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2020 AIME II Problems/Problem 4

Revision as of 16:41, 7 March 2021 by Timz2005 (talk | contribs) (→‎Video Solution)

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://www.youtube.com/watch?v=iJkNkSAmqhg

~North America Math Contest Go Go Go


Video Solution

https://youtu.be/atqPgGG0Ekk

~IceMatrix

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

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