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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.

## 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

Integers

Trigonometry

Algebra

## Check the Answer

Answer: is 777.

AIME I, 1996, Question 15

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

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

Second Hint

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 $\frac{sin2\theta}{sin\theta}$=$\frac{sin(180-5\theta)}{sin 2\theta}$

$\Rightarrow sin^{2}{2\theta}=sin{5\theta}sin{\theta}$

Final Step

$\Rightarrow 1-cos^{2}2\theta=\frac{cos4\theta-cos6\theta}{2}$

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

1. Arnab Chattopadhyay. says:

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 $\theta$?

2. Arnab Chattopadhyay. says:

Oh! Sorry then $3 \theta > 180^{\circ}$ which cannot be possible.

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