Difference between revisions of "2020 AIME II Problems/Problem 4"

Revision as of 10:01, 21 June 2020

Problem

Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ , $B(0,12)$ , $C(16,0)$ , $A'(24,18)$ , $B'(36,18)$ , $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ .

Solution

After sketching, it is clear a $90^{\circ}$ rotation is done about $(x,y)$ . Looking between $A$ and $A'$ , $x+y=18$ and $x-y=24$ . Solving gives $(x,y)\implies(21,-3)$ . Thus $90+21-3=\boxed{108}$ . ~mn28407

Solution 2 (Official MAA)

Because the rotation sends the vertical segment $\overline{AB}$ to the horizontal segment $\overline{A'B'}$ , the angle of rotation is $90^\circ$ degrees clockwise. For any point $(x,y)$ not at the origin, the line segments from $(0,0)$ to $(x,y)$ and from $(x,y)$ to $(x-y,y+x)$ are perpendicular and are the same length. Thus a $90^\circ$ clockwise rotation around the point $(x,y)$ sends the point $A(0,0)$ to the point $(x-y,y+x) = A'(24,18)$ . This has the solution $(x,y) = (21,-3)$ . The requested sum is $90+21-3=108$ .

Solution 3

A $90^\circ$ degree rotation is obvious. Let's look at $C$ and $C'$ . They are very close to each other. Let's join $C$ and $C'$ with a line. Then construct a perpendicular bisector to $\overline{CC'}$ with the midpoint being $M$ which is at $(20, 1)$ . We also draw a point $N$ on the perpendicular bisector such that $\angle CNC'$ is $90^\circ$ . That point $N$ is the same distance to $M$ as $M$ is to $C$ but it is on a line perpendicular to $\overline{CM}$ Therefore $N$ is at $(20+1, 1-4)$ . The sum is $90+20+1+1-4=108$ .

Solution 4

For the above reasons, the transformation is simply a $90^\circ$ rotation. Proceed with complex numbers on the points $C$ and $C'$ . Let $(x, y)$ be the origin. Thus, $C \rightarrow (16-x)+(-y)i$ and $C' \rightarrow (24-x)+(2-y)i$ . The transformation from $C'$ to $C$ is a multiplication of $i$ , which yields $(16-x)+(-y)i=(y-2)+(24-x)i$ . Equating the real and complex terms results in the equations $16-x=y-2$ and $-y=24-x$ . Solving, $(x, y) : (21, -3) \rightarrow 90+21-3=\boxed{108}$

~beastgert

Solution 5

We know that the rotation point $P$ has to be equidistant from both $A$ and $A'$ so it has to lie on the line that is on the midpoint of the segment $AA'$ and also the line has to be perpendicular to $AA'$ . Solving, we get the line is $y=\frac{-4}{3}x+25$ . Doing the same for $B$ and $B'$ , we get that $y=-6x+123$ . Since the point $P$ of rotation must lie on both of these lines, we set them equal, solve and get: $x=21$ , $y=-3$ . We can also easily see that the degree of rotation is $90$ since $AB$ 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 $21-3+90 = \boxed{108}$

Video Solution

https://youtu.be/atqPgGG0Ekk

~IceMatrix

@@ Line 17: / Line 17: @@
 ~beastgert
+==Solution 5==
+We know that the rotation point <math>P</math> has to be equidistant from both <math>A</math> and <math>A'</math> so it has to lie on the line that is on the midpoint of the segment <math>AA'</math> and also the line has to be perpendicular to <math>AA'</math>. Solving, we get the line is <math>y=\frac{-4}{3}x+25</math>. Doing the same for <math>B</math> and <math>B'</math>, we get that <math>y=-6x+123</math>. Since the point <math>P</math> of rotation must lie on both of these lines, we set them equal, solve and get: <math>x=21</math>,<math>y=-3</math>. We can also easily see that the degree of rotation is <math>90</math> since <math>AB</math> 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 <math>21-3+90 = \boxed{108}</math>
 ==Video Solution==

2020 AIME II (Problems • Answer Key • Resources)
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