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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

Number of points and planes – AIME I, 1999


Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 489
  • is 840
  • cannot be determined from the given information

Key Concepts


Number of points

Plane

Probability

Check the Answer


But try the problem first…

Answer: is 489.

Source
Suggested Reading

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

\(10 \choose 3\) sets of 3 points which form triangles,

Second Hint

fourth distinct segment excluding 3 segments of triangles=45-3=42

Final Step

Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)

where \({45 \choose 4}\) is choosing 4 segments from 45 segments

=\(\frac{16}{473}\) then m+n=16+473=489.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

Sequence and fraction – AIME I, 2000


A sequence of numbers \(x_1,x_2,….,x_{100}\) has the property that, for every integer k between 1 and 100, inclusive, the number \(x_k\) is k less than the sum of the other 99 numbers, given that \(x_{50}=\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 173
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


But try the problem first…

Answer: is 173.

Source
Suggested Reading

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

Let S be the sum of the sequence \(x_k\)

given that \(x_k=S-x_k-k\) for any k

taking k=1,2,….,100 and adding

\(100S-2(x_1+x_2+….+x_{100})=1+2+….+100\)

Second Hint

\(\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050\)

\(\Rightarrow S=\frac{2525}{49}\)

for \(k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}\)

Final Step

\(\Rightarrow x_{50}=\frac{75}{98}\)

\(\Rightarrow m+n\)=75+98

=173.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

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

Key Concepts


Functions

Trigonometry

Integers

Check the Answer


But try the problem first…

Answer: is 159.

Source
Suggested Reading

AIME I, 1996, Question 10

Plane Trigonometry by Loney

Try with Hints


First hint

\(\frac{cos96+sin96}{cos96-sin96}\)

=\(\frac{sin(90+96)+sin96}{sin(90+96)-sin96}\)

=\(\frac{sin186+sin96}{sin186-sin96}\)

=\(\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}\)

=\(\frac{2sin141cos45}{2cos141sin45}\)

=tan141

Second Hint

here \(tan(180+\theta)\)=\(tan\theta\)

\(\Rightarrow 19x=141+180n\) for some integer n is first equation

Final Step

multiplying equation with 19 gives

\(x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)\) [since 2679 divided by 180 gives remainder 159]

\(\Rightarrow x=159\).

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AIME I Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

Roots of Equation and Vieta’s formula – AIME I, 1996


Suppose that the roots of \(x^{3}+3x^{2}+4x-11=0\) are a,b and c and that the roots of \(x^{3}+rx^{2}+sx+t=0\) are a+b,b+c and c+a, find t.

  • is 107
  • is 23
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Roots of Equation

Vieta s formula

Check the Answer


But try the problem first…

Answer: is 23.

Source
Suggested Reading

AIME I, 1996, Question 5

Polynomials by Barbeau

Try with Hints


First hint

With Vieta s formula

\(f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0\)

\(\Rightarrow a+b+c=-3\), \(ab+bc+ca=4\) and \(abc=11\)

Second Hint

Let a+b+c=-3=p

here t=-(a+b)(b+c)(c+a)

\(\Rightarrow t=-(p-c)(p-a)(p-b)\)

\(\Rightarrow t=-f(p)=-f(-3)\)

Final Step

\(t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]\)

=23.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Tetrahedron Problem | AIME I, 1992 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.

Tetrahedron Problem – AIME I, 1992


Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron.

  • is 107
  • is 320
  • is 840
  • cannot be determined from the given information

Key Concepts


Area

Volume

Tetrahedron

Check the Answer


But try the problem first…

Answer: is 320.

Source
Suggested Reading

AIME I, 1992, Question 6

Coordinate Geometry by Loney

Try with Hints


First hint

Area BCD=80=\(\frac{1}{2} \times {10} \times {16}\),

where the perpendicular from D to BC has length 16.

Tetrahedron Problem

Second Hint

The perpendicular from D to ABC is 16sin30=8

[ since sin30=\(\frac{perpendicular}{hypotenuse}\) then height = perpendicular=hypotenuse \(\times\) sin30 ]

Final Step

or, Volume=\(\frac{1}{3} \times 8 \times 120\)=320.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

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.

Triangle and integers
  • is 107
  • is 616
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Triangle

Trigonometry

Check the Answer


But try the problem first…

Answer: is 616.

Source
Suggested Reading

AIME I, 1995, Question 9

Plane Trigonometry by Loney

Try with Hints


First hint

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]

Second Hint

\(\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx\)

solving we get, tanx=\(\frac{1}{2}\)

\(\Rightarrow CM=\frac{11}{2}\)

Final Step

\(\Rightarrow 2(AC+CM)\) where \(AC=\frac{11\sqrt {5}}{2}\) by Pythagoras formula

=\(\sqrt{605}+11\) then a+b=605+11=616.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

Sequence and greatest integer – AIME I, 2000


Let S be the sum of all numbers of the form \(\frac{a}{b}\),where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed \(\frac{S}{10}\).

  • is 107
  • is 248
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


But try the problem first…

Answer: is 248.

Source
Suggested Reading

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

We have 1000=(2)(2)(2)(5)(5)(5) and \(\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3\)

Second Hint

sum of all numbers of form \(\frac{a}{b}\) such that a and b are relatively prime positive divisors of 1000

=\((2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})\)

Final Step

\(\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times\) \(\frac{(5^{-3})(5^{7}-1)}{5-1}\)

=2480 + \(\frac{437}{1000}\)

\(\Rightarrow [\frac{s}{10}]\)=248.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

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


But try the problem first…

Answer: is 177.

Source
Suggested Reading

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.

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AIME I Geometry Math Olympiad USA Math Olympiad

Inscribed circle and perimeter | AIME I, 1999 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Inscribed circle and perimeter.

Inscribed circle and perimeter – AIME I, 1999


The inscribed circle of triangle ABC is tangent to AB at P, and its radius is 21 given that AP=23 and PB=27 find the perimeter of the triangle

Inscribed circle and perimeter
  • is 107
  • is 345
  • is 840
  • cannot be determined from the given information

Key Concepts


Inscribed circle

Perimeter

Triangle

Check the Answer


But try the problem first…

Answer: is 345.

Source
Suggested Reading

AIME I, 1999, Question 12

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Q tangency pt on AC, R tangency pt on BC AP=AQ=23 BP=BR=27 CQ=CR=x and

Second Hint

\(s \times r =A\) and \(s=\frac{27 \times 2+23 \times 2+x \times 2}{2}=50+x\) and A=\(({(50+x)(x)(23)(27)})\) then from these equations 441(50+x)=621x then x=\(\frac{245}{2}\)

Final Step

perimeter 2s=2(50+\(\frac{245}{2}\))=345.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.

Lcm and Integer – AIME I, 1998


Find the number of values of k in \(12^{12}\) the lcm of the positive integers \(6^{6}\), \(8^{8}\) and k.

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

Key Concepts


Lcm

Algebra

Integers

Check the Answer


But try the problem first…

Answer: is 25.

Source
Suggested Reading

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

here \(k=2^{a}3^{b}\) for integers a and b

\(6^{6}=2^{6}3^{6}\)

\(8^{8}=2^{24}\)

\(12^{12}=2^{24}3^{12}\)

Second Hint

lcm\((6^{6},8^{8})\)=\(2^{24}3^{6}\)

\(12^{12}=2^{24}3^{12}\)=lcm of \((6^{6},8^{6})\) and k

=\((2^{24}3^{6},2^{a}3^{b})\)

=\(2^{max(24,a)}3^{max(6,b)}\)

Final Step

\(\Rightarrow b=12, 0 \leq a \leq 24\)

\(\Rightarrow\) number of values of k=25.

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