Problem:
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
Answer Choices:
A.
B.
C.
D.
E.
Solution:
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, of the types of given quadrilaterals - the first, second, and fifth in the list - have the required property.
The required point exists if and only if the quadrilateral can be inscribed in a circle, in which case the point is the center of the circle. A square, rectangle, and isosceles trapezoid can each be inscribed in a circle, but a non-square rhombus and a non-rectangular parallelogram cannot. Therefore the required point exists for of the listed types of quadrilaterals.
The problems on this page are the property of the MAA's American Mathematics Competitions