AMC12 2000 A

AMC12 2000 A · Q17

AMC12 2000 A · Q17. It mainly tests Angle chasing, Triangles (properties).

A circle centered at $O$ has radius $1$ and contains the point $A$. The segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta$. If point $C$ lies on $\overline{OA}$ and $\overline{BC}$ bisects $\angle ABO$, then $OC =$

以$O$为圆心、半径为$1$的圆包含点$A$。线段$AB$在$A$点与圆相切，且$\angle AOB = \theta$。若点$C$在$\overline{OA}$上，且$\overline{BC}$平分$\angle ABO$，则$OC =$

(A) $\sec^2 \theta - \tan \theta$ $\sec^2 \theta - \tan \theta$

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

(D) $\frac{1}{1 + \sin \theta}$ $\frac{1}{1 + \sin \theta}$

(E) $\frac{\sin \theta}{\cos^2 \theta}$ $\frac{\sin \theta}{\cos^2 \theta}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. This means that $OA = 1$, $AB = \tan \theta$ and $OB = \sec \theta$. By the Angle Bisector Theorem, \[\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta\] We multiply both sides by $\cos \theta$ to simplify the trigonometric functions, \[AC=OC \sin \theta\] Since $AC + OC = 1$, $1 - OC = OC \sin \theta \Longrightarrow$ $OC = \dfrac{1}{1+\sin \theta}$. Therefore, the answer is $\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}$.

由于$\overline{AB}$与圆相切，$\triangle OAB$是直角三角形。这意味着$OA = 1$，$AB = \tan \theta$，且$OB = \sec \theta$。由角平分线定理，\[\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta\] 两边同乘$\cos \theta$以化简三角函数，\[AC=OC \sin \theta\] 由于$AC + OC = 1$，$1 - OC = OC \sin \theta \Longrightarrow$ $OC = \dfrac{1}{1+\sin \theta}$。因此答案为$\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}$。

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