Question

Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown, where $XY=YB+BC+CZ,$ $YB+BC+CZ=ZW,$ $ZW=WD+DA+AX,$ and $PQ\parallel AB.$ Find $AB$ (in cm) if $BC=19$ cm and $PQ=87$ cm.

Answer

Answer: 193

Concepts:rectangle system of equations trapezoid

Difficulty rating: 2270

Small Hint:

Call the common boundary length $t$ and the rectangle width $L$

Big Hint:

Equal areas above and below $PQ$ place $PQ$ halfway up the rectangle

Solution:

Put $AB=L$ and let the four equal boundary lengths be $t.$ Writing $AX=x$ and $DW=w,$ the left boundary condition gives $x+w=t-19.$ Substituting $XY=ZW=t$ into the right condition gives $2L+38=4t,$ so $t=\frac{L+19}{2}.$

The upper and lower central regions are trapezoids with the same bases $t$ and $87.$ Since their areas are equal, $PQ$ lies halfway up the $19$ -cm rectangle. Each central region therefore has area $\frac{19(t+87)}{4}.$ This is one quarter of the rectangle, $\frac{19L}{4},$ so $t+87=L.$ Combining with $t=\frac{L+19}{2}$ yields $L=193.$

1987 AIME Problem 6

Answer: 193

Difficulty rating: 2270

Small Hint:

Big Hint:

Solution:

Problem 6 in Other Years