# Difference between revisions of "2021 AMC 12B Problems/Problem 5"

The following problem is from both the 2021 AMC 10B #9 and 2021 AMC 12B #5, so both problems redirect to this page.

## Problem

The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$ $\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

## Solution

The final image of $P$ is $(-6,3)$. We know the reflection rule for reflecting over $y=-x$ is $(x,y) --> (-y, -x)$. So before the reflection and after rotation the point is $(-3,6)$.

By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$. The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$. This means the slope of $P$ and $(1,5)$ is $4$.

Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$.

Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$. The answer is $9-2 = 7 = \boxed{\textbf{(D)}}$.

--abhinavg0627

## Solution 2 (complex)

Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\cdot{i}$. A $90^\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\circ$, and then translating the origin back to the point $(1, 5)$. $a+b\cdot{i} \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} =$ $5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i$.

By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$. Hence, $6-b+(a+4)i = -3+6i \implies b=9, a=2.$ $b-a = 9-2 =7 = \boxed{\textbf{(D)}}$. ~ twotothetenthis1024

~ pi_is_3.14

## Video Solution by TheBeautyofMath

https://youtu.be/GYpAm8v1h-U?t=860 (for AMC 10B)

https://youtu.be/EMzdnr1nZcE?t=814 (for AMC 12B)

~IceMatrix

## Video Solution by Interstigation

~Interstigation

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 