Difference between revisions of "2021 AMC 12A Problems/Problem 11"
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Note that <math>\overline{AD}\parallel\overline{BE}.</math> Since the opposite sides are parallel, quadrilateral <math>ABED</math> is a parallelogram. From <math>\triangle BCF\cong\triangle ECF,</math> we get <math>BF=EF=2,</math> so <math>C=(2,0).</math> | Note that <math>\overline{AD}\parallel\overline{BE}.</math> Since the opposite sides are parallel, quadrilateral <math>ABED</math> is a parallelogram. From <math>\triangle BCF\cong\triangle ECF,</math> we get <math>BF=EF=2,</math> so <math>C=(2,0).</math> | ||
− | Let <math>B=(0,b).</math> | + | Let <math>B=(0,b).</math> We equate the slopes of <math>\overline{AB}</math> and <math>\overline{DC}:</math> <cmath>\frac{5-b}{3-0}=\frac{5-0}{7-2},</cmath> from which <math>b=2,</math> or <math>B=(0,2).</math> |
By the Distance Formula, we have <math>AB=3\sqrt2,BC=2\sqrt2,</math> and <math>CD=5\sqrt2.</math> The total distance that the beam will travel is <cmath>AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.</cmath> | By the Distance Formula, we have <math>AB=3\sqrt2,BC=2\sqrt2,</math> and <math>CD=5\sqrt2.</math> The total distance that the beam will travel is <cmath>AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.</cmath> |
Revision as of 21:13, 6 October 2021
Contents
Problem
A laser is placed at the point . The laser beam 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
Solution 1 (Reflections)
Let and Suppose that the beam hits and bounces off the -axis at then hits and bounces off the -axis at
When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent. Therefore, we straighten up the path of the beam by reflections:
- We reflect about the -axis to get
- We reflect about the -axis to get with then reflect about the -axis to get with
We obtain the following diagram: The total distance that the beam will travel is ~MRENTHUSIASM (Solution)
~JHawk0224 (Proposal)
Solution 2 (Parallelogram)
Define points and as Solution 1 does. Moreover, let be a point on such that is perpendicular to the -axis, and be a point on such that is perpendicular to the -axis, as shown below. When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent, from which and We conclude that by ASA, so It follows that by transitive, so by the Converse of the Alternate Interior Angles Theorem.
Note that Since the opposite sides are parallel, quadrilateral is a parallelogram. From we get so
Let We equate the slopes of and from which or
By the Distance Formula, we have and The total distance that the beam will travel is
Remark
When a straight line hits and bounces off a coordinate axis at point the ray entering and the ray leaving always have negative slopes. In this problem, and have negative slopes; and have negative slopes. So, and have the same slope, or
~MRENTHUSIASM
Solution 3 (Educated Guess)
Define points and as Solution 1 does.
Since choices and all involve we suspect that one of them is the correct answer. We take a guess in faith that and all form angles with the coordinate axes, from which and The given condition verifies our guess, as shown below. Following the last paragraph of Solution 2 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
Video Solution by TheBeautyofMath
~IceMatrix
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.