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 15:54, 11 February 2021

Problem

A laser is placed at the point $(3,5)$. The laser bean travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path?

$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5$

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 $(3, 5)$ and ends at $(-7, -5)$, so the path's length is $\sqrt{10^2+10^2}=\boxed{\textbf{(C)} 10\sqrt{2}}$ ~JHawk0224

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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