Difference between revisions of "2021 Fall AMC 10B Problems/Problem 17"
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== Solution 4 == | == 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> | + | 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> | <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>2 | + | 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>. |
==Video Solution== | ==Video Solution== |
Revision as of 00:02, 11 October 2022
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 (D) .
~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 .
Video Solution
~hurdler
Video Solution 2 (by Interstigation)
https://www.youtube.com/watch?v=KdrYlPmqqv0
~Interstigation
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.