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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Finding smallest positive Integer – AIME I, 1996


Find the smallest positive integer solution to \(tan19x=\frac{cos96+sin96}{cos96-sin96}\).

  • is 107
  • is 159
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Trigonometry

Integers

Check the Answer


Answer: is 159.

AIME I, 1996, Question 10

Plane Trigonometry by Loney

Try with Hints


First hint

\(\frac{cos96+sin96}{cos96-sin96}\)

=\(\frac{sin(90+96)+sin96}{sin(90+96)-sin96}\)

=\(\frac{sin186+sin96}{sin186-sin96}\)

=\(\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}\)

=\(\frac{2sin141cos45}{2cos141sin45}\)

=tan141

Second Hint

here \(tan(180+\theta)\)=\(tan\theta\)

\(\Rightarrow 19x=141+180n\) for some integer n is first equation

Final Step

multiplying equation with 19 gives

\(x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)\) [since 2679 divided by 180 gives remainder 159]

\(\Rightarrow x=159\).

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

Triangle and integers – AIME I, 1995


Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and \(\angle BDC\)=3\(\angle BAC\). then the perimeter of \(\Delta ABC\) may be written in the form \(a+\sqrt{b}\) where a and b are integers, find a+b.

Triangle and integers
  • is 107
  • is 616
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Triangle

Trigonometry

Check the Answer


Answer: is 616.

AIME I, 1995, Question 9

Plane Trigonometry by Loney

Try with Hints


First hint

Let x= \(\angle CAM\)

\(\Rightarrow \angle CDM =3x\)

\(\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}\)=11 [by trigonometry ratio property in right angled triangle]

Second Hint

\(\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx\)

solving we get, tanx=\(\frac{1}{2}\)

\(\Rightarrow CM=\frac{11}{2}\)

Final Step

\(\Rightarrow 2(AC+CM)\) where \(AC=\frac{11\sqrt {5}}{2}\) by Pythagoras formula

=\(\sqrt{605}+11\) then a+b=605+11=616.

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I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB Trigonometry

ISI MStat 2019 PSA Problem 15 | Trigonometry Problem

This is the problem from ISI MStat 2019 PSA Problem 15. First, try it yourself and then go through the sequential hints we provide.

Trigonometry – ISI MStat Year 2019 PSA Question 15


How many solutions does the equation \( cos ^{2} x+3 \sin x \cos x+1=0\) have for \( x \in[0,2 \pi) \) ?

  • 1
  • 3
  • 4
  • 2

Key Concepts


Trigonometry

Factorization

Check the Answer


Answer: is 4

ISI MStat 2019 PSA Problem 15

Precollege Mathematics

Try with Hints


Factorize and Solve.

\(\cos ^{2} x+3 \sin x \cos x + 1 = (2\cos x + \sin x)(\cos x +\sin x) = 0 \).
\( tanx = -2, tanx = -1 \).
Draw the graph.

Trigonometry problem graph - ISI MStat 2019 PSA Problem 15
Fig:1

So, if you see the figure you will find there are 4 such x for \( x \in[0,2 \pi) \).

Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Coordinate Geometry I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

Equation of a circle- ISI MStat Year 2016 PSA Question 9


Given \( \theta \) in the range \( 0 \leq \theta<\pi,\) the equation \( 2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0\) represents a circle for all \( \theta\) in the interval

  • \( 0 < \theta <\frac{\pi}{3} \)
  • \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)
  • \( 0 < \theta <\frac{\pi}{2} \)
  • \( 0 \le \theta <\frac{\pi}{2} \)

Key Concepts


Equation of a circle

Trigonometry

Basic Inequality

Check the Answer


Answer: is \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

Try with Hints


Complete the Square.

We get ,

\(2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3)) \)
\(6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2} \)

We are given that \( 0 \leq \theta<\pi,\) . So, \( {\sin^2 \theta} \geq \frac{1}{2} \) \( \Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4} \).

ISI MStat 2016 PSA Problem 9
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Algebra I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB Trigonometry

ISI MStat 2019 PSA Problem 12 | Domain of a function

This is a beautiful problem from ISI MStat 2019 PSA problem 12 based on finding the domain of the function. We provide sequential hints so that you can try.

Domain of a function- ISI MStat Year 2019 PSA Question 12


What is the set of numbers \(x\) in \( (0,2 \pi)\) such that \(\log \log (\sin x+\cos x)\) is well-defined?

  • \( [\frac{\pi}{8},\frac{3 \pi}{8}] \)
  • \( (0,\frac{\pi}{2}) \)
  • \( (0,\frac{ \pi}{4}] \)
  • \( (0,\pi) \cup (\frac{3 \pi}{2}, 2 \pi) \)

Key Concepts


Domain

Basic inequality

Trigonometry

Check the Answer


Answer: is \( (0,\frac{\pi}{2}) \)

ISI MStat 2019 PSA Problem 12

Pre-college Mathematics

Try with Hints


\(logx\) is defined for \( x \in (0,\infty)\).

\(sinx+cosx > 0\).
\(log(sinx+cosx) > 0 \Rightarrow sinx + cosx > 1\)
\( sin(x+\frac{\pi}{4}) > \frac{1}{\sqrt{2}}\)
For \(y\) in \( (0,2 \pi)\) , \(siny > \frac{1}{\sqrt{2}} \iff \frac{\pi}{4} < y < \frac{3\pi}{4 } \)

Hence we have \( 0< x < \frac{\pi}{2 } \) .

ISI MStat 2019 PSA Problem 12
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

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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Algebra Arithmetic Math Olympiad PRMO

Trigonometry & natural numbers | PRMO 2017 | Question 11

Try this beautiful problem from the Pre-RMO, 2017 based on Trigonometry & natural numbers.

Trigonometry & natural numbers – PRMO 2017


Let f(x) =\(sin\frac{x}{3}+cos\frac{3x}{10}\) for all real x, find the least natural number x such that \(f(n\pi+x)=f(x)\) for all real x.

  • is 107
  • is 60
  • is 840
  • cannot be determined from the given information

Key Concepts


Trigonometry

Least natural number

Functions

Check the Answer


Answer: is 60.

PRMO, 2017, Question 11

Plane Trigonometry by Loney

Try with Hints


First hint

here f(x) =\(sin\frac{x}{3}+cos\frac{3x}{10}\)

Second Hint

period of\(sin\frac{x}{3}\) is \(6\pi\)

period of \(cos\frac{3x}{10}\) is \(\frac{20\pi}{3}\)

Final Step

Lcm=\(\frac{60\pi}{3}\) \(\Rightarrow n=60\).

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Trigonometry and greatest integer | AIME I, 1997 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1997 based on Trigonometry and greatest integer.

Trigonometry and greatest integer – AIME I, 1997


Let x=\(\frac{\displaystyle\sum_{n=1}^{44}cos n}{\displaystyle\sum_{n=1}^{44}sin n}\), find greatest integer that does not exceed 100x.

  • is 107
  • is 241
  • is 840
  • cannot be determined from the given information

Key Concepts


Trigonometry

Greatest Integer

Algebra

Check the Answer


Answer: is 241.

AIME I, 1997, Question 11

Plane Trigonometry by Loney

Try with Hints


First hint

here \(\displaystyle\sum_{n=1}^{44}cosn+\displaystyle\sum_{n=1}^{44}sin n\)

=\(\displaystyle\sum_{n=1}^{44}sinn+\displaystyle\sum_{n=1}^{44}sin(90-n)\)

=\(2^\frac{1}{2}\displaystyle\sum_{n=1}^{44}cos(45-n)\)

=\(2^\frac{1}{2}\displaystyle\sum_{n=1}^{44}cosn\)

Second Hint

\(\displaystyle\sum_{n=1}^{44}sin n=(2^\frac{1}{2}-1)\displaystyle\sum_{n=1}^{44}cosn\)

\(\Rightarrow x=\frac{1}{2^\frac{1}{2}-1}\)

\(\Rightarrow x= 2^\frac{1}{2}+1\)

Final Step

\(\Rightarrow 100x=(100)(2^\frac{1}{2}+1)\)=241.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Trigonometry and positive integers | AIME I, 1995 | Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Trigonometry and positive integers.

Trigonometry and positive integers – AIME I, 1995


Given that (1+sint)(1+cost)=\(\frac{5}{4}\) and (1-sint)(1-cost)=\(\frac{m}{n}-k^\frac{1}{2}\) where k,m,n are positive integers with m and n relatively prime, find k+m+n.

  • is 107
  • is 27
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Algebra

Trigonometry

Check the Answer


Answer: is 27.

AIME I, 1995, Question 7

Plane Trigonometry by Loney

Try with Hints


First hint

Let (1-sint)(1-cost)=x

\(\Rightarrow \frac{5x}{4}=(1-sin^{2}t)(1-cos^{2}t)=sin^{2}tcos^{2}t\)

\(\Rightarrow\frac{(5x)^\frac{1}{2}}{2}=sintcost\)

from given equation sint+cost=\(\frac{5}{4}-(sin^{2}t+cos^{2}t)-sintcost\)=\(\frac{1}{4}-\frac{(5x)^\frac{1}{2}}{2}\)

Second Hint

again (1+sint)(1+cost)-2sintcost=x

\(\Rightarrow x=\frac{5}{4}-2(\frac{1}{4}-\frac{(5x)^\frac{1}{2}}{2})=\frac{3}{4}+(5x)^\frac{1}{2}\)

\(\Rightarrow (x-\frac{3}{4})^{2}=5x\)

\(\Rightarrow x=\frac{13}{4}-(10)^\frac{1}{2}\) for \(x \geq 0\)

Final Step

\(\Rightarrow 13+4+10=27\).

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Singapore Math Olympiad

Trigonometry Problem from SMO, 2008 | Problem No.17

Try this beautiful Problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

Problem – Trigonometry (SMO Test)


Find the value of \((log_{\sqrt 2}(cos 20^\circ) + log_{\sqrt 2} (cos 40^\circ) + log_{\sqrt 2}(cos 80^\circ))^2\)

  • 32
  • 15
  • 36
  • 20

Key Concepts


Trigonometry

Log Function

Check the Answer


Answer: 36

Singapore Mathematical Olympiad, 2008

Challenges and Thrills – Pre – College Mathematics

Try with Hints


This one is a very simple. We can start from here :

As all are in the function of log with \(\sqrt 2\) as base so we can take it as common such that

\(log_{\sqrt 2}(cos 20 ^\circ . cos 40 ^\circ . cos 80^\circ)\)

Now as you can see we dont know the exact value of \(cos 20^\circ\) or \(cos 40^\circ\) or \(cos 80^\circ\) values.

But theres a formula that we can use which is

cosA.cos B = \(\frac {1}{2} (cos (A+B) + cos (A-B))\)

Now try apply this formula in the above expression and try to solve………

Now those who did not get the answer yet try this:

If we apply the formula in the expression mentioned in the last hint :

\(log_{\sqrt 2}(cos 20 ^\circ . \frac {1}{2}(cos 120 ^\circ . cos 40^\circ)\)

\(log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{2} cos 40 ^\circ . cos 20^\circ)\)

\(log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{4} (cos 60 ^\circ + cos 20^\circ)\)

We need to do the rest of the calculation.Try to do that …………………..

Continue from the last hint:

\(log_{\sqrt 2} \frac {1}{8} = – \frac {log_{2} 8}{log_{2}(2^{\frac {1}{2}})} = -6 \)

So squaring this answer = \((-6)^2 = 36\) ……………………..(Answer)

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