Question

On top of a rectangular card with sides of length $1$ and $2+\sqrt3,$ an identical card is placed so that two of their diagonals line up, as shown ( $AC,$ in this case).

Two congruent rectangular cards sharing the diagonal AC, with the second card rotated.

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?

$6$

$8$

$10$

$12$

No new vertex will land on $B.$

Solution:

The diagonal of the card makes an angle $\theta$ with the long side where $\tan\theta=\dfrac{1}{2+\sqrt3}=2-\sqrt3=\tan15^\circ,$ so $\theta=15^\circ.$ All the cards share diagonals that are equal chords (diameters) of one common circle, and each newly added card is the previous one turned $15^\circ$ clockwise about the common center. A fresh vertex first coincides with $B$ once the accumulated rotation reaches $90^\circ,$ i.e. after $90^\circ/15^\circ=6$ cards. Since $15^\circ$ divides $90^\circ$ evenly, a vertex does land on $B.$ Thus, the correct answer is A.

2024 AMC 12A Problem 18

Solution:

Problem 18 in Other Years