2020 AIME II Problems/Problem 4
Contents
Problem
Triangles and lie in the coordinate plane with vertices , , , , , . A rotation of degrees clockwise around the point where , will transform to . Find .
Solution
After sketching, it is clear a rotation is done about . Looking between and , and . Solving gives . Thus . ~mn28407
Solution 2 (Official MAA)
Because the rotation sends the vertical segment to the horizontal segment , the angle of rotation is degrees clockwise. For any point not at the origin, the line segments from to and from to are perpendicular and are the same length. Thus a clockwise rotation around the point sends the point to the point . This has the solution . The requested sum is .
Solution 3
We first draw a diagram with the correct Cartesian coordinates and a center of rotation . Note that because lies on the perpendicular bisector of (it must be equidistant from and by properties of a rotation).
Since is vertical while is horizontal, we have that the angle of rotation must be , and therefore . Therefore, is a 45-45-90 right triangle, and .
We calculate to be . Since we translate right and up to get from point to point , we must translate right and down to get to point . This gives us . Our answer is then . ~Lopkiloinm & samrocksnature
Solution 4
For the above reasons, the transformation is simply a rotation. Proceed with complex numbers on the points and . Let be the origin. Thus, and . The transformation from to is a multiplication of , which yields . Equating the real and complex terms results in the equations and . Solving,
~beastgert
Solution 5
We know that the rotation point has to be equidistant from both and so it has to lie on the line that is on the midpoint of the segment and also the line has to be perpendicular to . Solving, we get the line is . Doing the same for and , we get that . Since the point of rotation must lie on both of these lines, we set them equal, solve and get: ,. We can also easily see that the degree of rotation is since is initially vertical, and now it is horizontal. Also, we can just sketch this on a coordinate plane and easily realize the same. Hence, the answer is
Video Solution
https://www.youtube.com/watch?v=iJkNkSAmqhg
~North America Math Contest Go Go Go
Video Solution
~IceMatrix
Solution 6
We make transformation of line into line using axes symmetry. Point is the crosspoint of this lines. Equation of line is maps into where
We make transform of the line into line using axes symmetry with respect to line The composition of two axial symmetries is a rotation through an angle twice as large as the angle between the axes around the point of their intersection .
vladimir.shelomovskii@gmail.com, vvsss
Solution 7 (Matrix and Transformations)
For a matrix to rotate a figure on a coordinate plane by degrees, it is written as:
We can translate the whole figure so that the centre of rotation is at , which is equivalent to subtracting and from all -coordinates and the -coordinates respectively of the given points.
We then record all the points , , in a matrix as follows:
and all the points , , in a matrix as follows:
Since is a rotation around of by , by the left multiplication rule, we can equate that:
We can obtain the follow equations:
From the first 2 equations, we get , substituting into the 3rd equation, we get .
Therefore,
~VitalsBat
Solution 8
It is clear that is a right triangle so . We use the angle formula, to find the slope of line . We know that line has slope and let , then plugging both values into the formula, we find that the slope of is . Also, has length . Create a right triangle where is parallel to the axis and is the hypotenuse. Then and and doing Pythagorean on gives . Therefore, we know that is a translation 3 units down and 5 units right from , from which we obtain . Adding the three variables, we obtain
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.