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May 11, 2020

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

Key Concepts


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

Parallelogram Problem

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. 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 $\theta$?

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