1995 AHSME Problems/Problem 21

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Problem

Two nonadjacent vertices of a rectangle are (4,3) and (-4,-3), and the coordinates of the other two vertices are integers. The number of such rectangles is


$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }$

Solution

The distance between (4,3) and (-4,-3) is $\sqrt{6^2+8^2}=10$. Therefore, if you circumscribe a circle around the rectangle, it has a center of (0,0) with a radius of 10/2=5. There are three cases:

Case 1: The point "above" the given diagonal is (4,-3).

Then the point "below" the given diagonal is (-4,3).

Case 2: The point "above" the given diagonal is (0,5).

Then the point "below" the given diagonal is (0,-5).

Case 3: The point "above" the given diagonal is (-5,0).

Then the point "below" the given diagonal is (5,0).


We have only three cases since there are 8 lattice points on the circle. $\Rightarrow \mathrm{(C)}$

See also