2010 AMC 8 Problem 6

A square has both diagonal and midpoint-to-midpoint symmetry lines.

Count reflection lines for each named shape.

First, each line of symmetry must go through the center of the shape. This would ensure that the center of the shape isn’t on only one side, as that would make it asymmetric.

Next, if a line goes through any side, it must go through its midpoint, to ensure that after a reflection over this line, the same amount of the line is on both sides. Moreover, it must be perpendicular, to ensure that the line when reflected stays on itself.

Similarly, if a line goes through any corner, it must go through its angle bisector, to ensure that after a reflection over this line, the angle of the line is the same after reflection. Moreover, it must have the same side length on both sides.

Now, let’s look at each of the shapes. An equilateral triangle has $3$ lines that intersect a corner or a midpoint, so it has at most $3$ symmetry lines. A non-square rhombus only has two symmetry lines, as only the lines that go through the corners work, but not the ones through the midpoints as they would not intersect perpendicularly. A non-square rectangle only has two symmetry lines as it has symmetry lines through the midpoints of opposite sides, but not through the corners since it doesn’t bisect the angle. An isosceles trapezoid only has $1$ symmetry line, that goes through the midpoints of opposite sides. A square has $4$ symmetry lines, which go through opposite corners and opposite midpoints.

A square therefore has the most symmetry lines.

Thus, the answer is E .