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

Functions

Trigonometry

Integers

## Check the Answer

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

## 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

Integers

Triangle

Trigonometry

## Check the Answer

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

## 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

Integers

Trigonometry

Algebra

## Check the Answer

AIME I, 1996, Question 15

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

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

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 $\frac{sin2\theta}{sin\theta}$=$\frac{sin(180-5\theta)}{sin 2\theta}$

$\Rightarrow sin^{2}{2\theta}=sin{5\theta}sin{\theta}$

$\Rightarrow 1-cos^{2}2\theta=\frac{cos4\theta-cos6\theta}{2}$

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 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

Trigonometry

Greatest Integer

Algebra

## Check the Answer

AIME I, 1997, Question 11

Plane Trigonometry by Loney

## Try with Hints

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

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

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

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

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

$\Rightarrow x=\frac{1}{2^\frac{1}{2}-1}$

$\Rightarrow x= 2^\frac{1}{2}+1$

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

Integers

Algebra

Trigonometry

## Check the Answer

AIME I, 1995, Question 7

Plane Trigonometry by Loney

## Try with Hints

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=$\frac{5}{4}-(sin^{2}t+cos^{2}t)-sintcost$=$\frac{1}{4}-\frac{(5x)^\frac{1}{2}}{2}$

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 $x \geq 0$

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

Categories

## Triangle and Trigonometry | AIME I, 1999 Question 14

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

## Triangle and Trigonometry – AIME 1999

Point P is located inside triangle ABC so that angles PAB,PBC and PCA are all congruent. The sides of the triangle have lengths AB=13, BC=14, CA=15, and the tangent of angle PAB is $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 463
• is 840
• cannot be determined from the given information

Triangles

Angles

Trigonometry

## Check the Answer

AIME, 1999, Question 14

Geometry Revisited by Coxeter

## Try with Hints

Let y be the angleOAB=angleOBC=angleOCA then from three triangles within triangleABC we have $b^{2}=a^{2}+169-26acosy$ $c^{2}=b^{2}+196-28bcosy$ $a^{2}=c^{2}+225-30ccosy$ adding these gives cosy(13a+14b+15c)=295

[ABC]=[AOB]+[BOC]+[COA]=$\frac{siny(13a+14b+15c)}{2}$=84 then (13a+14b+15c)siny=168

tany=$\frac{168}{295}$ then 168+295=463.

.

Categories

## Circles and Triangles | AIME I, 2012 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Circles and triangles.

## Circles and triangles – AIME I, 2012

Three concentric circles have radii 3,4 and 5. An equilateral triangle with one vertex on each circle has side length s. The largest possible area of the triangle can be written as $a+\frac{b}{c}d^\frac{1}{2}$ where a,b,c,d are positive integers b and c are relative prime and d is not divisible by the square of any prime, find a+b+c+d.

• is 107
• is 41
• is 840
• cannot be determined from the given information

Angles

Trigonometry

Triangles

## Check the Answer

AIME I, 2012, Question 13

Geometry Revisited by Coxeter

## Try with Hints

In triangle ABC AO=3, BO=4, CO=5 let AB-BC=CA=s [ABC]=$\frac{s^{2}3^\frac{1}{2}}{4}$

$s^{2}=3^{2}+4^{2}-2(3)(4)cosAOB$=25-24cosAOB then [ABC]=$\frac{25(3)^\frac{1}{2}}{4}-6(3)^\frac{1}{2}cosAOB$

of the required form for angle AOB=150 (in degrees) then [ABC]=$\frac{25(3)^\frac{1}{2}}{4}+9$ then a+b+c+d=25+3+4+9=41.

Categories

## Trigonometry Problem | AIME I, 2015 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Trigonometry.

## Trigonometry Problem – AIME 2015

With all angles measured in degrees, the product (\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n), where (m) and (n) are integers greater than 1. Find (m+n).

• is 107
• is 91
• is 840
• cannot be determined from the given information

Trigonometry

Sequence

Algebra

## Check the Answer

AIME, 2015, Question 13.

Plane Trigonometry by Loney .

## Try with Hints

Let (x = \cos 1^\circ + i \sin 1^\circ). Then from the identity[\sin 1 = \frac{x – \frac{1}{x}}{2i} = \frac{x^2 – 1}{2 i x},]we deduce that (taking absolute values and noticing (|x| = 1))[|2\sin 1| = |x^2 – 1|.]

But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ – z)$, if we let our product be $M$ then[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ][= \frac{1}{2^{90}} |x^2 – 1| |x^6 – 1| |x^{10} – 1| \dots |x^{354} – 1| |x^{358} – 1|]because $\sin$ is positive in the first and second quadrants.

Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z – x^2)(z – x^6)\dots (z – x^{358}) = z^{90} + 1$, and so[\frac{1}{M} = \dfrac{1}{2^{90}}|1 – x^2| |1 – x^6| \dots |1 – x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.]It is easy to see that $M = 2^{89}$ and that our answer is $2+89=91$.

Categories

## Smallest Perimeter of Triangle | AIME I, 2015 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle.

## Smallest Perimeter of Triangle – AIME 2015

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$..

• is 107
• is 108
• is 840
• cannot be determined from the given information

Inequalities

Trigonometry

Geometry

## Check the Answer

AIME, 2015, Question 11

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.Now let $BD=y$, $AB=x$, and $\angle IBD$ =$\frac{\angle ABD}{2}$ = $\theta$.Then $\mathrm{cos}{(\theta)} = \frac{y}{8}$and $\mathrm{cos}{(2\theta)} = \frac{y}{x} = 2\mathrm{cos^2}{(\theta)} – 1 = \frac{y^2-32}{32}$.

Cross-multiplying yields $32y = x(y^2-32)$.

Since $x,y>0$, $y^2-32$ must be positive, so $y > 5.5$.

Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y < 8$.

Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$.

However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\frac{32(6)}{(6)^2-32}=48$.

Thus the perimeter of $\triangle ABC$ must be $2(x+y) = {108}$.

Categories

## Area of a Triangle -AMC 8, 2018 – Problem 20

Try this beautiful problem from Geometry based on Area of a Triangle Using similarity

## Area of Triangle – AMC-8, 2018 – Problem 20

In $\triangle ABC$ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.

What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?

• $\frac{2}{3}$
• $\frac{4}{9}$
• $\frac{3}{5}$

Geometry

Area

similarity

## Check the Answer

Answer:$\frac{4}{9}$

AMC-8, 2018 problem 20

Pre College Mathematics

## Try with Hints

$\triangle ADE$ $\sim$ $\triangle ABC$

Can you now finish the problem ……….

$\triangle BEF$ $\sim$ $\triangle ABC$

can you finish the problem……..

Since $\triangle ADE$$\sim$ $\triangle ABC$

$\frac{ \text {area of} \triangle ADE}{ \text {area of} \triangle ABC}$=$\frac{AE^2}{AB^2}$

i.e $\frac{\text{area of} \triangle ADE}{\text{area of} \triangle ABC}$ =$\frac{(1)^2}{(3)^2}$=$\frac{1}{9}$

Again $\triangle BEF$ $\sim$ $\triangle ABC$

Therefore $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$=$\frac{BE^2}{AB^2}$

i.e $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$ =$\frac{(2)^2}{(3)^2}$=$\frac{4}{9}$

Therefore Area of quad. CDEF =$\frac {4}{9}$ of area $\triangle ABC$

i.e The ratio of the area of quad.CDEF to the area of $\triangle ABC$ is $\frac{4}{9}$