Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Parallelogram.
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.
Integers
Trigonometry
Algebra
Answer: is 777.
AIME I, 1996, Question 15
Geometry Vol I to IV by Hall and Stevens
Let \(\theta= \angle DBA\)
\(\angle CAB=\angle DBC=2 \theta\)
or, \(\angle AOB=180-3\theta, \angle ACB=180-5\theta\)
or, since ABCD parallelogram, OA=OC
by sine law on \(\Delta\)ABO, \(\Delta\)BCO
\(\frac{sin\angle CBO}{OC}\)=\(\frac{sin\angle ACB}{OB}\)
and \(\frac{sin\angle DBA}{OC}=\frac{sin\angle BAC}{OB}\)
here we divide and get \(\frac{sin2\theta}{sin\theta}\)=\(\frac{sin(180-5\theta)}{sin 2\theta}\)
\(\Rightarrow sin^{2}{2\theta}=sin{5\theta}sin{\theta}\)
\(\Rightarrow 1-cos^{2}2\theta=\frac{cos4\theta-cos6\theta}{2}\)
or, \(4 cos^{3}2\theta-4cos^{2}2\theta -3cos2\theta+3=(4cos^{2}2\theta-3)(cos2\theta-1)=0 [using cos3\theta=4cos^{3}\theta-3cos\theta]\)
or, \(cos 2\theta=\frac{\sqrt{3}}{2}\)
or, \(\theta\)=15
\([1000r]=[1000\frac{180-5\theta}{180-3\theta}]=[\frac{7000}{9}]\)=777.
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Why does $\cos^2 2 \theta = \frac 3 4 \implies \cos 2 \theta = \frac {\sqrt 3} {2}$? It may so happen that $\cos 2 \theta = - \frac {\sqrt 3} {2}$ in which case $2 \theta = 150^{\circ} \implies \theta = 75^{\circ}.$ Why don't we take this value of $\theta$?
Oh! Sorry then $3 \theta > 180^{\circ}$ which cannot be possible.