Difference between revisions of "2021 Fall AMC 10B Problems/Problem 17"
(→Solution 2) |
(→Solution 2) |
||
Line 16: | Line 16: | ||
==Solution 2== | ==Solution 2== | ||
− | 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 <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 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>. |
Revision as of 21:51, 25 November 2021
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
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 (D)
.
~hurdler
Solution 2
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
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.