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

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==Problem==
 
==Problem==
If (n+1)!+(n+2)! = n!*440, what is the sum of the digits of n?
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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>?
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<math>\textbf{(A) } </math> reflection in the <math>y</math>-axis
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<math>\textbf{(B) } </math> counterclockwise rotation around the origin by <math>90^{\circ}</math>
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<math>\textbf{(C) } </math> translation by 3 units to the right and 5 units down
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<math>\textbf{(D) } </math> reflection in the <math>x</math>-axis
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<math>\textbf{(E) } </math> clockwise rotation about the origin by <math>180^{\circ}</math>
  
 
==Solution==
 
==Solution==
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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>.
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~Dodgers66
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:54, 4 May 2019

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

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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

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