# Difference between revisions of "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 areas of the trapezoids $\displaystyle PQZW$ and $\displaystyle PQYX$ 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 $AB = 193$.