Difference between revisions of "2019 AMC 12B Problems/Problem 3"

Latest revision as of 01:55, 24 October 2023

Problem

Which of the following rigid transformations (isometries) maps the line segment $\overline{AB}$ onto the line segment $\overline{A'B'}$ so that the image of $A(-2, 1)$ is $A'(2, -1)$ and the image of $B(-1, 4)$ is $B'(1, -4)$ ?

$\textbf{(A) }$ reflection in the $y$ -axis

$\textbf{(B) }$ counterclockwise rotation around the origin by $90^{\circ}$

$\textbf{(C) }$ translation by 3 units to the right and 5 units down

$\textbf{(D) }$ reflection in the $x$ -axis

$\textbf{(E) }$ clockwise rotation about the origin by $180^{\circ}$

Solution

We can simply graph the points, or use coordinate geometry, to realize that both $A'$ and $B'$ are, respectively, obtained by rotating $A$ and $B$ by $180^{\circ}$ about the origin. Hence the rotation by $180^{\circ}$ about the origin maps the line segment $\overline{AB}$ to the line segment $\overline{A'B'}$ , so the answer is $\boxed{(\text{E})}$ .

~Dodgers66

Solution 2

Notice that the transformation is obtained by reflecting points across the origin. Only $B$ and $C$ involve the origin, and since obviously reflection across the origin is $180^\circ$ , the answer is $\boxed{(\text{E})}$ .

~Technodoggo

Video Solution 1

https://youtu.be/b-htQISQ7Oo

~Education, the Study of Everything

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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.

@@ Line 1: / Line 1: @@
 ==Problem==
-If (n+1)!+(n+2)! = n!*440, what is the sum of the digits of n?
+Which of the following rigid transformations (isometries) maps the line segment <math>\overline{AB}</math> onto the line segment <math>\overline{A'B'}</math> so that the image of <math>A(-2, 1)</math> is <math>A'(2, -1)</math> and the image of <math>B(-1, 4)</math> is <math>B'(1, -4)</math>?
-==Solution==
+<math>\textbf{(A) } </math> reflection in the <math>y</math>-axis
-n=19
-sum is 10 (SuperWill)
+<math>\textbf{(B) } </math> counterclockwise rotation around the origin by <math>90^{\circ}</math>
+<math>\textbf{(C) } </math> translation by 3 units to the right and 5 units down
+<math>\textbf{(D) } </math> reflection in the <math>x</math>-axis
+<math>\textbf{(E) } </math> clockwise rotation about the origin by <math>180^{\circ}</math>
-==Solution 2==
+==Solution==
+We can simply graph the points, or use coordinate geometry, to realize that both <math>A'</math> and <math>B'</math> are, respectively, obtained by rotating <math>A</math> and <math>B</math> by <math>180^{\circ}</math> about the origin. Hence the rotation by <math>180^{\circ}</math> about the origin maps the line segment <math>\overline{AB}</math> to the line segment <math>\overline{A'B'}</math>, so the answer is <math>\boxed{(\text{E})}</math>.
-Divide both sides by <math>n!</math>:
+~Dodgers66
-<math>(n+1)+(n+1)(n+2)=440</math>
+==Solution 2==
-factor out <math>(n+1)</math>:
+Notice that the transformation is obtained by reflecting points across the origin. Only <math>B</math> and <math>C</math> involve the origin, and since obviously reflection across the origin is <math>180^\circ</math>, the answer is <math>\boxed{(\text{E})}</math>.
-<math>(n+1)*(n+3)=440</math>
+~Technodoggo
+==Video Solution 1==
+https://youtu.be/b-htQISQ7Oo
-prime factorization of <math>440</math> and a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>\boxed{n=19}</math>.
+~Education, the Study of Everything
 ==See Also==
 {{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 {{MAA Notice}}

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