2024 AMC 12A Problems/Problem 18

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.

~minor Latex edits by eevee9406

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2024 AMC 12A Problems/Problem 18

Problem

Solution 1

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