AMC10 2010 A

AMC10 2010 A · Q22

AMC10 2010 A · Q22. It mainly tests Combinations, Circle theorems.

Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?

在圆周上选 8 个点，并连接每对点的弦。没有三条弦在圆内单一点相交。圆内顶点全在圆内的三角形有多少个？

(A) 28 28

(B) 56 56

(D) 84 84

(E) 140 140

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Three chords create a triangle if and only if they intersect pairwise inside the circle. Two chords intersect inside the circle if and only if their endpoints alternate in order around the circle. Therefore, if points $A, B, C, D, E,$ and $F$ are in order around the circle, then only the chords $\overline{AD}, \overline{BE}, \overline{CF}$ all intersect pairwise inside the circle. Thus every set of $6$ points determines a unique triangle, and there are $\binom{8}{6} = 28$ such triangles.

答案（A）：三条弦当且仅当它们在圆内两两相交时构成一个三角形。两条弦当且仅当它们的端点在圆周上的顺序交替出现时，才会在圆内相交。因此，若点 $A, B, C, D, E, F$ 按顺序位于圆周上，则只有弦 $\overline{AD}, \overline{BE}, \overline{CF}$ 会在圆内两两相交。于是，每一组 $6$ 个点确定一个唯一的三角形，这样的三角形共有 $\binom{8}{6} = 28$ 个。

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