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

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[[Image:AIME_1987_Problem_6.png]]
 
[[Image:AIME_1987_Problem_6.png]]
 
== Solution ==
 
== Solution ==
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 are <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>\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>.
  
 
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>.

Revision as of 21:52, 6 March 2007

Problem

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

AIME 1987 Problem 6.png

Solution

Since $XY = WZ$ and $PQ = PQ$ and the areas of the trapezoids $\displaystyle PQZW$ and $\displaystyle PQYX$ 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 (ProblemsAnswer KeyResources)
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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