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May 11, 2020

How to Pursue Mathematics after High School?

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

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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What to do to shape your Career in Mathematics after 12th? 

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

  • What are some of the best colleges for Mathematics that you can aim to apply for after high school?
  • How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
  • What are the best universities for MS, MMath, and Ph.D. Programs in India?
  • What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
  • How can you pursue a Ph.D. in Mathematics outside India?
  • What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

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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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