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Difference between revisions of "1995 AHSME Problems/Problem 21"

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==Solution==
 
==Solution==
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The distance between <math>(4,3)</math> and <math>(-4,-3)</math> is <math>\sqrt{6^2+8^2}=10</math>. Therefore, if you circumscribe a circle around the rectangle, it has a center of <math>(0,0)</math> with a [[radius]] of <math>10/2=5</math>. 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>(x,y)</math> and <math>(-x,-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{(C)}</math>
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*Case 1: The point "above" the given diagonal  is <math>(4,-3)</math>.
 
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Then the point "below" the given diagonal is <math>(-4,3)</math>.
 
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*Case 2: The point "above" the given diagonal is <math>(0,5)</math>.
 
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Then the point "below" the given diagonal is <math>(0,-5)</math>.
 
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*Case 3: The point "above" the given diagonal is <math>(-5,0)</math>.
 
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Then the point "below" the given diagonal is <math>(5,0)</math>.
 
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We have only three cases since there are <math>8</math> lattice points on the circle. <math>\Rightarrow \mathrm{(C)}</math>
 
  
 
==See also==
 
==See also==

Revision as of 20:51, 20 January 2010

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 $(x,y)$ and $(-x,-y)$, 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. $\Rightarrow \mathrm{(C)}$

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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