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2019 AMC 12B 第 15 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

15.

如图，线段 AD‾\overline{AD} 被点 BB 和 CC 三等分，使得 AB=BC=CD=2AB=BC=CD=2。三个半径为 11 的半圆 AEBAEB，BFCBFC，和 CGDCGD 的直径都在 AD‾\overline{AD} 上，并分别在 EE，FF，和 GG 处与直线 EGEG 相切。一个半径为 22 的圆以 FF 为圆心。图中阴影区域，即在该圆内但在三个半圆外的区域，其面积可表示为

ab⋅π−c+d， \dfrac{a}{b}\cdot\pi-\sqrt{c}+d\text{，}

其中 aa，bb，cc，和 dd 为正整数，且 aa 与 bb 互质。a+b+c+da+b+c+d 是多少？

As shown in the figure, line segment AD‾\overline{AD} is trisected by points BB and CC so that AB=BC=CD=2.AB=BC=CD=2. Three semicircles of radius 1,1, AEB,AEB, BFC,BFC, and CGD,CGD, have their diameters on AD‾,\overline{AD}, and are tangent to line EGEG at E,E, F,F, and G,G, respectively. A circle of radius 22 has its center on F.F. The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form

ab⋅π−c+d, \dfrac{a}{b}\cdot\pi-\sqrt{c}+d,

where a,a, b,b, c,c, and dd are positive integers and aa and bb are relatively prime. What is a+b+c+d?a+b+c+d?

1313

1414

1515

1616

1717

答案：E
知识点：圆面积面积分割坐标几何
难度评级：1830
小提示：

取 F=(3,1)F=(3,1)；半径为 22 的圆经过 EE 和 GG，且中间的半圆完全在它内部

Place F=(3,1);F=(3,1); the circle of radius 22 then passes through EE and G,G, and the middle semicircle lies entirely inside it

大提示：

从圆的面积中减去三个半圆落在该圆内部的部分

Subtract from the circle’s area the parts of the three semicircles that fall inside it

解答：

设 A=(0,0)A=(0,0)， B=(2,0)\ B=(2,0)， C=(4,0)\ C=(4,0)， D=(6,0)\ D=(6,0)，则三个半圆的圆心为 (1,0),(3,0),(5,0)(1,0),(3,0),(5,0)，顶点为 E=(1,1)E=(1,1)， F=(3,1)\ F=(3,1)， G=(5,1)\ G=(5,1)。该圆的圆心为 F=(3,1)F=(3,1)，半径为 22，所以它经过 EE 和 GG，面积为 4π4\pi。

中间半圆 BFCBFC 完全落在该圆内，去掉面积 π2\dfrac{\pi}{2}。对于左边的半圆，重叠部分从 EE 开始，包含它右侧的四分之一圆，但要除去落在大圆下方的那块区域。大圆与 xx 轴相交于 x=3−3x=3-\sqrt3。被除去的面积为I=∫13−31 dx−∫13−34−(x−3)2 dx=2−32−π3。 \begin{aligned} I &=\int_1^{3-\sqrt3}1\,dx\\ &\quad-\int_1^{3-\sqrt3} \sqrt{4-(x-3)^2}\,dx\\ &=2-\dfrac{\sqrt3}{2}-\dfrac{\pi}{3} \end{aligned}\text{。} 因此重叠部分的面积为R=π4−I=7π12−2+32。 \begin{aligned} R&=\dfrac{\pi}{4}-I\\ &=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2} \end{aligned}\text{。} 由对称性，右边的半圆贡献同样大的重叠面积。

阴影面积为4π−π2−2R=73π−3+4。 4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4\text{。} 因此 a=7, b=3, c=3, d=4a=7,\ b=3,\ c=3,\ d=4，所以 a+b+c+d=17a+b+c+d=17。

所以正确答案是 E。

Put A=(0,0),A=(0,0),  B=(2,0),\ B=(2,0),  C=(4,0),\ C=(4,0),  D=(6,0),\ D=(6,0), so the semicircles are centered at (1,0),(3,0),(5,0)(1,0),(3,0),(5,0) and their tops are E=(1,1),E=(1,1),  F=(3,1),\ F=(3,1),  G=(5,1).\ G=(5,1). The circle has center F=(3,1)F=(3,1) and radius 2,2, so it passes through EE and G,G, and has area 4π.4\pi.

The middle semicircle BFCBFC lies entirely inside the circle, removing area π2.\dfrac{\pi}{2}. For the left semicircle, the overlap starts at EE and includes its right-hand quarter-circle, except for the region below the large circle. The large circle meets the xx-axis at x=3−3.x=3-\sqrt3. The excluded area is I=∫13−31 dx−∫13−34−(x−3)2 dx=2−32−π3. \begin{aligned} I &=\int_1^{3-\sqrt3}1\,dx\\ &\quad-\int_1^{3-\sqrt3} \sqrt{4-(x-3)^2}\,dx\\ &=2-\dfrac{\sqrt3}{2}-\dfrac{\pi}{3}. \end{aligned} Therefore the overlap has area R=π4−I=7π12−2+32. \begin{aligned} R&=\dfrac{\pi}{4}-I\\ &=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2}. \end{aligned} By symmetry, the right semicircle contributes the same overlap.

The shaded area is 4π−π2−2R=73π−3+4. 4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4. Hence a=7, b=3, c=3, d=4,a=7,\ b=3,\ c=3,\ d=4, so a+b+c+d=17.a+b+c+d=17.

Thus, E is the correct answer.

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