Problem:
In parallelogram ABCD, let O be the intersection of diagonals AC and BD. Angles CAB and DBC are each twice as large as angle DBA, and angle ACB is r times as large as angle AOB. Find the greatest integer that does not exceed 1000r.
Solution:
The given data allows us to label x=OA=OC,y=OB, θ=∠OBA, and 2θ=∠OAB=∠OBC. Because angles CBO and CAB are congruent, triangles BCO and ACB are similar. Thus
CACB=CBCO=BAOB
or
2xCB=CBx=BAy
It follows that CB=x2, and that BA=y2. Now let P be the intersection of OB with the bisector of ∠OAB. Because angles OAP and OBA are congruent, triangles OPA and OAB are similar. Thus
BAAP=OBOA=OAOP
The first equation yields AP=x2. Because AP=PB, the second equation yields
y2−x2=xy2(1)
Apply the Law of Cosines to triangle COB to find that
cos3θ=2xyx2+y2−(x2)2=2xyy2−x2
which is 2/2, by equation (1). In other words, 3θ=45∘, so θ=15∘. It follows that ∠ACB=105∘ and ∠AOB=135∘, so r=105/135=7/9=0.7 and 1000r=777.7.
OR
Apply the Law of Sines to triangles BOC and ABC to find that
BC=sin2θOCsin3θ and BC=sin3θACsin2θ
respectively. Because 2⋅OC=AC, it follows that sin23θ=2sin22θ. Now use the identities sin2θ=2sinθcosθ and sin3θ=sinθ(4cos2θ−1) to produce the equation (4cos2θ−1)2=8cos2θ, then use the identity 2cos2θ=1+cos2θ to reduce it to (1+2cos2θ)2=4+4cos2θ. This is equivalent to 4cos22θ=3, hence cos2θ=±213. Because it is clear that 2θ is acute, only θ=15∘ is a possibility. Thus r=105/135=7/9 and 1000r=777.7.
