AMC12 2003 B

AMC12 2003 B · Q22

AMC12 2003 B · Q22. It mainly tests Triangles (properties), Coordinate geometry.

Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30$. Let $N$ be a point on $\overline{AB}$, and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$, respectively. Which of the following is closest to the minimum possible value of $PQ$?

设 $ABCD$ 是一个菱形，$AC = 16$，$BD = 30$。令 $N$ 为 $\overline{AB}$ 上的点，$P$ 和 $Q$ 分别为从 $N$ 到 $\overline{AC}$ 和 $\overline{BD}$ 的垂足。以下哪项最接近 $PQ$ 的最小可能值？

(A) 6.5 6.5

(B) 6.75 6.75

(D) 7.25 7.25

(E) 7.5 7.5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $\overline{AC}$ and $\overline{BD}$ intersect at $O$. Since $ABCD$ is a rhombus, then $\overline{AC}$ and $\overline{BD}$ are perpendicular bisectors. Thus $\angle POQ = 90^{\circ}$, so $OPNQ$ is a rectangle. Since the diagonals of a rectangle are of equal length, $PQ = ON$, so we want to minimize $ON$. It follows that we want $ON \perp AB$. Finding the area in two different ways, \[\frac{1}{2} AO \cdot BO = 60 = \frac{1}{2} ON \cdot AB = \frac{\sqrt{8^2 + 15^2}}{2} \cdot ON \Longrightarrow ON = \frac{120}{17} \approx 7.06 \Rightarrow \mathrm{(C)}\]

设 $\overline{AC}$ 与 $\overline{BD}$ 交于 $O$。由于 $ABCD$ 是菱形，$\overline{AC}$ 与 $\overline{BD}$ 互相垂直且互相平分。因此 $\angle POQ = 90^{\circ}$，所以 $OPNQ$ 是矩形。由于矩形的对角线等长，$PQ = ON$，因此我们要使 $ON$ 最小。由此可知应取 $ON \perp AB$。用两种不同方法求面积， \[\frac{1}{2} AO \cdot BO = 60 = \frac{1}{2} ON \cdot AB = \frac{\sqrt{8^2 + 15^2}}{2} \cdot ON \Longrightarrow ON = \frac{120}{17} \approx 7.06 \Rightarrow \mathrm{(C)}\]

Topics