = 0$ then $c$ can take on $11$ values, and if $a = c = d = 0$, then $b$ can take on $11$ values, but $a = b = c = d = 0$ is counted twice so this case has $11 + 11 - 1 = 21$ combinations.


Finally, we need to add the case where $y = x$, which occurs when $a = d$ and $b = c = 0$. $a$ can be any integer from $-5$ to $5$ except $0$, so this case has $10$ combinations. Since $y = -x$ occurs when $a = -d$ and $b = c = 0$, this case is already counted.


The answer is $1331 - 28 - 21 + 10 = \boxed{\textbf{B) }1292}$.

~babyhamster

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

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