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# Parallelogram Problem | AIME I, 1996 | Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Parallelogram Problem.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Parallelogram.

## Parallelogram Problem – AIME I, 1996

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.

• is 107
• is 777
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Trigonometry

Algebra

AIME I, 1996, Question 15

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

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.

## 2 replies on “Parallelogram Problem | AIME I, 1996 | Question 15”

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.