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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


Answer: is 173.

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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Arithmetic Geometry Math Olympiad USA Math Olympiad

Arithmetic and geometric mean | AIME I, 2000 Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Arithmetic and geometric mean with Algebra.

Arithmetic and geometric mean with Algebra – AIME 2000


Find the number of ordered pairs (x,y) of integers is it true that \(0 \lt y \lt 10^{6}\) and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y.

  • is 107
  • is 997
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Ordered pair

Check the Answer


Answer: is 997.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


First hint

 given that \(\frac{x+y}{2}=2+({xy})^\frac{1}{2}\) then solving we have \(y^\frac{1}{2}\)-\(x^\frac{1}{2}\)=+2 and-2

Second Hint

given that \(y \gt x\) then \(y^\frac{1}{2}\)-\(x^\frac{1}{2}\)=+2 and here maximum integer value of \(y^\frac{1}{2}\)=\(10^{3}-1\)=999 whose corresponding \(x^\frac{1}{2}\)=997 and decreases upto \(y^\frac{1}{2}\)=3 whose corresponding \(x^\frac{1}{2}\)=1

Final Step

then number of pairs (\(x^\frac{1}{2}\),\(y^\frac{1}{2}\))=number of pairs of (x,y)=997.

.

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Algebra Arithmetic Functional Equations Math Olympiad Math Olympiad Videos USA Math Olympiad

Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

Logarithms and Equations – AIME I 2000


\(log_{10}(2000xy)-log_{10}xlog_{10}y=4\) and \(log_{10}(2yz)-(log_{10}y)(log_{10}z)=1\) and \(log_{10}(zx)-(log_{10}z)(log_{10}x)=0\) has two solutions \((x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})\) find \(y_{1}+y_{2}\).

  • is 905
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Logarithms

Theory of Equations

Number Theory

Check the Answer


Answer: is 25.

AIME I, 2000, Question 9

Polynomials by Barbeau

Try with Hints


First hint

Rearranging equations we get \(-logxlogy+logx+logy-1=3-log2000\) and \(-logylogz+logy+logz-1=-log2\) and \(-logxlogz+logx+logz-1=-1\)

Second Hint

taking p, q, r as logx, logy and logz, \((p-1)(q-1)=log2\) and \((q-1)(r-1)=log2\) and \( (p-1)(r-1)=1\) which is first system of equations and multiplying the first three equations of the first system gives \((p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}\) gives \((p-1)(q-1)(r-1)=+-(log2)\) which is second equation

Final Step

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives \(y_{1}=20\),\(y_{2}=5\) then \(y_{1}+y_{2}=25\).

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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


Answer: is 159.

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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Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

Amplitude and Complex numbers – AIME 1996


Let P be the product of the roots of \(z^{6}+z^{4}+z^{2}+1=0\) that have a positive imaginary part and suppose that P=r(costheta+isintheta) where \(0 \lt r\) and \(0 \leq \theta \lt 360\) find \(\theta\)

  • is 107
  • is 276
  • is 840
  • cannot be determined from the given information

Key Concepts


Equations

Complex Numbers

Integers

Check the Answer


Answer: is 276.

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

here\(z^{6}+z^{4}+z^{2}+1\)=\(z^{6}-z+z^{4}+z^{2}+z+1\)=\(z(z^{5}-1)+\frac{(z^{5}-1)}{(z-1)}\)=\(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\) then \(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\)=0

Second Hint

gives \(z^{5}=1 for z\neq 1\) gives \(z=cis 72,144,216,288\) and \(z^{2}-z+1=0 for z \neq 1\) gives z=\(\frac{1+-(-3)^\frac{1}{2}}{2}\)=\(cis60,300\) where cis\(\theta\)=cos\(\theta\)+isin\(\theta\)

Final Step

taking \(0 \lt theta \lt 180\) for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276.

.

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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


Answer: is 23.

AIME I, 1996, Question 5

Polynomials by Barbeau

Try with Hints


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 \(abc=11\)

Let a+b+c=-3=p

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

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

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

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

=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


Answer: is 320.

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


Answer: is 616.

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


Answer: is 248.

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


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.

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