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Difference between revisions of "1995 AHSME Problems/Problem 21"
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==Problem==
 
==Problem==
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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
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Two nonadjacent vertices of a [[rectangle]] are <math>(4,3)</math> and <math>(-4,-3)</math>, and the [[coordinate]]s of the other two vertices are integers. The number of such rectangles is
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<math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }  </math>
 
<math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }  </math>
  
 
==Solution==
 
==Solution==
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The distance between (4,3) and (-4,-3) is <math>\sqrt{6^2+8^2}=10</math>. 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:
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The center of the rectangle is <math>(0,0)</math>, and the distance from the center to a corner is <math>\sqrt{4^2+3^2}=5</math>. 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 <math>(\pm x,\pm y)</math>, where <math>x^2+y^2=25</math>. This equation has six pairs of integer solutions: <math>(\pm 4, \pm 3)</math>, <math>(\pm 4, \mp 3)</math>, <math>(\pm 3, \pm 4)</math>, <math>(\pm 3, \mp 4)</math>, <math>(\pm 5, 0)</math>, and <math>(0, \pm 5)</math>. 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. <math>\Rightarrow \mathrm{(E)}</math>
  
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Case 1: The point "above" the given diagonal  is (4,-3).
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==See also==
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{{AHSME box|year=1995|num-b=20|num-a=22}}
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Then the point "below" the given diagonal is (-4,3).
 
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Case 2: The point "above" the given diagonal is (0,5).
 
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Then the point "below" the given diagonal is (0,-5).
 
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Case 3: The point "above" the given diagonal is (-5,0).
 
  
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Then the point "below" the given diagonal is (5,0).
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}
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We have only three cases since there are 8 lattice points on the circle. <math>\Rightarrow \mathrm{(C)}</math>
 
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==See also==
 
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{{Old AMC12 box|year=1995|num-b=19|num-a=21}}
 

Latest revision as of 17:30, 4 December 2020

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 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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