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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 ==
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Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]] 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 are equal to <math>\frac{19}{2}</math>.  Extending <math>PQ</math> to <math>AD</math> and <math>BC</math> at <math>P'</math> and <math>Q'</math>, we split the rectangle into two congruent parts, such that <math>\displaystyle DWPP' + CZQQ' = PQZW</math> (this can be proved through a quick subtraction of areas).<br /><br />
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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 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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Therefore, <math>\frac{1}{2} \cdot \frac{19}{2}(CZ + DW)(AB - PQ) = \frac{1}{2} \cdot \frac{19}{2}(PQ)(AB - (DW + CZ))</math>, which boils down to <math>CZ + DW = 87</math> (the same can reasoning can be repeated for the bottom half to yield <math>AX + BY = 87</math>). Notice that <math>AB = \frac{WZ + XY + (AX + BY)}{2} = \frac{(WD + DA + AX) + (YB + BC + CZ) + (CZ + DW) + (AX + BY)}{2} = \frac{87 \cdot 4 + 19 \cdot 2}{2} = 193</math>.
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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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Solving these two equations gives <math>AB = 193</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:31, 15 February 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 are $\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)$. 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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