Categories

## Length and Triangle | AIME I, 1987 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Length and Triangle.

## Length and Triangle – AIME I, 1987

Triangle ABC has right angle at B, and contains a point P for which PA=10, PB=6, and $\angle$APB=$\angle$BPC=$\angle$CPA. Find PC.

• is 107
• is 33
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Algebra

Triangles

AIME I, 1987, Question 9

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

Let PC be x, $\angle$APB=$\angle$BPC=$\angle$CPA=120 (in degrees)

Applying cosine law $\Delta$APB, $\Delta$BPC, $\Delta$CPA with cos120=$\frac{-1}{2}$ gives

$AB^{2}$=36+100+60=196, $BC^{2}$=36+$x^{2}$+6x, $CA^{2}$=100+$x^{2}$+10x

By Pathagorus Theorem, $AB^{2}+BC^{2}=CA^{2}$

or, $x^{2}$+10x+100=$x^{2}$+6x+36+196

or, 4x=132

or, x=33.

Categories

## Distance and Spheres | AIME I, 1987 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres.

## Distance and Sphere – AIME I, 1987

What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)?

• is 107
• is 137
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Algebra

Spheres

AIME I, 1987, Question 2

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

The distance between the center of the spheres is $\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$

=$\sqrt{14^{2}+18^{2}+21^{2}}$=31

The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.

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

## Angles and Triangles | AIME I, 2012 | Question 12

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

## Angles and Triangles – AIME I, 2012

Let triangle ABC be a right angled triangle with right angle at C. Let D and E be points on AB with D between A and E such that CD and CE trisect angle C. If $\frac{DE}{BE}$=$\frac{8}{15}$, then tan B can be written as $\frac{mp^\frac{1}{2}}{n}$ where m and n are relatively prime positive integers, and p is a positive integer not divisible by the square of any prime , find m+n+p.

• is 107
• is 18
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Algebra

Triangles

AIME I, 2012, Question 12

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

Let CD=2a,then with angle bisector theorem of triangle we have for triangle CDB $\frac{2a}{8}$=$\frac{CB}{15}$ then $CB=\frac{15a}{4}$

DF drawn perpendicular to BC gives CF=a, FD=$a \times 3^\frac{1}{2}$, FB= $\frac{11a}{4}$

then tan B = $\frac{a \times 3^\frac{1}{2}}{\frac{11a}{4}}$=$\frac{4 \times 3^\frac{1}{2}}{11}$ then m+n+p=4+3+11=18.

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

### Key Concepts

Triangles

Angles

Trigonometry

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

## Triangles and sides | AIME I, 2009 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Triangles and sides.

## Triangles and sides – AIME I, 2009

Triangle ABC AC=450 BC=300 points K and L are on AC and AB such that AK=CK and CL is angle bisectors of angle C. let P be the point of intersection of Bk and CL and let M be a point on line Bk for which K is the mid point of PM AM=180, find LP

• is 107
• is 72
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Triangles

Side Length

AIME I, 2009, Question 5

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

since K is mid point of PM and AC quadrilateral AMCP is a parallelogram which implies AM parallel LP and triangle AMB is similar to triangle LPB

then $\frac{AM}{LP}=\frac{AB}{LB}=\frac{AL+LB}{LB}=\frac{AL}{LB}+1$

from angle bisector theorem, $\frac{AL}{LB}=\frac{AC}{BC}=\frac{450}{300}=\frac{3}{2}$ then $\frac{AM}{LP}=\frac{AL}{LB}+1=\frac{5}{2}$

$\frac{180}{LP}=\frac{5}{2}$ then LP=72.

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

### Key Concepts

Angles

Trigonometry

Triangles

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.