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

解答：

若存在到四个顶点等距的点，则该点是外接圆圆心，四边形必须是圆内接四边形。

正方形和矩形都可以内接于圆。

非正方形菱形的两组对角分别相等，但不全是直角，因此对角之和不是 $180^{\circ}$ ，不能内接于圆。

一般平行四边形也同理；平行四边形只有在四个角都是 $90^{\circ}$ 时才能内接于圆，也就是必须为矩形。

等腰梯形总是圆内接四边形。

所以符合的有正方形、非正方形矩形和非平行四边形的等腰梯形，共三种。

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

解答：

其他年份的第 6 题