Question

对下面多少种四边形，存在一个位于该四边形所在平面内的点，它到四个顶点的距离都相等？

• 正方形

• 非正方形的矩形

• 非正方形的菱形

• 既不是矩形也不是菱形的平行四边形

• 不是平行四边形的等腰梯形

For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?

• a square

• a rectangle that is not a square

• a rhombus that is not a square

• a parallelogram that is not a rectangle or a rhombus

• an isosceles trapezoid that is not a parallelogram

$1$

$2$

$3$

$4$

$5$

答案：C

知识点：圆内接四边形外接圆、外心与外接圆半径

难度评级：1020

小提示：

到四个顶点等距意味着四边形可以内接于一个圆。

Equidistant from all four vertices means cyclic

大提示：

对四边形，检查对角是否互补。

For quadrilaterals, check whether opposite angles can be supplementary

解答：

注意，如果某个点到四个顶点的距离都相等，那么这个点就是该图形外接圆的圆心。

于是问题变成：这些图形中哪些是圆内接四边形（存在外接圆）。可以使用的一个条件是：对角互补。

显然，正方形和非正方形的矩形都可以（对角都是直角，和为 $180^{\circ}$ ）。

非正方形的菱形不行，因为它的对角虽然相等，但不是 $90^{\circ}$ 。

既不是矩形也不是菱形的平行四边形面临同样的问题，所以它同样不是圆内接四边形。

不是平行四边形的等腰梯形，按定义其对角互补，所以它是圆内接四边形。

所以正确答案是 C。

Note that if a point is equidistant from all the vertices, then that point is the center of the shape’s circumcircle.

The question then becomes which of these shapes is cyclic (has a circumcircle). One condition that we can use is that opposite angles are supplementary.

Clearly, a square and rectangle that is not a square work (opposite angles are right, adding up to $180^{\circ}$ ).

A rhombus that is not a square does not work, since opposite angles are equal, but they are not $90^{\circ}.$

A parallelogram that is not a rectangle or a rhombus faces the same problem as above, making it not cyclic as well.

An isosceles trapezoid that is not a parallelogram by definition has supplementary opposite angles, making it cyclic.

Thus, C is the correct answer.

2019 AMC 10A 第 6 题

答案：C

难度评级：1020

小提示：

大提示：

解答：

其他年份的第 6 题