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Difference between revisions of "1987 AIME Problems/Problem 6"

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== Problem ==
 
== Problem ==
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[[Rectangle]] <math>\displaystyle ABCD</math> is divided into four parts of equal [[area]] by five [[line segment | segments]] as shown in the figure, where <math>\displaystyle XY = YB + BC + CZ = ZW = WD + DA + AX</math>, and <math>\displaystyle PQ</math> is [[parallel]] to <math>\displaystyle AB</math>.  Find the [[length]] of <math>\displaystyle AB</math> (in cm) if <math>\displaystyle BC = 19</math> cm and <math>\displaystyle PQ = 87</math> cm.
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[[Rectangle]] <math>ABCD</math> is divided into four parts of equal [[area]] by five [[line segment | segments]] as shown in the figure, where <math>XY = YB + BC + CZ = ZW = WD + DA + AX</math>, and <math>PQ</math> is [[parallel]] to <math>AB</math>.  Find the [[length]] of <math>AB</math> (in cm) if <math>BC = 19</math> cm and <math>PQ = 87</math> cm.
  
 
[[Image:AIME_1987_Problem_6.png]]
 
[[Image:AIME_1987_Problem_6.png]]
 
== Solution ==
 
== Solution ==
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Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>\displaystyle PQZW</math> and <math>\displaystyle PQYX</math> are the same, the heights of the trapezoids are the same.  Thus both trapezoids have area <math>\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)</math>.  This number is also equal to one quarter the area of the entire rectangle, which is <math>\frac{19\cdot AB}{4}</math>, so we have <math>AB = XY + 87</math>.
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Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>PQZW</math> and <math>PQYX</math> are the same, the heights of the trapezoids are the same.  Thus both trapezoids have area <math>\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)</math>.  This number is also equal to one quarter the area of the entire rectangle, which is <math>\frac{19\cdot AB}{4}</math>, so we have <math>AB = XY + 87</math>.
  
 
In addition, we see that the [[perimeter]] of the rectangle is <math>2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY</math>, so <math>AB + 19 = 2XY</math>.
 
In addition, we see that the [[perimeter]] of the rectangle is <math>2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY</math>, so <math>AB + 19 = 2XY</math>.
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== See also ==
 
== See also ==
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* [[1987 AIME Problems]]
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{{AIME box|year=1987|num-b=5|num-a=7}}
  
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{{AIME box|year=1987|num-b=5|num-a=7}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 19:08, 23 October 2007

Problem

Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$, and $PQ$ is parallel to $AB$. Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm.

AIME 1987 Problem 6.png

Solution

Since $XY = WZ$ and $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, the heights of the trapezoids are the same. Thus both trapezoids have area $\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)$. This number is also equal to one quarter the area of the entire rectangle, which is $\frac{19\cdot AB}{4}$, so we have $AB = XY + 87$.

In addition, we see that the perimeter of the rectangle is $2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY$, so $AB + 19 = 2XY$.

Solving these two equations gives $AB = 193$.

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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