AMC12 2007 B

AMC12 2007 B · Q22

AMC12 2007 B · Q22. It mainly tests Triangles (properties), Transformations.

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

两个粒子沿正三角形 $\triangle ABC$ 的边按方向 \[A\Rightarrow B\Rightarrow C\Rightarrow A,\] 同时出发并以相同速度运动。一个从 $A$ 出发，另一个从 $\overline{BC}$ 的中点出发。连接这两个粒子的线段的中点所描出的轨迹围成一个区域 $R$。求 $R$ 的面积与 $\triangle ABC$ 的面积之比。

(A) $\frac{1}{16}$ $\frac{1}{16}$

(B) $\frac{1}{12}$ $\frac{1}{12}$

(D) $\frac{1}{6}$ $\frac{1}{6}$

(E) $\frac{1}{4}$ $\frac{1}{4}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

First, notice that each of the midpoints of $AB$,$BC$, and $CA$ are on the locus. Suppose after some time the particles have each been displaced by a short distance $x$, to new positions $A'$ and $M'$ respectively. Consider $\triangle ABM$ and drop a perpendicular from $M'$ to hit $AB$ at $Y$. Then, $BA'=1-x$ and $BM'=1/2+x$. From here, we can use properties of a $30-60-90$ triangle to determine the lengths $YA'$ and $YM'$ as monomials in $x$. Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. (A cheaper way to find the shape of the region is to look at the answer choices: if it were any sort of conic section then the ratio would not generally be rational.) Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by $R$ must be $\textbf{(A)} \frac{1}{16}$ of the total area.

首先注意到，$AB$、$BC$、$CA$ 的中点都在该轨迹上。设经过一段时间后，两粒子各自沿边前进了很短的距离 $x$，分别到达新位置 $A'$ 与 $M'$。考虑 $\triangle ABM$，从 $M'$ 向 $AB$ 作垂线，垂足为 $Y$。则 $BA'=1-x$ 且 $BM'=1/2+x$。由此可利用 $30-60-90$ 三角形的性质，将 $YA'$ 与 $YM'$ 表示为关于 $x$ 的单项式。因此，中点的轨迹在上述三个特殊点之间是线性的。于是轨迹就是以这三个点为顶点的唯一三角形。（另一种更省事的判断方式是看选项：若轨迹是某种圆锥曲线，则该比值一般不会是有理数。）比较这个“中点三角形”和原三角形的内切圆半径，可得区域 $R$ 的面积为总面积的 $\textbf{(A)} \frac{1}{16}$。

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