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

Latest revision as of 12:29, 19 January 2024

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 $A'$ , $x+y=18$ . 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

$[asy] /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.0451801958033, xmax = 47.246151591238494, ymin = -10.271454747548662, ymax = 21.426040258770957; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqffff = rgb(0,1,1); draw((16,0)--(0,0)--(0,12)--cycle, linewidth(2) + zzttqq); draw((24,2)--(24,18)--(36,18)--cycle, linewidth(2) + blue); draw((16,0)--(21,-3)--(24,2)--cycle, linewidth(2) + qqwuqq); draw((21.39134584768662,-2.3477569205223032)--(20.73910276820892,-1.9564110728356852)--(20.347756920522304,-2.608654152313382)--(21,-3)--cycle, linewidth(2) + qqffff); /* draw figures */ draw((16,0)--(0,0), linewidth(2) + zzttqq); draw((0,0)--(0,12), linewidth(2) + zzttqq); draw((0,12)--(16,0), linewidth(2) + zzttqq); draw((24,2)--(24,18), linewidth(2) + blue); draw((24,18)--(36,18), linewidth(2) + blue); draw((36,18)--(24,2), linewidth(2) + blue); draw((16,0)--(24,2), linewidth(2)); draw((16,0)--(21,-3), linewidth(2) + qqwuqq); draw((21,-3)--(24,2), linewidth(2) + qqwuqq); draw((24,2)--(16,0), linewidth(2) + qqwuqq); draw((21,-3)--(20,1), linewidth(2.8) + qqffff); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$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

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

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

2020 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2020 AIME II Problems/Problem 4"

Latest revision as of 12:29, 19 January 2024

Contents

Problem

Solution

Solution 2 (Official MAA)

Solution 3

Solution 4

Solution 5

Video Solution

Video Solution

Solution 6

Solution 7 (Matrix and Transformations)

Solution 8

See Also

@@ Line 3: / Line 3: @@
 ==Solution==
-After sketching, it is clear a <math>90^{\circ}</math> rotation is done about <math>(x,y)</math>. Looking between <math>A</math> and <math>A'</math>, <math>x+y=18</math> and <math>x-y=24</math>. Solving gives <math>(x,y)\implies(21,-3)</math>. Thus <math>90+21-3=\boxed{108}</math>.
+After sketching, it is clear a <math>90^{\circ}</math> rotation is done about <math>(x,y)</math>. Looking between <math>A</math> and <math>A'</math>, <math>x+y=18</math>. Thus <math>90+18=\boxed{108}</math>.
 ~mn28407
@@ 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>.
+<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 <math>P</math>. Note that <math>PC=PC'</math> because <math>P</math> lies on the perpendicular bisector of <math>CC'</math> (it must be equidistant from <math>C</math> and <math>C'</math> by properties of a rotation).
+Since <math>AB</math> is vertical while <math>A'B'</math> is horizontal, we have that the angle of rotation must be <math>90^{\circ}</math>, and therefore <math>\angle P = 90^{\circ}</math>. Therefore, <math>CPC'</math> is a 45-45-90 right triangle, and <math>CD=DP</math>.
+We calculate <math>D</math> to be <math>(20,1)</math>. Since we translate <math>4</math> right and <math>1</math> up to get from point <math>C</math> to point <math>D</math>, we must translate <math>1</math> right and <math>4</math> down to get to point <math>P</math>. This gives us <math>P(21,-3)</math>. Our answer is then <math>90+21-3=\boxed{108}</math>. ~Lopkiloinm & samrocksnature
 ==Solution 4==
-For the above reasons, the transformation is simply a <math>90^\circ</math> rotation. Proceed with complex numbers on the points <math>C</math> and <math>C'</math>. Let <math>(x, y)</math> be the origin. Thus, <math>C \rightarrow (16-x)+(-y)i</math> and <math>C' \rightarrow (24-x)+(2-y)i</math>. The transformation from <math>C'</math> to <math>C</math> is a multiplication of <math>i</math>, which yields <math>(16-x)+(-y)i=(y-2)+(24-x)i</math>. Equating the real and complex terms results in the equations <math>16-x=y-2</math> and <math>-y=24-x</math>. Solving yields the desired point to be <math>(21, -3) \rightarrow 90+21-3=\boxed{108}</math>
+For the above reasons, the transformation is simply a <math>90^\circ</math> rotation. Proceed with complex numbers on the points <math>C</math> and <math>C'</math>. Let <math>(x, y)</math> be the origin. Thus, <math>C \rightarrow (16-x)+(-y)i</math> and <math>C' \rightarrow (24-x)+(2-y)i</math>. The transformation from <math>C'</math> to <math>C</math> is a multiplication of <math>i</math>, which yields <math>(16-x)+(-y)i=(y-2)+(24-x)i</math>. Equating the real and complex terms results in the equations <math>16-x=y-2</math> and <math>-y=24-x</math>. Solving, <math>(x, y) : (21, -3) \rightarrow 90+21-3=\boxed{108}</math>
 ~beastgert
+==Solution 5==
+We know that the rotation point <math>P</math> has to be equidistant from both <math>A</math> and <math>A'</math> so it has to lie on the line that is on the midpoint of the segment <math>AA'</math> and also the line has to be perpendicular to <math>AA'</math>. Solving, we get the line is <math>y=\frac{-4}{3}x+25</math>. Doing the same for <math>B</math> and <math>B'</math>, we get that <math>y=-6x+123</math>. Since the point <math>P</math> of rotation must lie on both of these lines, we set them equal, solve and get: <math>x=21</math>,<math>y=-3</math>. We can also easily see that the degree of rotation is <math>90</math> since <math>AB</math> 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 <math>21-3+90 = \boxed{108}</math>
+==Video Solution==
+https://www.youtube.com/watch?v=iJkNkSAmqhg
+~North America Math Contest Go Go Go
 ==Video Solution==
@@ Line 22: / Line 79: @@
 ~IceMatrix
+==Solution 6==
+[[File:2020 AIME II 4.png|500px|right]]
+We make transformation of line <math>AB</math> into line <math>A'B'</math> using axes symmetry. Point <math>D(0,18)</math> is the crosspoint of this lines. Equation of line <math>DO</math> is
+<cmath>x + y = 18.</cmath>
+<math>\triangle ABC</math> maps into <math>\triangle A''B''C''</math> where <cmath>A''(18,18), B''(6,18), C''(18,2).</cmath>
+We make transform of the line <math>A''C''</math> into line <math>A'C'</math> using axes symmetry with respect to line
+<cmath>x  = \frac {A'' + A'}{2} = \frac {24 + 18}{2} = 21.</cmath>
+The composition of two axial symmetries is a rotation through an angle twice as large as the angle between the axes <math>(45^o)</math> around the point of their intersection <math>O(21, – 3).</math>
+<cmath>m + x + y = 90 + 21 – 3 = \boxed {108}</cmath>.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution 7 (Matrix and Transformations)==
+For a matrix to rotate a figure on a coordinate plane by <math>m</math> degrees, it is written as:
+<math>\left[ {\begin{array}{cc}
+ cos(m^{\circ}) & sin(m^{\circ})  \\
+ -sin(m^{\circ}) & cos(m^{\circ})  \\
+ \end{array} } \right]</math>
+We can translate the whole figure so that the centre of rotation is at <math>(0,0)</math>, which is equivalent to subtracting <math>x</math> and <math>y</math> from all <math>x</math>-coordinates and the <math>y</math>-coordinates respectively of the given points.
+We then record all the points <math>A</math>, <math>B</math>, <math>C</math> in a matrix as follows:
+<math>\left[ {\begin{array}{ccc}
+-x & 0-x & 16-x  \\
+-y & 12-y & 0-y  \\
+ \end{array} } \right]</math>
+and all the points <math>A'</math>, <math>B'</math>, <math>C'</math> in a matrix as follows:
+<math>\left[ {\begin{array}{ccc}
+-x & 36-x & 24-x  \\
+-y & 18-y & 2-y  \\
+ \end{array} } \right]</math>
+Since <math>\triangle A'B'C'</math> is a rotation around <math>(x,y)</math> of <math>\triangle ABC</math> by <math>m^{\circ}</math>, by the left multiplication rule, we can equate that:
+<math>\left[ {\begin{array}{cc}
+ cos(m^{\circ}) & sin(m^{\circ})  \\
+ -sin(m^{\circ}) & cos(m^{\circ})  \\
+ \end{array} } \right]</math>
+<math>\left[ {\begin{array}{ccc}
+-x & 0-x & 16-x  \\
+-y & 12-y & 0-y  \\
+ \end{array} } \right]</math>
+<math>=</math>
+<math>\left[ {\begin{array}{ccc}
+-x & 36-x & 24-x  \\
+-y & 18-y & 2-y  \\
+ \end{array} } \right]</math>
+We can obtain the follow equations:
+<math>\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}</math>
+From the first 2 equations, we get <math>m=90</math>, substituting into the 3rd equation, we get <math>x+y=18</math>.
+Therefore, <math>m+x+y=90+18=\boxed{108}</math>
+~VitalsBat
+==Solution 8==
+It is clear that <math>\bigtriangleup CPC'</math> is a <math>45-45-90</math> right triangle so <math>m=90</math>. We use the <math>\tan</math> angle formula, <math>\tan{(a-b)}=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}</math> to find the slope of line <math>CP</math>. We know that line <math>CC'</math> has slope <math>\frac{1}{4}</math> and let <math>b=-45^{\circ}</math>, then plugging both values into the formula, we find that the slope of <math>CP</math> is <math>\frac{-3}{5}</math>. Also, <math>CC'</math> has length <math>\sqrt{34}</math>. Create a right triangle <math>KCP</math> where <math>KP</math> is parallel to the <math>x</math> axis and <math>CP</math> is the hypotenuse. Then <math>CK=3x</math> and <math>KP=5x</math> and doing Pythagorean on <math>\bigtriangleup KCP</math> gives <math>x=1</math>. Therefore, we know that <math>P</math> is a translation 3 units down and 5 units right from <math>C(16,0)</math>, from which we obtain <math>P(21,-3)</math>. Adding the three variables, we obtain <math>90+21-3=\boxed{108}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]
 ==See Also==
 {{AIME box|year=2020|n=II|num-b=3|num-a=5}}
 [[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}