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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]]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>.
+
Since <math>XY = WZ</math> and <math>PQ = PQ</math> then 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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==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=193</math>
 
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=193</math>

Revision as of 14:10, 8 September 2013

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 $PQ = PQ$ then 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 $AB = 193$.


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 $AB=53*2+87=193$

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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