AMC12 2014 A

AMC12 2014 A · Q12

AMC12 2014 A · Q12. It mainly tests Triangles (properties), Circle theorems.

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?

两个圆相交于点 $A$ 和 $B$。在其中一个圆上，次弧 $AB$ 测 $30^\circ$，在另一个圆上测 $60^\circ$。较大圆的面积与较小圆的面积之比是多少？

(A) 2 2

(B) 1+\sqrt3 1+\sqrt3

(D) 2+\sqrt3 2+\sqrt3

(E) 4\qquad 4\qquad

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the radius of the larger and smaller circles be $x$ and $y$, respectively. Also, let their centers be $O_1$ and $O_2$, respectively. Then the ratio we need to find is \[\dfrac{\pi x^2}{\pi y^2} = \dfrac{x^2}{y^2}\] Draw the radii from the centers of the circles to $A$ and $B$. We can easily conclude that the $30^{\circ}$ belongs to the larger circle, and the $60$ degree arc belongs to the smaller circle. Therefore, $m\angle AO_1B = 30^{\circ}$ and $m\angle AO_2B = 60^{\circ}$. Note that $\Delta AO_2B$ is equilateral, so when chord AB is drawn, it has length $y$. Now, applying the Law of Cosines on $\Delta AO_1B$: \[y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2\] \[\dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}\]

设较大圆和小圆的半径分别为 $x$ 和 $y$，它们的圆心分别为 $O_1$ 和 $O_2$。我们需要找的比值为 \[\dfrac{\pi x^2}{\pi y^2} = \dfrac{x^2}{y^2}\] 画出从圆心到 $A$ 和 $B$ 的半径。我们可以轻易得出 $30^{\circ}$ 属于较大圆，$60^\circ$ 弧属于较小圆。因此，$m\angle AO_1B = 30^{\circ}$，$m\angle AO_2B = 60^{\circ}$。注意 $\Delta AO_2B$ 是等边三角形，因此当画出弦 $AB$ 时，它的长度为 $y$。现在，在 $\Delta AO_1B$ 上应用余弦定律： \[y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2\] \[\dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}\]

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