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

Revision as of 19:06, 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)

First, 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

@@ Line 66: / Line 66: @@
 ==Solution 2 (Rotation + Edge Cases)==
-Observe that the only linear functions that are symmetric about <math>y = x</math> are <math>y = x</math> and <math>y = -x</math>.
+First, 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

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

Page

Toolbox

Search

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

Revision as of 19:06, 9 November 2024

Contents

Problem

Solution 1

Solution 2 (Rotation + Edge Cases)

See also