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 2006. 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 2006=[P]=FX⋅YZ=zx, so z=2006/x. Thus
There are ⌊8023⌋=89 positive values of x that yield a positive square for the radicand, so there are 89 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,W=Y, let α=m∠AYX, let β=m∠XYW, and let z be the length of the sides of the rhombuses. Then β+2α=180∘, and the area of each of the four rhombuses is z2sinα=2006. Therefore
K=z2sinβ=z2sin2α=2z2sinαcosα=22006cosα.
Thus 1≤K<22006=8024, so 1≤K≤89, and there are 89 possible values for K.
