AMC12 2019 B

AMC12 2019 B · Q15

AMC12 2019 B · Q15. It mainly tests Area & perimeter, Geometry misc.

As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB = BC = CD = 2$. Three semicircles of radius 1, $\overline{AEB}$, $\overline{BFC}$, and $\overline{CGD}$, have their diameters on $\overline{AD}$, lie in the same halfplane determined by line $AD$, and are tangent to line $EG$ at $E$, $F$, and $G$, respectively. A circle of radius 2 has its center at $F$. The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form $\frac{a}{b} \cdot \pi - \sqrt{c} + d$, where $a, b, c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a + b + c + d$?

如图所示，线段 $\overline{AD}$ 被点 $B$ 和 $C$ 三等分，使得 $AB = BC = CD = 2$。三个半径为1的半圆 $\overline{AEB}$、$\overline{BFC}$ 和 $\overline{CGD}$，直径在 $\overline{AD}$ 上，位于线 $AD$ 确定同一半平面，且分别在 $E$、$F$、$G$ 处与线 $EG$ 相切。半径为2的圆以 $F$ 为中心。圆内但三个半圆外的阴影区域面积可表示为 $\frac{a}{b} \cdot \pi - \sqrt{c} + d$，其中 $a, b, c,$ 和 $d$ 为正整数且 $a$ 与 $b$ 互素。求 $a + b + c + d$？

(A) 13 13

(B) 14 14

(D) 16 16

(E) 17 17

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $H$ and $I$ be the intersections of $\overline{AD}$ with the circle centered at $F$, where $H$ lies between $A$ and $B$, and $I$ lies between $C$ and $D$; and let $K$ be the foot of the perpendicular line segment from $F$ to $\overline{AD}$. The specified region consists of three subregions: a semicircle of radius 2, a $4\times 1$ rectangle with 4 quarter circles of radius 1 removed, and the segment of the circle cut off by chord $\overline{HI}$, as shown in the figure below. The semicircle of radius 2 has area $2\pi$. The rectangle minus the 4 quarter circles has area $4-\pi$. Because $FK=1$ and $FI=2$, it follows that $\angle KFI$ has measure $60^\circ$, and therefore the segment of the circle is a third of the circle with $\triangle HFI$ removed. The area of the segment is $$ \frac{4}{3}\pi-\frac12\cdot 2\sqrt3\cdot 1=\frac{4}{3}\pi-\sqrt3. $$ Adding the areas of the three subregions gives $\frac{7}{3}\pi-\sqrt3+4$, and the requested sum is $7+3+3+4=17$.

答案（E）：设$H$与$I$为$\overline{AD}$与以$F$为圆心的圆的交点，其中$H$在$A$与$B$之间，$I$在$C$与$D$之间；并设$K$为从$F$到$\overline{AD}$的垂足。所求区域由三部分组成：半径为2的半圆、去掉4个半径为1的四分之一圆的$4\times 1$矩形，以及由弦$\overline{HI}$截出的圆弓形区域，如下图所示。半径为2的半圆面积为$2\pi$。矩形减去4个四分之一圆的面积为$4-\pi$。因为$FK=1$且$FI=2$，可得$\angle KFI=60^\circ$，因此该圆弓形面积等于整圆的三分之一再减去$\triangle HFI$的面积。圆弓形面积为 $$ \frac{4}{3}\pi-\frac12\cdot 2\sqrt3\cdot 1=\frac{4}{3}\pi-\sqrt3. $$ 三部分面积相加得到$\frac{7}{3}\pi-\sqrt3+4$，所求的和为$7+3+3+4=17$。

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