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

Answer: is 33.

AIME I, 1987, Question 9

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

Angles

Algebra

Spheres

## Check the Answer

Answer: is 137.

AIME I, 1987, Question 2

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

Angles

Triangles

Side Length

## Check the Answer

Answer: is 177.

AIME I, 2009, Question 5

Plane Trigonometry by Loney

## Try with Hints

First hint

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

Second Hint

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

Final Step

$=\frac{1-cos(175)}{sin175}$=$tan\frac{175}{2}$ 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

Answer: is 18.

AIME I, 2012, Question 12

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

## Area of Triangle and Integer | PRMO 2019 | Question 29

Try this beautiful problem from the PRMO, 2019 based on area of triangle and nearest integer.

## Area of triangle and integer – PRMO 2019

In a triangle ABC, the median AD (with D on BC) and the angle bisector BE (with E on AC) are perpendicular to each other, if AD=7 and BE=9, find the integer nearest to the area of triangle ABC

• is 107
• is 47
• is 840
• cannot be determined from the given information

Angles

Triangles

Integer

## Check the Answer

Answer: is 47.

PRMO, 2019, Question 29

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

let AD and BE meet at F angle ABF =angleFBD=$\frac{B}{2}$ angle AFB=angle BFD=90 (in degrees), BF is common with triangles ABF and BFD then triangle ABF is congruent to triangle BFD,AF=FD=$\frac{7}{2}$ where AF=FD

Second Hint

AB=BD then$\frac{AB}{BC}=\frac{AB}{BD+CD}=\frac{AB}{2BD}=\frac{1}{2}$ where AB=BD $\frac{AE}{EC}$=$\frac{AB}{BC}$=$\frac{1}{2}$

Final Step

$\frac{area triangle ABC}{area triangle ABE}=\frac{3}{1}$ then $area triangle ABC=3area triangle ABE$=$(3)(\frac{1}{2} \times AF \times BE)=\frac{3}{2} \times \frac{7}{2} \times 9$=47.25 then nearest integer=47.

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

Answer: is 463.

AIME, 1999, Question 14

Geometry Revisited by Coxeter

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

Answer: is 72.

AIME I, 2009, Question 5

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

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

Second Hint

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 $\frac{AM}{LP}=\frac{AL}{LB}+1=\frac{5}{2}$

Final Step

$\frac{180}{LP}=\frac{5}{2}$ 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

Angles

Trigonometry

Triangles

## Check the Answer

Answer: is 41.

AIME I, 2012, Question 13

Geometry Revisited by Coxeter

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

## Triangles and Internal bisectors | PRMO 2019 | Question 10

Try this beautiful problem from the PRMO, 2019 based on triangles and internal bisectors.

## Triangles and internal bisectors – PRMO 2019

Let ABC be a triangle and let D be its circumcircle, The internal bisectors of angles A,B and C intersect D at $A_1,B_1 and C_1$ the internal bisectors of $A_1,B_1,C_1$ of the triangle $A_1B_1C_1$ intersect D at $A_2,B_2,C_2$. If the smallest angle of triangle ABC is 40 find the magnitude of the smallest angle of triangle $A_2B_2C_2$ in degrees.

• is 107
• is 55
• is 840
• cannot be determined from the given information

Lines

Algebra

Angles

## Check the Answer

Answer: is 55.

PRMO, 2019, Question 10

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

angle $A_1B_1C_1=90 – \frac{ABC}{2}$ angle $A_1C_1B_1=90-\frac{ACB}{2}$

Second Hint

angle $B_1A_1C_1$=90-$\frac{BAC}{2}$

Final Step

then angle $A_2B_2C_2=90-\frac{90-\frac{ABC}{2}}{2}$=45+$\frac{ABC}{4}$=55.

Categories

## Lines and Angles | PRMO 2019 | Question 7

Try this beautiful problem from the PRMO, 2019 based on Lines and Angles.

## Lines and Angles – PRMO 2019

On a clock, there are two instants between 12 noon and 1 pm when the hour hand and the minute hand are at right angles. the difference in minutes between these two instants written as a+$\frac{b}{c}$ where a b c are positive integers with $b\lt c$ and $\frac{b}{c}$ in the reduced form. find a+b+c.

• is 107
• is 51
• is 840
• cannot be determined from the given information

Lines

Algebra

Angles

## Check the Answer

Answer: is 51.

PRMO, 2019, Question 7

Higher Algebra by Hall and Knight

## Try with Hints

First hint

Minute hand turns 6 (in degrees) in 1 min and hour hand turns half (in degree) in 1 min

Second Hint

x min after 12 is 6x -$\frac{x}{2}$=90 or 270 (in degrees) then x =16$\frac{4}{11}$ or 49$\frac{1}{11}$

Final Step

Difference 32$\frac{8}{11}$ then a+b+c=32+8+11=51.