AIME 2020 II · 第 4 题

Solution

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

We first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$. Note that $PC=PC'$ because $P$ lies on the perpendicular bisector of $CC'$ (it must be equidistant from $C$ and $C'$ by properties of a rotation).

Since $AB$ is vertical while $A'B'$ is horizontal, we have that the angle of rotation must be $90^{\circ}$, and therefore $\angle P = 90^{\circ}$. Therefore, $CPC'$ is a 45-45-90 right triangle, and $CD=DP$.

We calculate $D$ to be $(20,1)$. Since we translate $4$ right and $1$ up to get from point $C$ to point $D$, we must translate $1$ right and $4$ down to get to point $P$. This gives us $P(21,-3)$. Our answer is then $90+21-3=\boxed{108}$. ~Lopkiloinm & samrocksnature

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

Solution 6

We make transformation of line $AB$ into line $A'B'$ using axes symmetry. Point $D(0,18)$ is the crosspoint of this lines. Equation of line $DO$ is

$$x + y = 18.$$ $\triangle ABC$ maps into $\triangle A''B''C''$ where

$$A''(18,18), B''(6,18), C''(18,2).$$ We make transform of the line $A''C''$ into line $A'C'$ using axes symmetry with respect to line

$$x = \frac {A'' + A'}{2} = \frac {24 + 18}{2} = 21.$$ The composition of two axial symmetries is a rotation through an angle twice as large as the angle between the axes $(45^o)$ around the point of their intersection $O(21, – 3).$

$$m + x + y = 90 + 21 – 3 = \boxed {108}$$ .

vladimir.shelomovskii@gmail.com, vvsss

Solution 7 (Matrix and Transformations)

For a matrix to rotate a figure on a coordinate plane by $m$ degrees, it is written as: $\left[ {\begin{array}{cc} \cos(m^{\circ}) & \sin(m^{\circ}) \\ -\sin(m^{\circ}) & \cos(m^{\circ}) \\ \end{array} } \right]$

We can translate the whole figure so that the centre of rotation is at $(0,0)$, which is equivalent to subtracting $x$ and $y$ from all $x$-coordinates and the $y$-coordinates respectively of the given points.

We then record all the points $A$, $B$, $C$ in a matrix as follows: $\left[ {\begin{array}{ccc} 0-x & 0-x & 16-x \\ 0-y & 12-y & 0-y \\ \end{array} } \right]$

and all the points $A'$, $B'$, $C'$ in a matrix as follows: $\left[ {\begin{array}{ccc} 24-x & 36-x & 24-x \\ 18-y & 18-y & 2-y \\ \end{array} } \right]$

Since $\triangle A'B'C'$ is a rotation around $(x,y)$ of $\triangle ABC$ by $m^{\circ}$, by the left multiplication rule, we can equate that:

$\left[ {\begin{array}{cc} \cos(m^{\circ}) & \sin(m^{\circ}) \\ -\sin(m^{\circ}) & \cos(m^{\circ}) \\ \end{array} } \right]$ $\left[ {\begin{array}{ccc} 0-x & 0-x & 16-x \\ 0-y & 12-y & 0-y \\ \end{array} } \right]$ $=$ $\left[ {\begin{array}{ccc} 24-x & 36-x & 24-x \\ 18-y & 18-y & 2-y \\ \end{array} } \right]$

We can obtain the follow equations: $\begin{cases} -x\cos(m^{\circ})-y\sin(m^{\circ})=24-x \\ -x\cos(m^{\circ})-y\sin(m^{\circ})+12\sin(m^{\circ})=36-x \\ x\sin(m^{\circ})-y\cos(m^{\circ})=24-x \end{cases}$

From the first 2 equations, we get $m=90$, substituting into the 3rd equation, we get $x+y=18$.

Therefore, $m+x+y=90+18=\boxed{108}$

~VitalsBat ~small latex fixes by Glowworm

Solution 8

It is clear that $\bigtriangleup CPC'$ is a $45-45-90$ right triangle so $m=90$. We use the $\tan$ angle formula, $\tan{(a-b)}=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$ to find the slope of line $CP$. We know that line $CC'$ has slope $\frac{1}{4}$ and let $b=-45^{\circ}$, then plugging both values into the formula, we find that the slope of $CP$ is $\frac{-3}{5}$. Also, $CC'$ has length $\sqrt{34}$. Create a right triangle $KCP$ where $KP$ is parallel to the $x$ axis and $CP$ is the hypotenuse. Then $CK=3x$ and $KP=5x$ and doing Pythagorean on $\bigtriangleup KCP$ gives $x=1$. Therefore, we know that $P$ is a translation 3 units down and 5 units right from $C(16,0)$, from which we obtain $P(21,-3)$. Adding the three variables, we obtain $90+21-3=\boxed{108}$

~Magnetoninja