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2020 AIME II Problems/Problem 4

Revision as of 05:25, 22 May 2023 by Vitalsbat (talk | contribs)

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

[asy] /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.0451801958033, xmax = 47.246151591238494, ymin = -10.271454747548662, ymax = 21.426040258770957; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqffff = rgb(0,1,1); draw((16,0)--(0,0)--(0,12)--cycle, linewidth(2) + zzttqq); draw((24,2)--(24,18)--(36,18)--cycle, linewidth(2) + blue); draw((16,0)--(21,-3)--(24,2)--cycle, linewidth(2) + qqwuqq); draw((21.39134584768662,-2.3477569205223032)--(20.73910276820892,-1.9564110728356852)--(20.347756920522304,-2.608654152313382)--(21,-3)--cycle, linewidth(2) + qqffff); /* draw figures */ draw((16,0)--(0,0), linewidth(2) + zzttqq); draw((0,0)--(0,12), linewidth(2) + zzttqq); draw((0,12)--(16,0), linewidth(2) + zzttqq); draw((24,2)--(24,18), linewidth(2) + blue); draw((24,18)--(36,18), linewidth(2) + blue); draw((36,18)--(24,2), linewidth(2) + blue); draw((16,0)--(24,2), linewidth(2)); draw((16,0)--(21,-3), linewidth(2) + qqwuqq); draw((21,-3)--(24,2), linewidth(2) + qqwuqq); draw((24,2)--(16,0), linewidth(2) + qqwuqq); draw((21,-3)--(20,1), linewidth(2.8) + qqffff); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$A$", (-0.6228029714727868,0.12704474547474198), NE * labelscalefactor); dot((0,12),dotstyle); label("$B$", (0.1301918194013232,12.354245873478124), NE * labelscalefactor); dot((16,0),dotstyle); label("$C$", (16.15822379657881,0.34218611429591583), NE * labelscalefactor); dot((24,18),dotstyle); label("$A'$", (24.154311337765787,18.342347305667463), NE * labelscalefactor); dot((24,2),dotstyle); label("$C'$", (23.186175178070503,1.95574638045472), NE * labelscalefactor); dot((36,18),dotstyle); label("$B'$", (36.13051420214449,18.342347305667463), NE * labelscalefactor); dot((21,-3),dotstyle); label("$P$", (21.35747354309052,-3.458644734878156), NE * labelscalefactor); dot((20,1),linewidth(4pt) + dotstyle); label("$D$", (20.13833911977053,1.2744653791876692), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

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

2020 AIME II 4.png

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: $\begin{cases} -xcos(m^{\circ})-ysin(m^{\circ})=24-x \\ -xcos(m^{\circ})-ysin(m^{\circ})+12sin(m^{\circ})=36-x \\ xsin(m^{\circ})-ycos(m^{\circ})=24-x \end{cases}$

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

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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