2020 AIME II Problems/Problem 4

Revision as of 23:47, 10 April 2021 by Samrocksnature (talk | contribs) (Solution 3)


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 $m+x+y$.


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


/* 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 */


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, $(x, y) : (21, -3) \rightarrow 90+21-3=\boxed{108}$


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


~North America Math Contest Go Go Go

Video Solution



See Also

2020 AIME II (ProblemsAnswer KeyResources)
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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