2024 AMC 12A Problem 18

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

Small hint:

The diagonal makes angle

\arctan\dfrac{1}{2+\sqrt3}=15^\circ

with the long side, since

\dfrac{1}{2+\sqrt3}=2-\sqrt3

Big hint:

The two diagonals meet at

30^\circ;

track the unused diagonal as five new cards are added

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.