1953 AMC 12 第 32 题
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所有题目均经美国数学协会（MAA）官方合法授权使用。
32.
将一个矩形的每个角三等分。与同一条边相邻的两条三等分线分别相交，所得四个交点总是构成：
Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
正方形
a square
矩形
a rectangle
邻边不等的平行四边形
a parallelogram with unequal sides
菱形
a rhombus
没有特殊性质的四边形
a quadrilateral with no special properties
答案：D
小提示：
利用矩形的水平和竖直对称轴
Use the horizontal and vertical symmetry axes of the rectangle
大提示：
四个交点所成四边形的两条对角线互相垂直且互相平分
The four intersection points have perpendicular diagonals that bisect each other
解答：
对于矩形的每一条边，与它相邻的两条三等分线在该边的垂直平分线上相交。这样得到的相对交点关于矩形中心对称，因此所得四边形的两条对角线互相平分。一条对角线位于水平对称轴上，另一条位于竖直对称轴上，所以它们互相垂直。
对角线互相平分的四边形是平行四边形；若其对角线还互相垂直，则四条边相等。因此这个四边形是菱形。由于两条对角线不一定等长，它不一定是正方形。
因此，正确答案是 D。
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.
其他年份的第 32 题
1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12 · 1968 AMC 12 · 1969 AMC 12 · 1970 AMC 12 · 1971 AMC 12 · 1972 AMC 12 · 1973 AMC 12