2019 AMC 10A Problem 6

Question

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

Answer: C

Concepts:cyclic quadrilateral circumcircle, circumcenter, and circumradius

Difficulty rating: 1020

Small hint:

Equidistant from all four vertices means cyclic

Big hint:

For quadrilaterals, check whether opposite angles can be supplementary

Solution:

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.