1987 AIME Problems/Problem 6

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.

Solution

Since $XY = WZ$ and $PQ = PQ$ and the area of the trapezoids $\displaystyle PQZW$ and $\displaystyle PQYX$ are the same, the heights of the trapezoids are the same, or $\frac{19}{2}$ . Extending $PQ$ to $AD$ and $BC$ at $P'$ and $Q'$ , we split the rectangle into two congruent parts, such that $\displaystyle DWPP' + CZQQ' = PQZW$ (this can be proved through a quick subtraction of areas).

Therefore, $\frac{1}{2} \cdot \frac{19}{2}(CZ + DW)(AB - PQ) = \frac{1}{2} \cdot \frac{19}{2}(PQ)(AB - (DW + CZ))$ , which boils down to $CZ + DW = 87$ (the same can reasoning can be repeated for the bottom half to yield $AX + BY = 87$ ). Notice that $AB = \frac{WZ + XY + (AX + BY)}{2} = \frac{(WD + DA + AX) + (YB + BC + CZ) + (CZ + DW) + (AX + BY)}{2} = \frac{87 \cdot 4 + 19 \cdot 2}{2} = 193$ .

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

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