AMC10 2019 B

AMC10 2019 B · Q20

AMC10 2019 B · Q20. 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 $\overline{CGD}$，其直径在 $\overline{AD}$ 上，位于线 $AD$ 确定的同一半平面，并分别在 $E$、$F$ 和 $G$ 处与线 $EG$ 相切。以 $F$ 为圆心、半径为 2 的圆。圆内但三个半圆外的阴影区域的面积可表示为 $$\frac{a}{b} \cdot \pi - \sqrt{c} + d，$$ 其中 $a, b, c,$ 和 $d$ 是正整数，且 $a$ 和 $b$ 互质。$a + b + c + d$ 是多少？

(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 $$ \frac{4}{3}\pi-\frac{1}{2}\cdot 2\sqrt{3}\cdot 1=\frac{4}{3}\pi-\sqrt{3}. $$ Adding the areas of the three subregions gives $\frac{7}{3}\pi-\sqrt{3}+4$, and the requested sum is $7+3+3+4=17$.

答案（E）：设 $H$ 与 $I$ 为线段 $\overline{AD}$ 与以 $F$ 为圆心的圆的交点，其中 $H$ 位于 $A$ 与 $B$ 之间，$I$ 位于 $C$ 与 $D$ 之间；并设 $K$ 为从 $F$ 到 $\overline{AD}$ 的垂线段的垂足。所求区域由三个子区域组成：半径为 $2$ 的半圆、一个 $4\times 1$ 的长方形去掉 $4$ 个半径为 $1$ 的四分之一圆后的部分，以及被弦 $\overline{HI}$ 截下的圆弓形，如下图所示。半径为 $2$ 的半圆面积为 $2\pi$。长方形减去 $4$ 个四分之一圆的面积为 $4-\pi$。因为 $FK=1$ 且 $FI=2$，可得 $\angle KFI$ 的度数为 $60^\circ$，因此该圆弓形等于全圆的三分之一再减去 $\triangle HFI$。圆弓形的面积为 $$ \frac{4}{3}\pi-\frac{1}{2}\cdot 2\sqrt{3}\cdot 1=\frac{4}{3}\pi-\sqrt{3}. $$ 三部分面积相加得到 $\frac{7}{3}\pi-\sqrt{3}+4$，所求的和为 $7+3+3+4=17$。

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