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

$\frac{cos96+sin96}{cos96-sin96}$

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

=$\frac{sin186+sin96}{sin186-sin96}$

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

=$\frac{2sin141cos45}{2cos141sin45}$

=tan141

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

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

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

## Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

## Amplitude and Complex numbers – AIME 1996

Let P be the product of the roots of $z^{6}+z^{4}+z^{2}+1=0$ that have a positive imaginary part and suppose that P=r(costheta+isintheta) where $0 \lt r$ and $0 \leq \theta \lt 360$ find $\theta$

• is 107
• is 276
• is 840
• cannot be determined from the given information

### Key Concepts

Equations

Complex Numbers

Integers

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\frac{(z^{5}-1)}{(z-1)}$=$\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0

gives $z^{5}=1 for z\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \neq 1$ gives z=$\frac{1+-(-3)^\frac{1}{2}}{2}$=$cis60,300$ where cis$\theta$=cos$\theta$+isin$\theta$

taking $0 \lt theta \lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276.

.

Categories

## Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

## Roots of Equation and Vieta’s formula – AIME I, 1996

Suppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t.

• is 107
• is 23
• is 840
• cannot be determined from the given information

### Key Concepts

Functions

Roots of Equation

Vieta s formula

AIME I, 1996, Question 5

Polynomials by Barbeau

## Try with Hints

With Vieta s formula

$f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0$

$\Rightarrow a+b+c=-3$, $ab+bc+ca=4$ and $abc=11$

Let a+b+c=-3=p

here t=-(a+b)(b+c)(c+a)

$\Rightarrow t=-(p-c)(p-a)(p-b)$

$\Rightarrow t=-f(p)=-f(-3)$

$t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]$

=23.

Categories

## Tetrahedron Problem | AIME I, 1992 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.

## Tetrahedron Problem – AIME I, 1992

Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron.

• is 107
• is 320
• is 840
• cannot be determined from the given information

### Key Concepts

Area

Volume

Tetrahedron

AIME I, 1992, Question 6

Coordinate Geometry by Loney

## Try with Hints

Area BCD=80=$\frac{1}{2} \times {10} \times {16}$,

where the perpendicular from D to BC has length 16.

The perpendicular from D to ABC is 16sin30=8

[ since sin30=$\frac{perpendicular}{hypotenuse}$ then height = perpendicular=hypotenuse $\times$ sin30 ]

or, Volume=$\frac{1}{3} \times 8 \times 120$=320.

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

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]

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

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

$\Rightarrow CM=\frac{11}{2}$

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

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

Categories

## Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

## Sequence and greatest integer – AIME I, 2000

Let S be the sum of all numbers of the form $\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\frac{S}{10}$.

• is 107
• is 248
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

## Try with Hints

We have 1000=(2)(2)(2)(5)(5)(5) and $\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$

sum of all numbers of form $\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000

=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$

$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$ $\frac{(5^{-3})(5^{7}-1)}{5-1}$

=2480 + $\frac{437}{1000}$

$\Rightarrow [\frac{s}{10}]$=248.

Categories

## Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

## Series and sum – AIME I, 1999

given that $\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}$ where angles are measured in degrees, m and n are relatively prime positive integer that satisfy $\frac{m}{n} \lt 90$, find m+n.

• is 107
• is 177
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Triangles

Side Length

AIME I, 2009, Question 5

Plane Trigonometry by Loney

## Try with Hints

s=$\displaystyle\sum_{k=1}^{35}sin5k$

s(sin5)=$\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\frac{1+cos5}{sin5}$

$=\frac{1-cos(175)}{sin175}$=$tan\frac{175}{2}$ then m+n=175+2=177.

Categories

## LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.

## Lcm and Integer – AIME I, 1998

Find the number of values of k in $12^{12}$ the lcm of the positive integers $6^{6}$, $8^{8}$ and k.

• is 107
• is 25
• is 840
• cannot be determined from the given information

### Key Concepts

Lcm

Algebra

Integers

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

here $k=2^{a}3^{b}$ for integers a and b

$6^{6}=2^{6}3^{6}$

$8^{8}=2^{24}$

$12^{12}=2^{24}3^{12}$

lcm$(6^{6},8^{8})$=$2^{24}3^{6}$

$12^{12}=2^{24}3^{12}$=lcm of $(6^{6},8^{6})$ and k

=$(2^{24}3^{6},2^{a}3^{b})$

=$2^{max(24,a)}3^{max(6,b)}$

$\Rightarrow b=12, 0 \leq a \leq 24$

$\Rightarrow$ number of values of k=25.

Categories

## Integers | AIME I, 1993 Problem | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

## Integer – AIME I, 1993

Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

• is 107
• is 870
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Algebra

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

## Try with Hints

Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)

taking first two solutions a<b<c<d<500

or,$1 \leq c-93, c+1 \leq 499$

or, $94 \leq c \leq 498$ gives 405 solutions

and $1 \leq c-31, c+3 \leq 499$

or, $32 \leq c \leq 496$ gives 465 solutions

or, 405+465=870 solutions.

Categories

## Greatest Positive Integer | AIME I, 1996 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Greatest Positive Integer.

## Positive Integer – AIME I, 1996

For each real number x, Let [x] denote the greatest integer that does not exceed x,find number of positive integers n is it true that $n \lt 1000$ and that $[log_{2}n]$ is a positive even integer.

• is 107
• is 340
• is 840
• cannot be determined from the given information

### Key Concepts

Inequality

Greatest integer

Integers

AIME I, 1996, Question 2

Elementary Number Theory by Sierpinsky

## Try with Hints

here Let $[log_{2}n]$=2k for k is an integer

$\Rightarrow 2k \leq log_{2}n \lt 2k+1$

$\Rightarrow 2^{2k} \leq n \lt 2^{2k+1}$ and $n \lt 1000$

$\Rightarrow 4 \leq n \lt 8$

$16 \leq n \lt 32$

$64 \leq n \lt 128$

$256 \leq n \lt 512$

$\Rightarrow 4+16+64+256$=340.