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

(→‎Solution 1)
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
 
Let the midpoint of <math>AC</math> be <math>P</math>.
 
Let the midpoint of <math>AC</math> be <math>P</math>.
 +
 
We see that no matter how many moves we do, <math>P</math> stays where it is.
 
We see that no matter how many moves we do, <math>P</math> stays where it is.
 +
 
Now we can find the angle of rotation (<math>\angle APB</math>) per move with the following steps:
 
Now we can find the angle of rotation (<math>\angle APB</math>) per move with the following steps:
 +
 
<cmath>AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}</cmath>
 
<cmath>AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}</cmath>
 
<cmath>1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB</cmath>
 
<cmath>1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB</cmath>

Revision as of 22:07, 8 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.

~lptoggled, minor Latex edits by eevee9406

Solution 2

AC intersect BD at O, since $cot(15^\circ) = 2+ \sqrt{3}$, $\angle CBD = \angle BCA = 15^\circ$ 

$\angle AOB  = 60^\circ , n = 360^\circ / 60^\circ = 6  \fbox{(A) 6}$

~luckuso

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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