1987 AIME Problems/Problem 6

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.

Solution

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 $AB=53*2+87=193$

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
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1987 AIME Problems/Problem 6

Contents

Problem

Solution

Solution 2

See also