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2007 AMC 12B 第 22 题

先试着解答 2007 AMC 12B 第 22 题，然后核对你的答案与精心整理的解答，解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2007 AMC 12B 解答，或核对答案。

所有题目均经美国数学协会（MAA）官方合法授权使用。

22.

两个粒子沿等边 △ABC\triangle ABC 的边按方向 A→B→C→A，A\to B\to C\to A\text{，} 同时同速运动。一个从 AA 出发，另一个从 BC‾\overline{BC} 的中点出发。连接两个粒子的线段的中点所走路径围成区域 RR。求 RR 的面积与 △ABC\triangle ABC 面积之比。

Two particles move along the edges of equilateral △ABC\triangle ABC in the direction A→B→C→A,A\to B\to C\to A, starting simultaneously and moving at the same speed. One starts at A,A, and the other starts at the midpoint of BC‾.\overline{BC}. The midpoint of the line segment joining the two particles traces out a path that encloses a region R.R. What is the ratio of the area of RR to the area of △ABC?\triangle ABC?

116\dfrac{1}{16}

112\dfrac{1}{12}

19\dfrac{1}{9}

16\dfrac{1}{6}

14\dfrac{1}{4}

答案：A
知识点：相似面积比重心
难度评级：2220
小提示：

因为两个粒子都作线性运动，中点在关键时刻之间也作线性运动；把它描出的三角形记为 XYZXYZ

Because both particles move linearly, the midpoint also moves linearly between key instants; call the traced triangle XYZXYZ

大提示：

该三角形以重心 OO 为中心；比较它的外接半径 OZOZ 与 OCOC。

That triangle is centered at the centroid O;O; compare its circumradius OZOZ to OCOC

解答：

设 D,E,FD,E,F 分别为 BC,CA,ABBC,CA,AB 的中点，设 X,Y,ZX,Y,Z 分别为 AD,BE,CFAD,BE,CF 的中点。追踪一个始终位于两个粒子正中间的第三点：当两个粒子位于 A,DA,D 时，它位于 XX；当它们位于 F,CF,C 时，它位于 ZZ；当它们位于 B,EB,E 时，它位于 YY。

在这些时刻之间，两个粒子的位置都线性变化，所以它们的中点依次描出线段 XZ,ZY,YXXZ,ZY,YX。因此围成的轨迹是等边三角形 XYZXYZ，由对称性，它与 △ABC\triangle ABC 有同一个中心 OO。因为 ZZ 是中线 CFCF 的中点，OZ=OC−ZC=23CF−12CF=16CF， \begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF \end{aligned}\text{，}而 OC=23CFOC=\dfrac23 CF。

所以外接圆半径之比为 OZOC=14\dfrac{OZ}{OC}=\dfrac14，面积之比为 (14)2=116\left(\dfrac14\right)^2=\dfrac{1}{16}。

所以正确答案是 A。

Let D,E,FD,E,F be the midpoints of BC,CA,AB,BC,CA,AB, respectively, and let X,Y,ZX,Y,Z be the midpoints of AD,BE,CF,AD,BE,CF, respectively. Track a third point halfway between the two particles. It is at XX when the particles are at A,D;A,D; at ZZ when they are at F,C;F,C; and at YY when they are at B,E.B,E.

Between these instants both particle positions vary linearly, so their midpoint traces the segments XZ,ZY,YX.XZ,ZY,YX. Thus the enclosed path is the equilateral triangle XYZ,XYZ, which by symmetry shares the center OO of △ABC.\triangle ABC. Because ZZ is the midpoint of the median CF,CF, OZ=OC−ZC=23CF−12CF=16CF, \begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF, \end{aligned} while OC=23CF.OC=\dfrac23 CF.

So the ratio of circumradii is OZOC=14,\dfrac{OZ}{OC}=\dfrac14, and the area ratio is (14)2=116.\left(\dfrac14\right)^2=\dfrac{1}{16}.

Thus, the correct answer is A.

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