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Difference between revisions of "2024 AMC 12A Problems/Problem 18"

(→‎Solution 3 (In case you have no time and that's what I did))
(→‎Solution 6 (the simplest solution ever))
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~EaZ_Shadow
 
~EaZ_Shadow
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==Solution 7==
 +
[[File:Rotation of rectangle.png|250px|right]]
 +
Process is the rotation around the center of the card point <math>O</math> at the angle <math>\alpha = \angle AOB.</math>
 +
<cmath>AO = BO = R, BD^2 = 4R^2 = AB^2 + AD^2 = 4 \cdot (2+\sqrt{3}).</cmath>
 +
By applying the Law of Cosines, we get
 +
<cmath>2R^2 (1-\cos \alpha) = AB^2 \implies \cos \alpha = \frac {\sqrt{3}}{2} \implies</cmath>
 +
<cmath>\alpha = 30^\circ \implies \boxed{\textbf{(A) or 6}}. </cmath>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2024|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:02, 14 November 2024

Problem

On top of a rectangular card with sides of length $1$ and $2+\sqrt{3}$, an identical card is placed so that two of their diagonals line up, as shown ($\overline{AC}$, in this case).

[asy] defaultpen(fontsize(12)+0.85); size(150); real h=2.25; pair C=origin,B=(0,h),A=(1,h),D=(1,0),Dp=reflect(A,C)*D,Bp=reflect(A,C)*B; pair L=extension(A,Dp,B,C),R=extension(Bp,C,A,D); draw(L--B--A--Dp--C--Bp--A); draw(C--D--R); draw(L--C^^R--A,dashed+0.6); draw(A--C,black+0.6); dot("$C$",C,2*dir(C-R)); dot("$A$",A,1.5*dir(A-L)); dot("$B$",B,dir(B-R)); [/asy]

Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled $B$ in the figure?

$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{No new vertex will land on }B.$

Solution 1

Let the midpoint of $AC$ be $P$.

We see that no matter how many moves we do, $P$ stays where it is.

Now we can find the angle of rotation ($\angle APB$) per move with the following steps:

\[AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}\] \[1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB\] \[1=2(2+\sqrt{3})(1-\cos\angle APB)\] \[\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\] \[\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot\frac{4-2\sqrt{3}}{4-2\sqrt{3}}\] \[\cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\] \[\angle APB=30^\circ\] Since Vertex $C$ is the closest one and \[\angle BPC=360-180-30=150\]

Vertex C will land on Vertex B when $\frac{150}{30}+1=\fbox{(A) 6}$ cards are placed.

(someone insert diagram maybe)

~lptoggled, minor Latex edits by eevee9406

Solution 2

2024 amc12A p18.png

AC intersects BD at O,

we want to find $\angle AOB$

since $tan(75^\circ) = 2+ \sqrt{3} =\frac{AD}{AB}$, $\angle CBD = \angle BCA = 15^\circ$ 

\[\angle AOB  = \angle CBD  + \angle BCA  =30^\circ ,\]

so each time we rotate BD to AC for $30^\circ$, and we need to rotate n = $180^\circ / 30^\circ = 6$ times to overlap with B (from one of A,B,C,D) ( should not be n = $360^\circ / 30^\circ = 12)$ \[answer {(A) 6}\]

note: if you don't remember $tan(75^\circ)$ \[tan(75^\circ) = \frac{tan(45^\circ) + tan(30^\circ)}{ 1 - tan(45^\circ)*tan(30^\circ)}\] \[= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1* \frac{1}{\sqrt{3}}}\] \[= \frac{(\sqrt{3}+1)^2 }{ (\sqrt{3})^2-1} = 2+ \sqrt{3}\]

~luckuso

Solution 3(In case you have no time and that's what I did)

$\tan{15}=\frac{\sin{15}}{\cos{15}}=\frac{1}{2+\sqrt3}$ and it eliminates all options except $6$ and $12$. After one rotation it has turned $30^{\circ}$, so to satisfy the problem, divide $\frac{180}{30}$ and get $\boxed{\textbf{A. }6}$.

Solution 4 (cheese core)

This problem is, of course, infinitely cheesable: it is easy to see that the answer will either be $6$ rotations or no valid rotations whatsoever (A or E). In general, the answer is almost never "none of the above" (or the like), so it makes sense to go with $\boxed{\textbf{(A) }6}.$


Solution 5

Memorize that in a 15-75-90 right triangle, the ratios of the side lengths are $1$, $2+\sqrt3$, and $\sqrt2+\sqrt6$. So, we have that the diagonal of the rectangle forms a 15 degree angle. Drawing it out we see the answer is $\boxed{\textbf{(A) }6}$, and this makes sense because 15 times 6 is 90, thus rotating a vertex back to B.

Solution 6 (the simplest solution ever)

Look at the picture and draw the next one and continue draw down the line and then when you first hit B, count how many rectangles you’ve drawn (excluding the first which hasn’t been rotated), and you should get $\boxed{\textbf{(A) or 6}}$ as the answer.

~EaZ_Shadow

Solution 7

Rotation of rectangle.png

Process is the rotation around the center of the card point $O$ at the angle $\alpha = \angle AOB.$ \[AO = BO = R, BD^2 = 4R^2 = AB^2 + AD^2 = 4 \cdot (2+\sqrt{3}).\] By applying the Law of Cosines, we get \[2R^2 (1-\cos \alpha) = AB^2 \implies \cos \alpha = \frac {\sqrt{3}}{2} \implies\] \[\alpha = 30^\circ \implies \boxed{\textbf{(A) or 6}}.\] vladimir.shelomovskii@gmail.com, vvsss

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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