AMC10 2013 B

AMC10 2013 B · Q7

AMC10 2013 B · Q7. It mainly tests Triangles (properties), Circle theorems.

Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?

圆周上等间距地有六个点，半径为1。其中三个点构成一个既非等边也非等腰的三角形。这个三角形的面积是多少？

(A) $\frac{\sqrt{3}}{3}$ $\frac{\sqrt{3}}{3}$

(B) $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{2}$

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

(E) 2 2

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The six points divide the circle into six arcs each measuring $60^\circ$. By the Inscribed Angle Theorem, the angles of the triangle can only be $30^\circ$, $60^\circ$, $90^\circ$, and $120^\circ$. Because the angles of the triangle are pairwise distinct the triangle must be a $30-60-90^\circ$ triangle. Therefore the hypotenuse of the triangle is the diameter of the circle, and the legs have lengths $1$ and $\sqrt{3}$. The area of the triangle is $\frac{1}{2}\cdot 1\cdot \sqrt{3}=\frac{\sqrt{3}}{2}$.

答案（B）：这六个点把圆分成六段弧，每段弧的度数都是 $60^\circ$。由圆周角定理可知，这个三角形的内角只能是 $30^\circ$、$60^\circ$、$90^\circ$ 和 $120^\circ$。因为三角形的三个角两两不同，所以该三角形必须是一个 $30-60-90^\circ$ 三角形。因此三角形的斜边是圆的直径，两条直角边的长度分别为 $1$ 和 $\sqrt{3}$。三角形的面积为 $\frac{1}{2}\cdot 1\cdot \sqrt{3}=\frac{\sqrt{3}}{2}$。

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