2018 AMC 10B 第 24 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。

24.

ABCDEFABCDEF 是边长为 11 的正六边形。令 XXYYZZ 分别为边 ABABCDCDEFEF 的中点。某个凸六边形的内部恰好是 ACE\triangle ACEXYZ\triangle XYZ 的内部交集。这个凸六边形的面积是多少?

Let ABCDEFABCDEF be a regular hexagon with side length 1.1. Denote by X,X, Y,Y, and ZZ the midpoints of sides AB,AB, CD,CD, and EF,EF, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of ACE\triangle ACE and XYZ?\triangle XYZ?

383\dfrac{3}{8}\sqrt{3}

7163\dfrac{7}{16}\sqrt{3}

15323\dfrac{15}{32}\sqrt{3}

123\dfrac{1}{2}\sqrt{3}

9163\dfrac{9}{16}\sqrt{3}

答案:C
知识点:正多边形等边三角形面积分割
难度评级:2470
小提示:

ACE\triangle ACE 是边长为 3\sqrt{3} 的等边三角形;XYZ\triangle XYZ 是边长为 32\tfrac{3}{2} 的等边三角形。

ACE\triangle ACE is equilateral with side 3;\sqrt{3}; XYZ\triangle XYZ is equilateral with side 32\tfrac{3}{2}

大提示:

以正六边形的中心为公共中心,这两个三角形同心且相差 3030^\circ;交集是 XYZ\triangle XYZ 去掉三个伸出的角三角形。

Centered at the hexagon’s center the two triangles are concentric and rotated 30;30^\circ; the overlap is XYZ\triangle XYZ minus its three protruding corner triangles

解答:

三角形 XYZXYZ 是边长为 32\frac{3}{2} 的等边三角形,所以面积为 34(32)2=9316\dfrac{\sqrt3}{4}\left(\dfrac32\right)^2=\dfrac{9\sqrt3}{16}\text{。}

三角形 ACEACEXYZXYZ 同心,且彼此旋转 3030^\circ。在 XYZXYZ 的每个顶点处,ACEACE 的两边截出一个 3030-6060-9090 三角形,其斜边是半边长线段 AX=12AX=\frac{1}{2}。两条直角边分别为 14\frac{1}{4}34\frac{\sqrt3}{4},所以每个角的面积为 332\frac{\sqrt3}{32}

去掉三个角后,面积为 93163332=15332\dfrac{9\sqrt3}{16}-3\cdot\dfrac{\sqrt3}{32}=\dfrac{15\sqrt3}{32}\text{。}因此答案是 C

The triangle XYZXYZ is equilateral with side 32,\frac{3}{2}, so its area is 34(32)2=9316.\dfrac{\sqrt3}{4}\left(\dfrac32\right)^2=\dfrac{9\sqrt3}{16}.

The triangles ACEACE and XYZXYZ are concentric and rotated 3030^\circ from each other. At each vertex of XYZ,XYZ, the sides of ACEACE cut off a 3030-6060-9090 triangle whose hypotenuse is the half-side segment AX=12.AX=\frac{1}{2}. Its legs are 14\frac{1}{4} and 34,\frac{\sqrt3}{4}, so each corner has area 332.\frac{\sqrt3}{32}.

Removing the three corners gives 93163332=15332.\dfrac{9\sqrt3}{16}-3\cdot\dfrac{\sqrt3}{32}=\dfrac{15\sqrt3}{32}. Therefore, the answer is C.

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