Difference between revisions of "2021 AMC 12A Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | A laser is placed at the point <math>(3,5)</math>. The laser | + | A laser is placed at the point <math>(3,5)</math>. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the <math>y</math>-axis, then hit and bounce off the <math>x</math>-axis, then hit the point <math>(7,5)</math>. What is the total distance the beam will travel along this path? |
<math>\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</math> | <math>\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</math> | ||
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Graph in Desmos: https://www.desmos.com/calculator/lgfiiqgqc2 | Graph in Desmos: https://www.desmos.com/calculator/lgfiiqgqc2 | ||
− | Let <math>B=(0,b).</math> | + | Let <math>B=(0,b).</math> In parallelogram <math>ABED,</math> we get <math>E=(4,b).</math> By symmetry, we obtain <math>C=(2,0).</math> |
− | Applying the slope formula | + | Applying the slope formula to <math>\overline{AB}</math> and <math>\overline{DC}</math> gives <cmath>m=\frac{5-b}{3-0}=\frac{5-0}{7-2}.</cmath> Equating the last two expressions gives <math>b=2.</math> |
By the Distance Formula, <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, <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 19:58, 7 May 2021
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1
- 4 Solution 2 (Detailed Explanation of Solution 1)
- 5 Solution 3 (Slopes and Parallelogram)
- 6 Solution 4 (Answer Choices and Educated Guesses)
- 7 Video Solution by OmegaLearn (Using Reflections and Distance Formula)
- 8 Video Solution by Hawk Math
- 9 Video Solution by TheBeautyofMath
- 10 See also
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
Graph in Desmos: https://www.desmos.com/calculator/bsiulzrjrn
~MRENTHUSIASM
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 about the -axis gives Finally, reflecting about the -axis gives as shown below.
Graph in Desmos: https://www.desmos.com/calculator/lxjt0ewbou
It follows that The total distance that the beam will travel is
~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, these two rays coincide when reflected about the line perpendicular to that coordinate axis, creating 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 from which is a parallelogram, as shown below.
Graph in Desmos: https://www.desmos.com/calculator/lgfiiqgqc2
Let In parallelogram we get By symmetry, we obtain
Applying the slope formula to and gives Equating the last two expressions gives
By the Distance Formula, and The total distance that the beam will travel is
~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 all form angles with the coordinate axes, from which and The given condition 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
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.