1987 AIME Problems/Problem 6

Revision as of 21:52, 6 March 2007 by Mathgeek2006 (talk | contribs) (Solution)

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.

AIME 1987 Problem 6.png

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 $AB = 193$.

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
Invalid username
Login to AoPS