2019 AMC 10A Problem 6

If a point is the same distance from all four corners, then a circle centered at that point passes through all four corners. So the question is: which of these shapes can have all four corners on one circle?

There is a simple test for that: a four-sided shape fits in a circle exactly when each pair of opposite angles adds to $180^\circ$.

Check the five shapes one by one.

1. Square. Every angle is $90^\circ$, and $90^\circ + 90^\circ = 180^\circ$. It fits. (The center is where the diagonals cross.)
2. Rectangle. Every angle is $90^\circ$ again. It fits, with the center where the diagonals cross.
3. Rhombus that is not a square. In a rhombus, opposite angles are equal. Two equal angles add to $180^\circ$ only if each is $90^\circ$, and then it would be a square. It does not fit.
4. Parallelogram that is not a rectangle. Same reason: opposite angles are equal but not $90^\circ$, so they do not add to $180^\circ$. It does not fit.
5. Isosceles trapezoid. Say the angles along the bottom are both $70^\circ$. Because the top and bottom are parallel, each top angle is $180^\circ - 70^\circ = 110^\circ$. Each bottom angle is opposite a top angle, and $70^\circ + 110^\circ = 180^\circ$. The same works for any base angle. It fits.

Three types qualify: the square, the rectangle and the isosceles trapezoid.

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

"Same distance from every corner" means "center of a circle through the corners". For a four-sided shape, the quick test is that opposite angles add to $180^\circ$. Parallelograms have equal opposite angles, so they pass only when those angles are right angles. The isosceles trapezoid always passes because its matching base angles make each opposite pair add to $180^\circ$.

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 $\boxed{\textbf{(C)}\ 3}$.