AMC12 2024 A

AMC12 2024 A · Q18

AMC12 2024 A · Q18. It mainly tests Circle theorems, Transformations.

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). 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?

在一张长宽为 $1$ 和 $2+\sqrt{3}$ 的矩形卡片上方，放置一张相同的卡片，使它们的对角线对齐（如图中的 $\overline{AC}$）。继续这个过程，在第二张上添加第三张，依此类推，每次顺时针旋转后对齐连续的对角线。总共需要使用多少张卡片，直到一张新卡片的顶点恰好落在图中标记为 $B$ 的顶点上？

(A) 6 6

(B) 8 8

(D) 12 12

(E) \text{No new vertex will land on }B. \text{No new vertex will land on }B.

Answer

Correct choice: (A)

正确答案：（A）

Solution

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.

设 $AC$ 的中点为 $P$。我们看到无论做多少次移动，$P$ 都保持在原位。现在可以通过以下步骤找到每次移动的旋转角（$\angle APB$）： \[AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}\] \[1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB\] \[1=2(2+\sqrt{3})(1-\cos\angle APB)\] \[\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\] \[\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot\frac{4-2\sqrt{3}}{4-2\sqrt{3}}\] \[\cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\] \[\angle APB=30^\circ\] 由于顶点 $C$ 是最近的，且 \[\angle BPC=360-180-30=150\] 顶点 $C$ 将落在顶点 $B$ 上时，需要 $\frac{150}{30}+1=\fbox{(A) 6}$ 张卡片。

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