Difference between revisions of "2021 AMC 12A Problems/Problem 11"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 4 (System of | + | ==Solution 4 (System of Linear Equations)== |
− | Denote <math>(3, 5)</math> as point <math>P</math> and <math>(7, 5)</math> as point S. Define the point <math>Q</math> on the <math>y</math>-axis that the laser hits as <math>(0, y_1)</math> and the point <math>R</math> on the <math>x</math>-axis that the laser hits as <math>(x_1,0)</math>. The laser will bounce off of the two axes at right angles, so we have that line <math>PQ \parallel RS</math>, meaning they have the same slope which we will denote as <math>k</math>. Line <math>QR</math> will be perpendicular to both <math>PQ</math> and <math>RS</math>, meaning it will have slope <math>-\frac{1}{k}</math>. | + | Denote <math>(3, 5)</math> as point <math>P</math> and <math>(7, 5)</math> as point <math>S</math>. Define the point <math>Q</math> on the <math>y</math>-axis that the laser hits as <math>(0, y_1)</math> and the point <math>R</math> on the <math>x</math>-axis that the laser hits as <math>(x_1,0)</math>. The laser will bounce off of the two axes at right angles, so we have that line <math>PQ \parallel RS</math>, meaning they have the same slope which we will denote as <math>k</math>. Line <math>QR</math> will be perpendicular to both <math>PQ</math> and <math>RS</math>, meaning it will have slope <math>-\frac{1}{k}</math>. |
− | |||
We can write the equations for lines <math>PQ</math> and <math>RS</math> in point-slope form, and then plug in points <math>Q</math> and <math>R</math> into those respective equations to get equations in terms of our three variables. We can also write the equation for line <math>QR</math> in slope-intercept form and plug in point <math>R</math>. Doing this yields | We can write the equations for lines <math>PQ</math> and <math>RS</math> in point-slope form, and then plug in points <math>Q</math> and <math>R</math> into those respective equations to get equations in terms of our three variables. We can also write the equation for line <math>QR</math> in slope-intercept form and plug in point <math>R</math>. Doing this yields | ||
− | \begin{align*} | + | <cmath>\begin{align*} |
− | y_1-5&=k(0-3) | + | y_1-5&=k(0-3)\\ |
− | 0-5&=k(x_1-7) | + | 0-5&=k(x_1-7)\\ |
− | 0&=-\frac{1}{k}x_1+y_1 | + | 0&=-\frac{1}{k}x_1+y_1. |
− | \end{align*} | + | \end{align*}</cmath> |
− | We can simplify this into the following cubic equation in terms of k: | + | We can simplify this into the following cubic equation in terms of <math>k</math>: |
− | <cmath>-3k^3+5k^2-7k+5=0</cmath> | + | <cmath>-3k^3+5k^2-7k+5=0.</cmath> |
− | Luckily, we notice that the coefficients sum up to zero, so <math>k=1</math>. Now we can solve for the other two, yielding <math>x_1=y_1=2</math>. Now we simply have three <math> | + | Luckily, we notice that the coefficients sum up to zero, so <math>k=1</math>. Now we can solve for the other two, yielding <math>x_1=y_1=2</math>. Now we simply have three <math>45^{\circ}</math>-<math>45^{\circ}</math>-<math>90^{\circ}</math> triangles with leg lengths of <math>3, 2,</math> and <math>5</math>, so our final distance is <cmath>3\sqrt{2}+2\sqrt{2}+5\sqrt{2}=\boxed{\textbf{(C) }10\sqrt2}.</cmath> |
+ | ~Mooshiros | ||
== Video Solution by OmegaLearn (Using Reflections and Distance Formula) == | == Video Solution by OmegaLearn (Using Reflections and Distance Formula) == |
Latest revision as of 01:28, 1 November 2023
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Reflections)
- 4 Solution 2 (Parallelogram)
- 5 Solution 3 (Educated Guess)
- 6 Solution 4 (System of Linear Equations)
- 7 Video Solution by OmegaLearn (Using Reflections and Distance Formula)
- 8 Video Solution by Hawk Math
- 9 Video Solution by TheBeautyofMath
- 10 Video Solution (Quick and Easy)
- 11 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
~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
Solution 4 (System of Linear Equations)
Denote as point
and
as point
. Define the point
on the
-axis that the laser hits as
and the point
on the
-axis that the laser hits as
. The laser will bounce off of the two axes at right angles, so we have that line
, meaning they have the same slope which we will denote as
. Line
will be perpendicular to both
and
, meaning it will have slope
.
We can write the equations for lines and
in point-slope form, and then plug in points
and
into those respective equations to get equations in terms of our three variables. We can also write the equation for line
in slope-intercept form and plug in point
. Doing this yields
We can simplify this into the following cubic equation in terms of
:
Luckily, we notice that the coefficients sum up to zero, so
. Now we can solve for the other two, yielding
. Now we simply have three
-
-
triangles with leg lengths of
and
, so our final distance is
~Mooshiros
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
Video Solution (Quick and Easy)
~Education, the Study of Everything
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.