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

Check the Answer

Answer: is 33.

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.

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

Check the Answer

Answer: is 177.

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.

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

Check the Answer

Answer: is 18.

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.

Number of triangles in Polygon | TOMATO B.Stat Objective 105

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on number of triangles in a Polygon.

Number of triangles in Polygons ( B.Stat Objective Question )

For a regular polygon with n sides (n>5) the number of triangles whose vertices are joining non adjacent vertices of the polygon is

\({n \choose 3}\)
\(\frac{n(n-4)(n-5)}{6}\)
0
none of these

Key Concepts

Polygons

Triangles

Combinations

Check the Answer

Answer: \(\frac{n(n-4)(n-5)}{6}\).

B.Stat Objective Problem 105

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

here total triangles \({n \choose 3}\)

\({n \choose 3}\) -n(n-4) for one side common triangles -n for two side common triangles

=\({n \choose 3}\)-n(n-4)-n=\(\frac{n(n-4)(n-5)}{6}\).

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

Key Concepts

Squares

Triangles

Algebra

Check the Answer

Answer: is 185.

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.

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

Key Concepts

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

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

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

\(\frac{area triangle ABC}{area triangle ABE}=\frac{3}{1}\) 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.

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

Key Concepts

Incentre

Triangles

Algebra

Check the Answer

Answer: is 923.

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.

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

Check the Answer

Answer: is 463.

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.

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

Check the Answer

Answer: is 72.

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.

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

Key Concepts

Squares

Trapezium

Triangles

Check the Answer

Answer: is 25.

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.