AMC10 2000 A

AMC10 2000 A · Q16

AMC10 2000 A · Q16. It mainly tests Coordinate geometry, Counting in geometry (lattice points).

The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment AB meets segment CD at E. Find the length of segment AE.

图中显示了28个格点，每个点与其最近邻点相距一单位。线段AB与线段CD在点E相交。求线段AE的长度。

(A) $4\sqrt{5}/3$ $4\sqrt{5}/3$

(B) $5\sqrt{5}/3$ $5\sqrt{5}/3$

(D) $2\sqrt{5}$ $2\sqrt{5}$

(E) $5\sqrt{65}/9$ $5\sqrt{65}/9$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Extend $\overline{DC}$ to $F$. Triangle $FAE$ and $DBE$ are similar with ratio $5:4$. Thus $AE = 5\cdot AB/9$, $AB = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$, and $AE = 5(3\sqrt{5})/9 = 5\sqrt{5}/3$.

答案（B）：将$\overline{DC}$延长至点$F$。三角形$FAE$与$DBE$相似，相似比为$5:4$。因此$AE = 5\cdot AB/9$，$AB = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$，且$AE = 5(3\sqrt{5})/9 = 5\sqrt{5}/3$。

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