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

Angles

Algebra

Triangles

## Check the Answer

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

## 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

Angles

Triangles

Side Length

## Check the Answer

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

Angles

Algebra

Triangles

## Check the Answer

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

## Squares and Triangles | AIME I, 1999 | Question 4

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

## Squares and triangles – AIME I, 1999

The two squares share the same centre O and have sides of length 1, The length of AB is $\frac{43}{99}$ and the area of octagon ABCDEFGH is $\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n.

• is 107
• is 185
• is 840
• cannot be determined from the given information

Squares

Triangles

Algebra

## Check the Answer

AIME I, 1999, Question 4

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

Triangle AOB, triangleBOC, triangleCOD, triangleDOE, triangleEOF, triangleFOG, triangleGOH, triangleHOA are congruent triangles

with each area =$\frac{\frac{43}{99} \times \frac{1}{2}}{2}$

then the area of all 8 of them is (8)$\frac{\frac{43}{99} \times \frac{1}{2}}{2}$=$\frac{86}{99}$ then 86+99=185.

Categories

## Incentre and Triangle | AIME I, 2001 | Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2001 based on Incentre and triangle.

## Incentre and Triangle – AIME I, 2001

Triangle ABC has AB=21 AC=22 BC=20 Points D and E are located on AB and AC such that DE parallel to BC and contains the centre of the inscribed circle of triangle ABC then $DE=\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n

• is 107
• is 923
• is 840
• cannot be determined from the given information

Incentre

Triangles

Algebra

## Check the Answer

AIME I, 2001, Question 7

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

F is incentre, BF and CF are angular bisectors of angle ABC and angle ACB DE drawn parallel to BC then angle BFD=angle FBC= angle FBD

triangle BDF is isosceles then same way triangle CEF is isosceles then perimeter triangle ADE=AD+DE+AE=AB+AC=43 perimeter triangle ABC=63 and triangle ABC similar to triangle ADE

DE=$BC \times \frac{43}{63}$=$20 \times \frac{43}{63}$=$\frac{860}{63}$ then m+n=860+63=923.

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

## 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

Angles

Triangles

Side Length

## Check the Answer

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

## Squares and Triangles | AIME I, 2008 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Squares and Triangles.

## Squares, triangles and Trapezium – AIME I, 2008

Square AIME has sides of length 10 units, isosceles triangle GEM has base EM, and the area common to triangle GEM and square AIME is 80 square units.Find the length of the altitude to EM in triangle GEM.

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

Squares

Trapezium

Triangles

## Check the Answer

AIME I, 2008, Question 2

Geometry Revisited by Coxeter

## Try with Hints

let X and Y be points where the triangle intersects the square and [AXE]=[YIM]=$\frac{100-80}{2}$=10 then AX=YI=2 units then XY=10-4=6 units

triangle GXY is similar to triangle GEM where h=height of triangle GXY then by similarity $\frac{h+10}{10}=\frac{h}{6}$

then h=15 and h+10=25.

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

## Complex Numbers and Triangles | AIME I, 2012 | Question 14

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

## Complex numbers and triangles – AIME I, 2012

Complex numbers a,b and c are zeros of a polynomial P(z)=$z^{3}+qz+r$ and $|a|^{2}+|b|^{2}+|c|^{2}$=250, The points corresponding to a,b,.c in a complex plane are the vertices of right triangle with hypotenuse h, find $h^{2}$.

• is 107
• is 375
• is 840
• cannot be determined from the given information

Complex Numbers

Algebra

Triangles

## Check the Answer

AIME I, 2012, Question 14

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

here q ,r real a real b,c complex and conjugate pair x+iy,x-iy then a+b+c=0 gives a=-2x and by given condition a-x=y then y=-3x

$|a|^{2}+|b|^{2}+|c|^{2}$=250 then 24$x^{2}$=250

h distance between b and c h=2y=-6x then $h^{2}=36x^{2}$=36$\frac{250}{24}$=375.