Categories

## 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

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]

$\Rightarrow x=159$.

Categories

## 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.

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

### Key Concepts

Integers

Triangle

Trigonometry

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.

Categories

## 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

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.

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

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## 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

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, ${\sin^2 \theta} \geq \frac{1}{2}$ $\Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4}$.

Categories

## 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

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 $0< x < \frac{\pi}{2 }$ .

Categories

## 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

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.

Categories

## 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

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=$\frac{60\pi}{3}$ $\Rightarrow n=60$.

Categories

## 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

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.

Categories

## 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

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

$\Rightarrow 13+4+10=27$.

Categories

## 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

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)