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 $B.$

Answer: A

Concepts:transformation trigonometry

Difficulty rating: 2010

Solution:

A diagonal makes an angle $\theta$ with a long side, where $\tan\theta=\dfrac{1}{2+\sqrt3}$ $=2-\sqrt3=\tan15^\circ.$ Thus the acute angle between the two diagonals of a card is $2\theta=30^\circ.$

Each new card shares one diagonal with the previous card, and its other diagonal is the next line obtained by turning $30^\circ$ clockwise. All these equal diagonals have the same midpoint and are diameters of one common circle. The line through the original card's other diagonal, which contains $B,$ is $30^\circ$ counterclockwise from $AC.$ As an unoriented line, this is the same as $150^\circ$ clockwise from $AC.$ Five additions advance the unused diagonal by $5\cdot30^\circ=150^\circ,$ so the sixth card is the first new card with a vertex at $B.$

Thus, the correct answer is A.

2024 AMC 12A Problem 18

Answer: A

Difficulty rating: 2010

Solution:

Problem 18 in Other Years