Question

两个圆交于点 $A$ 和 $B$ 。其中一个圆上的小弧 $AB$ 为 $30^\circ$ ，另一个圆上的小弧 AB 为 $60^\circ$ 。大圆面积与小圆面积之比是多少？

Two circles intersect at points $A$ and $B.$ The minor arcs $AB$ measure $30^\circ$ on one circle and $60^\circ$ on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?

$2$

$1+\sqrt3$

$3$

$2+\sqrt3$

$4$

答案：D

知识点：弦三角学面积比

难度评级：1630

解答：

设对应 $30^\circ$ 弧的圆半径为 $R$ ，对应 $60^\circ$ 弧的圆半径为 $r$ 。公共弦的长度满足 $2R\sin15^\circ=2r\sin30^\circ$ ，所以 $\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}$ 。

较小的圆心角对应较大的半径，因此 $R\gt r$ 。所求面积比为 $\begin{gathered} \left(\dfrac{R}{r}\right)^2\\ =\dfrac{1}{4\sin^2 15^\circ}\\ =\dfrac{1}{2(1-\cos30^\circ)}\\ =\dfrac{1}{2-\sqrt3}=2+\sqrt3. \end{gathered}$

所以正确答案是 D。

Let the circles have radii $R$ (with the $30^\circ$ arc) and $r$ (with the $60^\circ$ arc). The common chord has length $2R\sin15^\circ=2r\sin30^\circ,$ so $\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}.$

The smaller central angle gives the larger radius, so $R\gt r.$ The area ratio is $\begin{gathered} \left(\dfrac{R}{r}\right)^2\\ =\dfrac{1}{4\sin^2 15^\circ}\\ =\dfrac{1}{2(1-\cos30^\circ)}\\ =\dfrac{1}{2-\sqrt3}=2+\sqrt3. \end{gathered}$

Thus, the correct answer is D.

2014 AMC 12A 第 12 题

答案：D

难度评级：1630

解答：

其他年份的第 12 题