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

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===Solution 2===
 
===Solution 2===
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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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>
  
 
===Solution 3===
 
===Solution 3===

Revision as of 21:54, 25 November 2016

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 (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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