Problem:
Hexagon ABCDEF is divided into five rhombuses, P,Q,R,S, and T, as shown. Rhombuses P,Q,R, and S are congruent, and each has area 2006β. Let K be the area of rhombus T. Given that K is a positive integer, find the number of possible values for K.
Solution:
Draw the diagonals of rhombus T. Let Z be their point of intersection, and let X and Y be the shared vertices of rhombuses P and T, with Y on AB. Let YZ=x and XY=z. Consequently 2006β=[P]=FXβ YZ=zx, so z=2006β/x. Thus
There are β8023ββ=89 positive values of x that yield a positive square for the radicand, so there are 89β possible values for K.
OR
Define X,Y, and Z as in the first solution, and let W be the shared vertex of rhombuses Q and T,Wξ =Y, let Ξ±=mβ AYX, let Ξ²=mβ XYW, and let z be the length of the sides of the rhombuses. Then Ξ²+2Ξ±=180β, and the area of each of the four rhombuses is z2sinΞ±=2006β. Therefore
K=z2sinΞ²=z2sin2Ξ±=2z2sinΞ±cosΞ±=22006βcosΞ±.
Thus 1β€K<22006β=8024β, so 1β€Kβ€89, and there are 89β possible values for K.
