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

Revision as of 13:16, 13 July 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 $90+20+1+1-4=\boxed{108}$ . ~Lopkiloinm

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 10: / Line 10: @@
 ==Solution 3==
-A <math>90^\circ</math> degree rotation is obvious. Let's look at <math>C</math> and <math>C'</math>. They are very close to each other. Let's join <math>C</math> and <math>C'</math> with a line. Then construct a perpendicular bisector to <math>\overline{CC'}</math> with the midpoint being <math>M</math> which is at <math>(20, 1)</math>. We also draw a point <math>N</math> on the perpendicular bisector such that <math>\angle CNC'</math> is <math>90^\circ</math>. That point <math>N</math> is the same distance to <math>M</math> as <math>M</math> is to <math>C</math> but it is on a line perpendicular to <math>\overline{CM}</math> Therefore <math>N</math> is at <math>(20+1, 1-4)</math>. The sum is <math>90+20+1+1-4=108</math>.
+A <math>90^\circ</math> degree rotation is obvious. Let's look at <math>C</math> and <math>C'</math>. They are very close to each other. Let's join <math>C</math> and <math>C'</math> with a line. Then construct a perpendicular bisector to <math>\overline{CC'}</math> with the midpoint being <math>M</math> which is at <math>(20, 1)</math>. We also draw a point <math>N</math> on the perpendicular bisector such that <math>\angle CNC'</math> is <math>90^\circ</math>. That point <math>N</math> is the same distance to <math>M</math> as <math>M</math> is to <math>C</math> but it is on a line perpendicular to <math>\overline{CM}</math> Therefore <math>N</math> is at <math>(20+1, 1-4)</math>. The sum is <math>90+20+1+1-4=\boxed{108}</math>. ~Lopkiloinm
 ==Solution 4==

2020 AIME II (Problems • Answer Key • Resources)
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