Difference between revisions of "2021 Fall AMC 10B Problems/Problem 17"
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<math>(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0</math> | <math>(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0</math> | ||
+ | ==Solution 1== | ||
+ | Denote <math>O</math> as the origin. | ||
+ | |||
+ | Even though the problem is phrased as a coordinate bash, that looks disgusting. Instead, let's try to phrase this problem in terms of Euclidean geometry, using the observation that <math>\angle POP'' = 90^{\circ}</math>, and that both <math>\ell</math> and <math>m</math> must pass through <math>O</math> in order to preserve the distance from <math>P</math> to the origin. | ||
+ | <asy> | ||
+ | unitsize(1.4cm); | ||
+ | draw((0,3)--(0,0)--(3,0), dashed); | ||
+ | dot((0,3)); | ||
+ | dot((3,0)); | ||
+ | label("$P$", (0,3), W); | ||
+ | label("$P''$", (3,0), S); | ||
+ | |||
+ | draw((0,0)--(1.5,4.5)); | ||
+ | label("$\ell$", (1.5,4.5), N); | ||
+ | draw((0,0)--(4,2)); | ||
+ | label("$m$", (4,2), E); | ||
+ | |||
+ | dot((1.8,2.4)); | ||
+ | label("$P'$", (1.8,2.4), N); | ||
+ | label("$O$",(0,0)); | ||
+ | |||
+ | dot((1,3)); dot((2.5,1.25)); | ||
+ | label("$A$", (1,3), E); label("$B$", (2.5,1.25), N); | ||
+ | </asy> | ||
+ | (<math>A</math> and <math>B</math> are just defined as points on lines <math>\ell</math> and <math>m</math>.) | ||
+ | Because of how reflections work, we have that <math>\angle AOP' = \angle POA</math> and <math>\angle P'OB = \angle BOP''</math>; adding these two equations together and using angle addition, we have that <math>\angle AOB = \angle POA + \angle BOP''</math>. Since the sum of both sides combined must be <math>90^{\circ}</math> by angle addition, | ||
+ | <cmath>\angle AOB = 45^{\circ}.</cmath> | ||
+ | This is helpful! We can now return to using coordinates, with this piece of information in mind: | ||
+ | <asy> | ||
+ | unitsize(0.2cm); | ||
+ | markscalefactor = 0.08; | ||
+ | import graph; | ||
+ | Label f; | ||
+ | f.p=fontsize(9); | ||
+ | xaxis(-2,6,Ticks(f, 2.0)); | ||
+ | yaxis(-1,6,Ticks(f, 2.0)); | ||
+ | dot((-1,4)); | ||
+ | label("$P$", (-1,4), W); | ||
+ | dot((4,1)); | ||
+ | label("$P''$", (4,1), W); | ||
+ | |||
+ | draw((0,0)--(1.2,6)); | ||
+ | label("$\ell$", (1.2,6), N); | ||
+ | dot((0.5,2.5));label("$(0.5,2.5)$", (0.5,2.5), E);label("$A$", (0.5,2.5), W); | ||
+ | dot((3,2));label("$B$", (3,2), E); | ||
+ | draw((0.5,2.5)--(3,2), dashed); | ||
+ | |||
+ | draw((0,0)--(6,4)); | ||
+ | label("$m$", (6,4), E); | ||
+ | |||
+ | draw(anglemark((6, 4), (0, 0), (1, 5))); | ||
+ | label("$45^{\circ}$", (0.54,0.75)); | ||
+ | </asy> | ||
+ | The <math>45^{\circ}</math> angle is a little bit unwieldy in the coordinate plane, so we should try to make a <math>45-45-90</math> triangle. Let <math>A</math> be a point on <math>\ell</math>; to make <math>A</math> fit nicely in the diagram, let it be <math>(0.5,2.5)</math>. Now, let's draw a perpendicular to <math>\ell</math> through point <math>A</math>, intersecting <math>m</math> at point <math>B</math>. <math>OAB</math> is a <math>45-45-90</math> triangle, so <math>B</math> is a <math>90</math> degree counterclockwise rotation from <math>O</math> about <math>A</math>. Therefore, the coordinates of <math>B</math> are | ||
+ | <cmath>(0.5+2.5,2.5-0.5) = (3,2).</cmath> | ||
+ | So, <math>(3,2)</math> is a point on line <math>m</math>, which we already know passes through the origin; therefore, <math>m</math>'s equation is <math>y=\frac{2x}{3} \implies \boxed{\textbf{(D) } 2x-3y = 0}.</math> | ||
+ | |||
+ | ~ihatemath123 | ||
+ | |||
+ | (We never actually had to use the information of the exact coordinates of <math>P</math>; as long as <math>\angle POP'' = 90^{\circ}</math>, when we move <math>P</math> around, this will not affect <math>m</math>'s equation.) | ||
− | ==Solution | + | ==Solution 2== |
It is well known that the composition of 2 reflections , one after another, about two lines <math>l</math> and <math>m</math>, respectively, that meet at an angle <math>\theta</math> is a rotation by <math>2\theta</math> around the intersection of <math>l</math> and <math>m</math>. | It is well known that the composition of 2 reflections , one after another, about two lines <math>l</math> and <math>m</math>, respectively, that meet at an angle <math>\theta</math> is a rotation by <math>2\theta</math> around the intersection of <math>l</math> and <math>m</math>. | ||
Now, we note that <math>(4,1)</math> is a 90 degree rotation clockwise of <math>(-1,4)</math> about the origin, which is also where <math>l</math> and <math>m</math> intersect. So <math>m</math> is a 45 degree rotation of <math>l</math> about the origin clockwise. | Now, we note that <math>(4,1)</math> is a 90 degree rotation clockwise of <math>(-1,4)</math> about the origin, which is also where <math>l</math> and <math>m</math> intersect. So <math>m</math> is a 45 degree rotation of <math>l</math> about the origin clockwise. | ||
− | To rotate <math>l</math> 90 degrees clockwise, we build a square with adjacent vertices <math>(0,0)</math> and <math>(1,5)</math>. The other two vertices are at <math>(5,-1)</math> and <math>(6,4)</math>. The center of the square is at <math>(3,2)</math>, which is the midpoint of <math>(1,5)</math> and <math>(5,-1)</math>. The line <math>m</math> passes through the origin and the center of the square we built, namely at <math>(0,0)</math> and <math>(3,2)</math>. Thus the line is <math>y = \frac{2}{3} x</math>. The answer is | + | To rotate <math>l</math> 90 degrees clockwise, we build a square with adjacent vertices <math>(0,0)</math> and <math>(1,5)</math>. The other two vertices are at <math>(5,-1)</math> and <math>(6,4)</math>. The center of the square is at <math>(3,2)</math>, which is the midpoint of <math>(1,5)</math> and <math>(5,-1)</math>. The line <math>m</math> passes through the origin and the center of the square we built, namely at <math>(0,0)</math> and <math>(3,2)</math>. Thus the line is <math>y = \frac{2}{3} x</math>. The answer is <math>\boxed{\textbf{(D) } 2x-3y = 0}</math>. |
~hurdler | ~hurdler | ||
− | ==Solution | + | ~minor edits by nightshade2526 |
+ | |||
+ | ==Solution 3== | ||
− | We know that the equation of line <math>\ell</math> is <math>y = 5x</math>. This means that <math>P'</math> is <math>(-1,4)</math> reflected over the line <math>y = 5x</math>. This means that the line with <math>P</math> and <math>P'</math> is perpendicular to <math>\ell</math>, so it has slope <math>\frac{ | + | We know that the equation of line <math>\ell</math> is <math>y = 5x</math>. This means that <math>P'</math> is <math>(-1,4)</math> reflected over the line <math>y = 5x</math>. This means that the line with <math>P</math> and <math>P'</math> is perpendicular to <math>\ell</math>, so it has slope <math>-\frac{1}{5}</math>. Then the equation of this perpendicular line is <math>y = -\frac{1}{5}x + c</math>, and plugging in <math>(-1,4)</math> for <math>x</math> and <math>y</math> yields <math>c = \frac{19}{5}</math>. |
− | The midpoint of <math>P'</math> and <math>P</math> lies at the intersection of <math>y = 5x</math> and <math>y = \frac{ | + | The midpoint of <math>P'</math> and <math>P</math> lies at the intersection of <math>y = 5x</math> and <math>y = -\frac{1}{5}x + \frac{19}{5}</math>. Solving, we get the x-value of the intersection is <math>\frac{19}{26}</math> and the y-value is <math>\frac{95}{26}</math>. Let the x-value of <math>P'</math> be <math>x'</math> - then by the midpoint formula, <math>\frac{x' - 1}{2} = \frac{19}{26} \implies x' = \frac{32}{13}</math>. We can find the y-value of <math>P'</math> the same way, so <math>P' = (\frac{32}{13},\frac{43}{13})</math>. |
− | Now we have to reflect < | + | Now we have to reflect <math>P'</math> over <math>m</math> to get to <math>(4,1)</math>. The midpoint of <math>P'</math> and <math>P''</math> will lie on <math>m</math>, and this midpoint is, by the midpoint formula, <math>(\frac{42}{13},\frac{28}{13})</math>. <math>y = mx</math> must satisfy this point, so <math>m = \frac{\frac{28}{13}}{\frac{42}{13}} = \frac{28}{42} = \frac{2}{3}</math>. |
Now the equation of line <math>m</math> is <math>y = \frac{2}{3}x \implies 2x-3y = 0 = \boxed{D}</math> | Now the equation of line <math>m</math> is <math>y = \frac{2}{3}x \implies 2x-3y = 0 = \boxed{D}</math> | ||
Line 26: | Line 88: | ||
~KingRavi | ~KingRavi | ||
+ | == Solution 4 == | ||
+ | |||
+ | First, use Solution 1's method to get <math>\angle POP'' = 90^\circ</math> and that the angle between lines <math>\ell</math> and <math>m</math> is <math>45^\circ</math>. From here, note that the slope of line <math>m</math> is less than that of line <math>\ell</math> as otherwise <math>P''</math> wouldn't even be close to <math>(4, 1)</math>. Thus, line <math>m</math> is a <math>45^\circ</math> clockwise rotation of line <math>\ell</math>. Line <math>\ell</math> makes an angle of <math>\tan^{-1}(5)</math> with the positive x axis. Thus, line <math>m</math> makes an angle of <math>\tan^{-1}(5) - 45^\circ</math> with the positive x axis. Thus, the slope of line <math>m</math> is | ||
+ | <cmath> \tan (\tan^{-1}(5) - 45^\circ) = \frac{5 - 1}{1 + 5\cdot 1} = \frac{2}{3},</cmath> | ||
+ | by the tangent addition formula. Since the slope of line <math>m</math> is <math>\frac{2}{3}</math>, its equation is <math>y = \frac{2}{3}x \implies 2x - 3y = 0</math>, which is choice <math>\boxed{\textbf{D}}</math>. | ||
+ | |||
+ | == Solution 5(cheese) == | ||
+ | When we graph all the lines and points with a ruler, you can see that a slope of <math>\frac{5}{3}</math> is too big while <math>\frac{1}{3}</math> is too small. We also see that the slope cannot be negative, therefore the answer is <math>\boxed{D}.</math> | ||
+ | ~ agentdabber | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | [https://youtu.be/bqCacR8aXmc|Video Solution 2021 Fall 10B #17] | ||
+ | |||
+ | ~hurdler | ||
+ | ==Video Solution 2 (by Interstigation)== | ||
+ | https://www.youtube.com/watch?v=KdrYlPmqqv0 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/Wwzqihd3cUg | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/YOpyq7Zu_hA | ||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=PgFX55o6h1g | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=18|num-b=16}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=18|num-b=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:44, 16 December 2023
Contents
Problem
Distinct lines and
lie in the
-plane. They intersect at the origin. Point
is reflected about line
to point
, and then
is reflected about line
to point
. The equation of line
is
, and the coordinates of
are
. What is the equation of line
Solution 1
Denote as the origin.
Even though the problem is phrased as a coordinate bash, that looks disgusting. Instead, let's try to phrase this problem in terms of Euclidean geometry, using the observation that , and that both
and
must pass through
in order to preserve the distance from
to the origin.
(
and
are just defined as points on lines
and
.)
Because of how reflections work, we have that
and
; adding these two equations together and using angle addition, we have that
. Since the sum of both sides combined must be
by angle addition,
This is helpful! We can now return to using coordinates, with this piece of information in mind:
The
angle is a little bit unwieldy in the coordinate plane, so we should try to make a
triangle. Let
be a point on
; to make
fit nicely in the diagram, let it be
. Now, let's draw a perpendicular to
through point
, intersecting
at point
.
is a
triangle, so
is a
degree counterclockwise rotation from
about
. Therefore, the coordinates of
are
So,
is a point on line
, which we already know passes through the origin; therefore,
's equation is
~ihatemath123
(We never actually had to use the information of the exact coordinates of ; as long as
, when we move
around, this will not affect
's equation.)
Solution 2
It is well known that the composition of 2 reflections , one after another, about two lines and
, respectively, that meet at an angle
is a rotation by
around the intersection of
and
.
Now, we note that is a 90 degree rotation clockwise of
about the origin, which is also where
and
intersect. So
is a 45 degree rotation of
about the origin clockwise.
To rotate 90 degrees clockwise, we build a square with adjacent vertices
and
. The other two vertices are at
and
. The center of the square is at
, which is the midpoint of
and
. The line
passes through the origin and the center of the square we built, namely at
and
. Thus the line is
. The answer is
.
~hurdler
~minor edits by nightshade2526
Solution 3
We know that the equation of line is
. This means that
is
reflected over the line
. This means that the line with
and
is perpendicular to
, so it has slope
. Then the equation of this perpendicular line is
, and plugging in
for
and
yields
.
The midpoint of and
lies at the intersection of
and
. Solving, we get the x-value of the intersection is
and the y-value is
. Let the x-value of
be
- then by the midpoint formula,
. We can find the y-value of
the same way, so
.
Now we have to reflect over
to get to
. The midpoint of
and
will lie on
, and this midpoint is, by the midpoint formula,
.
must satisfy this point, so
.
Now the equation of line is
~KingRavi
Solution 4
First, use Solution 1's method to get and that the angle between lines
and
is
. From here, note that the slope of line
is less than that of line
as otherwise
wouldn't even be close to
. Thus, line
is a
clockwise rotation of line
. Line
makes an angle of
with the positive x axis. Thus, line
makes an angle of
with the positive x axis. Thus, the slope of line
is
by the tangent addition formula. Since the slope of line
is
, its equation is
, which is choice
.
Solution 5(cheese)
When we graph all the lines and points with a ruler, you can see that a slope of is too big while
is too small. We also see that the slope cannot be negative, therefore the answer is
~ agentdabber
Video Solution
~hurdler
Video Solution 2 (by Interstigation)
https://www.youtube.com/watch?v=KdrYlPmqqv0
~Interstigation
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=PgFX55o6h1g
~IceMatrix
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.