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Geometry Math Olympiad USA Math Olympiad

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

Key Concepts


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


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

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

The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

Arithmetic Mean of Number Theory – AIME 2015


Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is \(\frac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p + q\).

  • is 107
  • is 431
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequalities

Algebra

Number Theory

Check the Answer


Answer: is 431.

AIME, 2015, Question 12

Elementary Number Theory by David Burton

Try with Hints


Each 1000-element subset \({ a_1, a_2,a_3,…,a_{1000}}\) of \({1,2,3,…,2015}\) with \(a_1<a_2<a_3<…<a_{1000}\) contributes \(a_1\) to sum of least element of each subset and set \({a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\). \(a_1\) ways to choose a positive integer \(k\) such that \(k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1\) (\(k\) can be anything from \(1\) to \(a_1\) inclusive

Thus, the number of ways to choose the set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is equal to the sum. But choosing a set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is same as choosing a 1001-element subset from \({1,2,3,…,2016}\)!

average =\(\frac{2016}{1001}\)=\(\frac{288}{143}\). Then \(p+q=288+143={431}\)

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Distance Time | AIME I, 2012 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.

Distance Time – AIME 2012


When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes \(\frac{1}{6}\) hours per mile, Butch takes \(\frac{1}{4}\) hours per mile, and Sundance takes \(\frac{2}{5}\) hours per mile.

  • is 107
  • is 279
  • is 840
  • cannot be determined from the given information

Key Concepts


Time

Distance

Speed

Check the Answer


Answer: is 279.

AIME, 2012, Question 4

Problem Solving Strategies by Arther Engel

Try with Hints


After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.

Then
\(\frac{a}{6} + \frac{n-a}{4}\) = \(\frac{n-a}{6} + \frac{2a}{5}\) implies that \(a = \frac{5}{19}n\)

Then integral value of n is 19 and a = 5 and \(t = \frac{13}{3}\) hours that is 260 minutes. Then \(19 + 260 = {279}\).

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Algebra Combinatorics Math Olympiad USA Math Olympiad

Algebra and Combination | AIME I, 2000 Question 3

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

Algebra and combination – AIME 2000


In expansion \((ax+b)^{2000}\) where a and b are relatively prime positive integers the coefficient of \(x^{2}\) and \(x^{3}\) are equal, find a+b

  • is 107
  • is 667
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Combination

Check the Answer


Answer: is 667.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


 here coefficient of \(x^{2}\)= coefficient of \(x^{3}\) in the same expression

then \({2000 \choose 1998}a^{2}b^{1998}\)=\({2000 \choose 1997}a^{3}b^{1997}\)

then \(b=\frac{1998}{3}\)a=666a where a and b are relatively prime that is a=1,b=666 then a+b=666+1=667.

.

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Algebra Functional Equations Math Olympiad USA Math Olympiad

Algebraic Equation | AIME I, 2000 Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

Algebraic Equation – AIME 2000


Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, \(x+\frac{1}{z}=5\) and \(y+\frac{1}{x}=29\) then \(z+\frac{1}{y}\)=\(\frac{m}{n}\) where m and n are relatively prime, find m+n

  • is 107
  • is 5
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Integers

Check the Answer


Answer: is 5.

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

Try with Hints


 here \(x+\frac{1}{z}=5\) then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and \(y=(29-\frac{1}{x}\)) together gives 5-x=x\((29-\frac{1}{x}\)) then x=\(\frac{1}{5}\)

then y=29-5=24 and z=\(\frac{1}{5-x}\)=\(\frac{5}{24}\)

\(z+\frac{1}{y}\)=\(\frac{1}{4}\) then 1+4=5.

.

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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


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

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

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

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

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

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

\(\Rightarrow m+n\)=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


 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

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

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


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

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

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


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

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

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

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

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

=tan141

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

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

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


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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