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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
−
 
 
  
 
<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 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:
 +
 
 +
*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 1: The point "above" the given diagonal  is (4,-3).
 
  
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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 <math>(0,5)</math>.
  
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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 <math>(0,-5)</math>.
  
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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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*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 (5,0).
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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 8 lattice points on the circle. <math>\Rightarrow \mathrm{(C)}</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==
 
{{Old AMC12 box|year=1995|num-b=19|num-a=21}}
 
{{Old AMC12 box|year=1995|num-b=19|num-a=21}}
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[[Category:Introductory Geometry Problems]]

Revision as of 22:26, 9 January 2008

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

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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