1995 AMC 12 Problem 21

Attempt Problem 21 of the 1995 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 1995 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

Two nonadjacent vertices of a rectangle are (4,3)(4,3) and (−4,−3),(-4,-3), and the coordinates of the other two vertices are integers. The number of such rectangles is
11
22
33
44
55
Answer: E
Concepts:rectangleslattice pointsdiagonals
Small hint:
A rectangle’s diagonals have equal length and the same midpoint
Big hint:
Enumerate integer vectors of length 55 from the origin, identifying antipodal pairs
Solution:
The given diagonal has midpoint (0,0)(0,0) and half-length 5.5. The other diagonal must therefore have endpoints (u,v)(u,v) and (−u,−v)(-u,-v) with u2+v2=25.u^2+v^2=25. Conversely, two equal diagonals with the same midpoint form a rectangle. There are 1212 integer points on this circle: (±5,0),(0,±5),(±3,±4),(±4,±3). \begin{gathered} (\pm5,0),(0,\pm5),\\ (\pm3,\pm4),(\pm4,\pm3). \end{gathered} Antipodal points determine the same diagonal, giving 66 possibilities. One is the given diagonal itself, which is degenerate, so 6−1=56-1=5 rectangles remain. Thus the correct answer is E.
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