Question

在一张边长分别为 $1$ 和 $2+\sqrt3$ 的矩形卡片上，放置一张相同的卡片，使得两张卡片的两条对角线重合，如图所示（本图中为 $AC$ ）。

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

继续这个过程，在第二张卡片上加第三张卡片，依此类推，每次顺时针旋转后让相邻的对角线重合。总共必须使用多少张卡片，才会使一张新卡片的某个顶点正好落在图中标为 $B$ 的顶点上？

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?

$6$

$8$

$10$

$12$

不会有新顶点落在 $B$ 上。

No new vertex will land on $B.$

答案：A

知识点：变换三角学

难度评级：2010

小提示：

对角线与长边所成角为 $\arctan\dfrac{1}{2+\sqrt3}=15^\circ$ ，因为 $\dfrac{1}{2+\sqrt3}=2-\sqrt3$ 。

The diagonal makes angle $\arctan\dfrac{1}{2+\sqrt3}=15^\circ$ with the long side, since $\dfrac{1}{2+\sqrt3}=2-\sqrt3$

大提示：

两条对角线成 $30^\circ$ 角；追踪加入五张新卡片的过程中未被共用的那条对角线。

The two diagonals meet at $30^\circ;$ track the unused diagonal as five new cards are added

解答：

卡片的对角线与长边所成角 $\theta$ 满足 $\tan\theta=\dfrac{1}{2+\sqrt3}$ $=2-\sqrt3=\tan15^\circ$ ，所以 $2\theta=30^\circ$ 。

每张新卡片与前一张共用一条对角线，而它的另一条对角线正是再顺时针转 $30^\circ$ 得到的下一条直线。这些等长的对角线有共同的中点，都是同一个圆的直径。原卡片另一条对角线所在的直线经过 $B$ ，它相对 $AC$ 逆时针转了 $30^\circ$ ；作为不带方向的直线，这等同于相对 $AC$ 顺时针转了 $150^\circ$ 。每加一张卡片，未被共用的对角线就前进一步，加五张后共前进 $5\cdot30^\circ=150^\circ$ ，所以第六张卡片是第一张有顶点落在 $B$ 上的新卡片。

因此正确答案是 A。

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 第 18 题

答案：A

难度评级：2010

小提示：

大提示：

解答：

其他年份的第 18 题