AMC10 2005 A

AMC10 2005 A · Q23

AMC10 2005 A · Q23. It mainly tests Triangles (properties), Circle theorems.

Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?

令$\overline{AB}$为圆的直径，$C$为$\overline{AB}$上的点且$2 \cdot AC = BC$。$D,E$为圆上的点使得$DC \perp AB$且$DE$为另一条直径。$ riangle DCE$的面积与$ riangle ABD$的面积之比是多少？

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

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

(D) $\frac{1}{2}$ $\frac{1}{2}$

(E) $\frac{2}{3}$ $\frac{2}{3}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Let $O$ be the center of the circle. Each of $\triangle DCE$ and $\triangle ABD$ has a diameter of the circle as a side. Thus the ratio of their areas is the ratio of the two altitudes to the diameters. These altitudes are $DC$ and the altitude from $C$ to $DO$ in $\triangle DCE$. Let $F$ be the foot of this second altitude. Since $\triangle CFO$ is similar to $\triangle DCO$, $$ \frac{CF}{DC}=\frac{CO}{DO}=\frac{AO-AC}{DO}=\frac{\frac{1}{2}AB-\frac{1}{3}AB}{\frac{1}{2}AB}=\frac{1}{3}, $$ which is the desired ratio.

（C）设 $O$ 为圆心。$\triangle DCE$ 和 $\triangle ABD$ 都以圆的一条直径为边。因此，它们面积之比等于对应高与直径之比。这两条高分别为 $DC$，以及在 $\triangle DCE$ 中从点 $C$ 向 $DO$ 作的高。设 $F$ 为这第二条高的垂足。由于 $\triangle CFO$ 与 $\triangle DCO$ 相似， $$ \frac{CF}{DC}=\frac{CO}{DO}=\frac{AO-AC}{DO}=\frac{\frac{1}{2}AB-\frac{1}{3}AB}{\frac{1}{2}AB}=\frac{1}{3}, $$ 这就是所求的比值。

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