AMC10 2019 A

AMC10 2019 A · Q6

AMC10 2019 A · Q6. It mainly tests Circle theorems, Symmetry.

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

在以下哪几种四边形中，存在一个在四边形平面内的点，该点到四边形四个顶点的距离相等？ * 正方形 * 非正方形的矩形 * 非正方形的菱形 * 既非矩形也非菱形的平行四边形 * 非平行四边形的等腰梯形

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): In a square or non-square rectangle, the diagonals are congruent and bisect each other, and their point of intersection is equidistant from all four vertices. This point also lies on the perpendicular bisectors of all four sides. In a non-square rhombus or a parallelogram that is not a rectangle or rhombus, the perpendicular bisectors of parallel sides do not meet, so no point could be equidistant from all four vertices. Finally, in an isosceles trapezoid that is not a parallelogram, the perpendicular bisectors of the parallel sides are the same line, and the perpendicular bisectors of the nonparallel sides meet at a point on this line; that point is equidistant from all four vertices. In summary, 3 of the types of given quadrilaterals—the first, second, and fifth in the list—have the required property.

答案（C）：在正方形或非正方形的矩形中，两条对角线全等并且互相平分，它们的交点到四个顶点的距离相等。这个点也位于四条边的垂直平分线上。在非正方形的菱形，或既不是矩形也不是菱形的平行四边形中，平行边的垂直平分线不会相交，因此不存在一个点能到四个顶点等距。最后，在不是平行四边形的等腰梯形中，两条平行边的垂直平分线是同一条直线，而两条不平行边的垂直平分线在这条直线上相交于一点；该点到四个顶点的距离相等。总之，给出的四边形类型中有 3 种——列表中的第 1、2、5 种——具有所需性质。

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