Maths Olympiad Prep

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Geometry Difficulty 3.2 AMC 10/12 Find the answer

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
$\mathrm{

Pick one

Solution

The center of the rectangle is (0,0)(0,0), and the distance from the center to a corner is 42+32=5\sqrt{4^2+3^2}=5. The remaining two vertices of the rectangle must be another pair of points opposite each other on the circle of radius 5 centered at the origin. Let these points have the form (±x,±y)(\pm x,\pm y), where x2+y2=25x^2+y^2=25. This equation has six pairs of integer solutions: (±4,±3)(\pm 4, \pm 3), (±4,3)(\pm 4, \mp 3), (±3,±4)(\pm 3, \pm 4), (±3,4)(\pm 3, \mp 4), (±5,0)(\pm 5, 0), and (0,±5)(0, \pm 5). The first pair of solutions are the endpoints of the given diagonal, and the other diagonal must span one of the other five pairs of points. (E)\Rightarrow \mathrm{(E)}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.