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Difference between revisions of "2015 AMC 10B Problems/Problem 8"
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==Solution==
 
==Solution==
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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{(abcdEfghijklmnopqrstuvwxyz)}}</math>.
  
 
==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 09:20, 5 September 2016

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 $\boxed{\mathbf{(abcdEfghijklmnopqrstuvwxyz)}}$.

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
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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