2019 AMC 10A Problem 6

A point equidistant from all four vertices is the center of a circle through them, so the question asks which quadrilaterals can be inscribed in a circle. A quadrilateral is cyclic exactly when each pair of opposite angles sums to $180^\circ$.

• Square: all angles $90^\circ$; cyclic. (The center is where the diagonals meet.)
• Rectangle: all angles $90^\circ$; cyclic, center at the diagonal intersection.
• Rhombus that is not a square: opposite angles are equal, so they sum to $180^\circ$ only if each is $90^\circ$, i.e. a square. Not cyclic.
• Parallelogram that is not a rectangle: same reasoning, opposite angles equal and not $90^\circ$. Not cyclic.
• Isosceles trapezoid: the two base angles on the longer base are equal, and each is supplementary to a base angle on the shorter base, so opposite angles are supplementary. Cyclic.

Three types qualify.

The answer is $\boxed{\textbf{(C)}\ 3}$.

"Equidistant from the vertices" translates to "center of the circumscribed circle." For quadrilaterals the clean test is supplementary opposite angles. Parallelograms have equal opposite angles, so they are cyclic only in the right-angled case; isosceles trapezoids always pass because their symmetry makes the angle pairs supplementary.

• Confusing 'equidistant from the vertices' with 'equidistant from the sides,' which would make the rhombus qualify instead of the trapezoid.
• Assuming a rhombus is cyclic because its diagonals meet at a center of symmetry; that point is equidistant from the sides, not the vertices.