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.
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
Publishing original scientific research in a peer-reviewed journal is an achievement many scholars work toward for years. Achieving this feat early in one's academic journey requires exceptional intellectual rigor, creativity, and persistence. Aaditya Punatar and Muskan Sadana have reached this remarkable milestone, successfully publishing their collaborative research paper in the prestigious Journal of Emerging Investigators […]
Her project on quantum-accelerated Steiner trees for 15-minute cities earned a place among the finalists at the world's largest pre-college STEM competition. Prishaa Shrimali of Marymount School of New York, was selected as a Grand Finalist at the 2026 Regeneron International Science and Engineering Fair (ISEF) in Phoenix, Arizona. Her project was Optimizing Urban Accessibility: […]
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's M.Stat Entrance. They ranked within the first 50 in the entire country in these entrances. I.S.I. M.Stat Entrance
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's B.Stat Entrance and Chennai Mathematical Institute's B.Sc. Math Entrance. They ranked within the first 200 in the entire country in these entrances. Most of these students attended the problem solving workshops regularly, which happen 5 days every week. CMI B.Sc. Math Entrance […]
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.