1995 AHSME Problems/Problem 21

Revision as of 21:06, 20 January 2010 by Beej175560 (talk | contribs) (Solution)

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 center of the rectangle is $(0,0)$, and the distance from the center to a corner is $\sqrt{4^2+3^2}=5$. 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 $(\pm x,\pm y)$, where $x^2+y^2=25$. This equation has six pairs of integer solutions: $(\pm 4, \pm 3)$, $(\pm 4, \mp 3)$, $(\pm 3, \pm 4)$, $(\pm 3, \mp 4)$, $(\pm 5, 0)$, and $(0, \pm 5)$. 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. $\Rightarrow \mathrm{(E)}$

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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