Difference between revisions of "2021 AMC 12A Problems/Problem 11"
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Every time the laser bounces off a wall, instead we can imagine it going straight by reflecting it about the wall. Thus, the laser starts at <math>(3, 5)</math> and ends at <math>(-7, -5)</math>, so the path's length is <math>\sqrt{10^2+10^2}=\boxed{\textbf{(C)} 10\sqrt{2}}</math> | Every time the laser bounces off a wall, instead we can imagine it going straight by reflecting it about the wall. Thus, the laser starts at <math>(3, 5)</math> and ends at <math>(-7, -5)</math>, so the path's length is <math>\sqrt{10^2+10^2}=\boxed{\textbf{(C)} 10\sqrt{2}}</math> | ||
~JHawk0224 | ~JHawk0224 | ||
| + | |||
| + | ==Video Solution by Hawk Math== | ||
| + | https://www.youtube.com/watch?v=AjQARBvdZ20 | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2021|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:54, 11 February 2021
Problem
A laser is placed at the point 
. The laser bean travels in a straight line. Larry wants the beam to hit and bounce off the 
-axis, then hit and bounce off the 
-axis, then hit the point 
. What is the total distance the beam will travel along this path?
Solution
Every time the laser bounces off a wall, instead we can imagine it going straight by reflecting it about the wall. Thus, the laser starts at 
and ends at 
, so the path's length is 
~JHawk0224
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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. 
