AMC12 2018 B

AMC12 2018 B · Q13

AMC12 2018 B · Q13. It mainly tests Polygons, Geometry misc.

Square $ABCD$ has side length 30. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, and $\triangle DAP$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?

正方形$ABCD$边长为30。点$P$位于正方形内部，使得$AP = 12$，$BP = 26$。$\triangle ABP$、$\triangle BCP$、$\triangle CDP$和$\triangle DAP$的质心构成一个凸四边形的顶点。该四边形的面积是多少？

(A) $100\sqrt{2}$ $100\sqrt{2}$

(B) $100\sqrt{3}$ $100\sqrt{3}$

(D) $200\sqrt{2}$ $200\sqrt{2}$

(E) $200\sqrt{3}$ $200\sqrt{3}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $E$ and $F$ be the midpoints of sides $\overline{BC}$ and $\overline{CD}$, respectively. Let $G$ and $H$ be the centroids of $\triangle BCP$ and $\triangle CDP$, respectively. Then $G$ is on $\overline{PE}$, a median of $\triangle BCP$, a distance $\frac{2}{3}$ of the way from $P$ to $E$. Similarly, $H$ is on $\overline{PF}$ a distance $\frac{2}{3}$ of the way from $P$ to $F$. Thus $\overline{GH}$ is parallel to $\overline{EF}$ and $\frac{2}{3}$ the length of $\overline{EF}$. Because $BC = 30$, it follows that $EC = 15$, $EF = 15\sqrt{2}$, and $GH = 10\sqrt{2}$. The midpoints of $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$ form a square, so the centroids of $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, and $\triangle DAP$ also form a square, and that square has side length $10\sqrt{2}$. The requested area is $(10\sqrt{2})^2 = 200$.

答案（C）：设 $E$ 和 $F$ 分别为边 $\overline{BC}$ 与 $\overline{CD}$ 的中点。设 $G$ 和 $H$ 分别为 $\triangle BCP$ 与 $\triangle CDP$ 的重心。则 $G$ 在 $\overline{PE}$ 上，且 $\overline{PE}$ 是 $\triangle BCP$ 的一条中线；$G$ 位于从 $P$ 到 $E$ 距离的 $\frac{2}{3}$ 处。同理，$H$ 在 $\overline{PF}$ 上，且位于从 $P$ 到 $F$ 距离的 $\frac{2}{3}$ 处。因此 $\overline{GH}\parallel \overline{EF}$，且 $\overline{GH}$ 的长度为 $\overline{EF}$ 的 $\frac{2}{3}$。因为 $BC=30$，可得 $EC=15$，$EF=15\sqrt{2}$，并且 $GH=10\sqrt{2}$。$\overline{AB}$、$\overline{BC}$、$\overline{CD}$、$\overline{DA}$ 的中点构成一个正方形，所以 $\triangle ABP$、$\triangle BCP$、$\triangle CDP$、$\triangle DAP$ 的重心也构成一个正方形，该正方形的边长为 $10\sqrt{2}$。所求面积为 $(10\sqrt{2})^2=200$。

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