Difference between revisions of "2015 AMC 10B Problems/Problem 8"

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==Solution==
 
==Solution==
 
The first rotation moves the base of the <math>F</math> to the negative <math>y</math>-axis, and the stem to the positive <math>x</math>-axis. The reflection then moves the stem to the negative <math>x</math>-axis, with the base unchanged. Then the half turn moves the stem to the positive <math>x</math> axis and the base to the positive <math>y</math>-axis, choice <math>\boxed{\mathbf{(E)}}</math>.
 
The first rotation moves the base of the <math>F</math> to the negative <math>y</math>-axis, and the stem to the positive <math>x</math>-axis. The reflection then moves the stem to the negative <math>x</math>-axis, with the base unchanged. Then the half turn moves the stem to the positive <math>x</math> axis and the base to the positive <math>y</math>-axis, choice <math>\boxed{\mathbf{(E)}}</math>.
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==Video Solution==
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https://youtu.be/mYMfgQ0EuiY
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2015|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:27, 16 July 2020

Problem

The letter F shown below is rotated $90^\circ$ clockwise around the origin, then reflected in the $y$-axis, and then rotated a half turn around the origin. What is the final image?

Foriginal.png

Fproblem.png

Solution

The first rotation moves the base of the $F$ to the negative $y$-axis, and the stem to the positive $x$-axis. The reflection then moves the stem to the negative $x$-axis, with the base unchanged. Then the half turn moves the stem to the positive $x$ axis and the base to the positive $y$-axis, choice $\boxed{\mathbf{(E)}}$.

Video Solution

https://youtu.be/mYMfgQ0EuiY

~savannahsolver

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 10 Problems and Solutions

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