Difference between revisions of "2021 AMC 12A Problems/Problem 11"
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Define points <math>A,B,C,</math> and <math>D</math> as Solution 2 does. | Define points <math>A,B,C,</math> and <math>D</math> as Solution 2 does. | ||
− | Since choices <math>\textbf{(B)}, \textbf{(C)},</math> and <math>\textbf{(D)}</math> all involve <math>\sqrt2,</math> we suspect that one of them is the correct answer. We take a guess in faith that <math>\overleftrightarrow{AB}</math> and <math>\overleftrightarrow{BC}</math> form <math>45^\circ</math> angles with the coordinate axes | + | Since choices <math>\textbf{(B)}, \textbf{(C)},</math> and <math>\textbf{(D)}</math> all involve <math>\sqrt2,</math> we suspect that one of them is the correct answer. We take a guess in faith that <math>\overleftrightarrow{AB}</math> and <math>\overleftrightarrow{BC}</math> form <math>45^\circ</math> angles with the coordinate axes, then we get that <math>B=(0,2)</math> and <math>C=(2,0).</math> This result verifies our guess. Following the penultimate paragraph of Solution 3 gives the answer <math>\boxed{\textbf{(C) }10\sqrt2}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 14:10, 15 February 2021
Contents
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?
Diagram
~MRENTHUSIASM (by Desmos: https://www.desmos.com/calculator/y6wyqok8gm)
Solution 1
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
Solution 2 (Detailed Explanation of Solution 1)
Let be the point where the beam hits the
-axis, and
be the point where the beam hits the
-axis.
Reflecting about the
-axis gives
Then, reflecting
over the
-axis gives
Finally, reflecting
about the
-axis gives
as shown below.
It follows that The total distance that the beam will travel is
Graph in Desmos: https://www.desmos.com/calculator/lqatf2r9bu
~MRENTHUSIASM
Solution 3 (Slopes and Parallelogram)
Define points and
as Solution 2 does.
When a line segment hits and bounces off a coordinate axis at point the ray entering
and the ray leaving
have negative slopes. Geometrically, the rays coincide when reflected about the line perpendicular to that coordinate axis, creating a line symmetry. Let the slope of
be
It follows that the slope of
is
and the slope of
is
Here, we conclude that
Next, we locate on
such that
thus
is a parallelogram, as shown below.
Let By the property of slopes, we get
By symmetry, we obtain
Applying the slope formula on and
gives
Equating the last two expressions gives
By the Distance Formula, and
The total distance that the beam will travel is
Graph in Desmos: https://www.desmos.com/calculator/cp9uauhl6k
~MRENTHUSIASM
Solution 4 (Answer Choices and Educated Guesses)
Define points and
as Solution 2 does.
Since choices and
all involve
we suspect that one of them is the correct answer. We take a guess in faith that
and
form
angles with the coordinate axes, then we get that
and
This result verifies our guess. Following the penultimate paragraph of Solution 3 gives the answer
~MRENTHUSIASM
Video Solution by OmegaLearn (Using Reflections and Distance Formula)
~ pi_is_3.14
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.