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

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This problem is, of course, infinitely cheesable: it is easy to see that the answer will either be <math>6</math> 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 <math>\boxed{\textbf{(A) }6}.</math>  
 
This problem is, of course, infinitely cheesable: it is easy to see that the answer will either be <math>6</math> 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 <math>\boxed{\textbf{(A) }6}.</math>  
  
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==Solution 5==
 +
Memorize that in a 15-75-90 right triangle, the ratios of the side lengths are <math>1</math>, <math>2+\sqrt3</math>, and <math>\sqrt2+\sqrt6</math>. So, we have that the diagonal of the rectangle forms a 15 degree angle. Drawing it out we see the answer is <math>\boxed{\textbf{(A) }6}</math>, and this makes sense because 15 times 6 is 90, thus rotating a vertex back to B.
 
==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 18:16, 9 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=sin15/cos15=1/(2+sqrt3) and it eliminates all options except 6 and 12. After one rotation it has turned 30degrees, so to satisfy the problem, divide 180 by 30 and you get 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.

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

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