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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 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 of the trapezoids <math>\displaystyle PQZW</math> and <math>\displaystyle PQYX</math> are the same, the heights of the trapezoids are the same, or <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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===Solution 1===
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Since <math>XY = WZ</math>, <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>PQZW</math> and <math>PQYX</math> are the same, then 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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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 = \boxed{193}</math>.
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===Solution 2===
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Let <math>YB=a</math>, <math>CZ=b</math>, <math>AX=c</math>, and <math>WD=d</math>. First we drop a perpendicular from <math>Q</math> to a point <math>R</math> on <math>BC</math> so <math>QR=h</math>. 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 both <math>\frac{19}{2}</math>.From here, we have that <math>[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}</math>. We are told that this area is equal to <math>[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}</math>. Setting these equal to each other and solving gives <math>h=53</math>. In the same way, we find that the perpendicular from <math>P</math> to <math>AD</math> is <math>53</math>. So <math>AB=53*2+87=\boxed{193}</math>
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===Solution 3===
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Since <math>XY = YB + BC + CZ = ZW = WD + DA + AX</math>. Let <math>a = AX + DW = BY + CZ</math>. Since <math>2AB - 2a = XY = WZ</math>, then <math>XY = AB - a</math>.Let <math>S</math> be the midpoint of <math>DA</math>, and <math>T</math> be the midpoint of <math>CB</math>. Since the area of <math>PQWZ</math> and <math>PQYX</math> are the same, then their heights are the same, and so <math>PQ</math> is [[equidistant]] from <math>AB</math> and <math>CD</math>. This means that <math>PS</math> is perpendicular to <math>DA</math>, and <math>QT</math> is perpendicular to <math>BC</math>. Therefore, <math>PSCW</math>, <math>PSAX</math>, <math>QZCT</math>, and <math>QYTB</math> are all trapezoids, and <math>QT = (AB - 87)/</math>2. This implies that <cmath>((a + 2((AB - 87)/2)/2) \cdot 19 = (((AB - a) + 87)/2) \cdot 19</cmath> <cmath>(a + AB - 87) = AB - a + 87</cmath> <cmath>2a = 174</cmath> <cmath>a = 87</cmath> Since <math>a + CB = XY</math>, <math>XY = 19 + 87 = 106</math>, and <math>AB = 106 + 87 = \boxed{193}</math>.
  
 
== 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]]
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{{MAA Notice}}

Latest revision as of 06:46, 1 June 2018

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

Solution 1

Since $XY = WZ$, $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, then 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 $AB = \boxed{193}$.


Solution 2

Let $YB=a$, $CZ=b$, $AX=c$, and $WD=d$. First we drop a perpendicular from $Q$ to a point $R$ on $BC$ so $QR=h$. Since $XY = WZ$ and $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, the heights of the trapezoids are both $\frac{19}{2}$.From here, we have that $[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}$. We are told that this area is equal to $[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}$. Setting these equal to each other and solving gives $h=53$. In the same way, we find that the perpendicular from $P$ to $AD$ is $53$. So $AB=53*2+87=\boxed{193}$

Solution 3

Since $XY = YB + BC + CZ = ZW = WD + DA + AX$. Let $a = AX + DW = BY + CZ$. Since $2AB - 2a = XY = WZ$, then $XY = AB - a$.Let $S$ be the midpoint of $DA$, and $T$ be the midpoint of $CB$. Since the area of $PQWZ$ and $PQYX$ are the same, then their heights are the same, and so $PQ$ is equidistant from $AB$ and $CD$. This means that $PS$ is perpendicular to $DA$, and $QT$ is perpendicular to $BC$. Therefore, $PSCW$, $PSAX$, $QZCT$, and $QYTB$ are all trapezoids, and $QT = (AB - 87)/$2. This implies that \[((a + 2((AB - 87)/2)/2) \cdot 19 = (((AB - a) + 87)/2) \cdot 19\] \[(a + AB - 87) = AB - a + 87\] \[2a = 174\] \[a = 87\] Since $a + CB = XY$, $XY = 19 + 87 = 106$, and $AB = 106 + 87 = \boxed{193}$.

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
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Problem 7
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