1953 AMC 12 Problem 32
Attempt Problem 32 of the 1953 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1953 AMC 12 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
32.
Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
a square
a rectangle
a parallelogram with unequal sides
a rhombus
a quadrilateral with no special properties
Answer: D
Small hint:
Use the horizontal and vertical symmetry axes of the rectangle
Big hint:
The four intersection points have perpendicular diagonals that bisect each other
Solution:
For each side, the two trisectors adjacent to it meet on that side’s perpendicular bisector. Opposite such intersection points are reflections across the center of the rectangle, so the diagonals of the resulting quadrilateral bisect each other. One diagonal lies on the horizontal symmetry axis and the other on the vertical symmetry axis, so they are perpendicular.
A quadrilateral whose diagonals bisect each other is a parallelogram; if those diagonals are perpendicular, its four sides are equal. Hence the quadrilateral is a rhombus. It need not be a square because the two diagonals need not have equal length.
Thus, the correct answer is D.