1956 AMC 12 Problem 29
Attempt Problem 29 of the 1956 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 1956 AMC 12 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
29.
The points of intersection of and are joined in succession. The resulting figure is:
a straight line
an equilateral triangle
a parallelogram
a rectangle
a square
Answer: D
Small hint:
Use to find the possible sums
Big hint:
The four intersection points are permutations and negatives of
Solution:
At an intersection, so Together with this gives the four points In their cyclic order around the circle, adjacent side vectors are perpendicular, so the quadrilateral is a rectangle. Its adjacent side lengths are unequal, so it is not a square.
Thus, the correct answer is D.
Problem 29 in other years
- 1950 AMC 12
- 1951 AMC 12
- 1952 AMC 12
- 1953 AMC 12
- 1954 AMC 12
- 1955 AMC 12
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