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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Trapezoid Problem | AIME I, 1992 | Question 9

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

Trapezoid – AIME I, 1992


Trapezoid ABCD has sides AB=92, BC=50,CD=19,AD=70 with AB parallel to CD. A circle with centre P on AB is drawn tangent to BC and AD. Given that AP=\(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 164
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Trapezoid

Angle Bisectors

Check the Answer


Answer: is 164.

AIME I, 1992, Question 9

Coordinate Geometry by Loney

Try with Hints


Let AP=y or, PB=92-y

extending AD and BC to meet at Y

and YP bisects angle AYB

Trapezoid Problem

Let F be point on CD where it meets

Taking angle bisector theorem,

let YB=z(92-y), YA=zy for some z

YD=zy-70, YC=z(92-y)-50

\(\frac{yz-79}{z(92-y)-50}=\frac{YD}{YC}=\frac{FD}{FC}=\frac{AP}{PB}=\frac{y}{42-y}\)

solving we get 120y=(70)(92)

or, AP=y=\(\frac{161}{3}\)

or, 161+3=164.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Digits and Numbers | AIME I, 2012 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on digits and numbers.

Digits and numbers – AIME I, 2012


Let S be set of all perfect squares whose rightmost three digits in base 10 are 256. T be set of numbers of form \(\frac{x-256}{1000}\) where x is in S, find remainder when 10th smallest element of T is divided by 1000.

  • is 107
  • is 170
  • is 840
  • cannot be determined from the given information

Key Concepts


Digits

Algebra

Numbers

Check the Answer


Answer: is 170.

AIME I, 2012, Question 10

Elementary Number Theory by David Burton

Try with Hints


x belongs to S so perfect square, Let x=\(y^{2}\), here \(y^{2}\)=1000a+256 \(y^{2}\) element in S then RHS being even y=2\(y_1\) then \(y_1^{2}=250a+64\) again RHS being even \(y_1=2y_2\) then \(y_2^{2}\)=125\(\frac{a}{2}\)+16 then both sides being integer a=2\(a_1\) then \(y_2^{2}=125a_1+16\)

\(y_2^{2}-16=125a_1\) then \((y_2-4)(y_2+4)=125a_1\)

or, one of \((y_2+4)\) and \((y_2-4)\) contains a non negative multiple of 125 then listing smallest possible values of \(y_2\)

or, \(y_2+4=125\) gives \(y_2=121\) or, \(y_2-4=125\) gives \(y_2=129\) and so on

or, \(y_2=4,121,129,upto ,621\) tenth term 621

\(y=4y_2\)=2484 then \(\frac{2483^{2}-256}{1000}\)=170.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Problem on Fibonacci sequence | AIME I, 1988 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.

Fibonacci sequence Problem – AIME I, 1988


Find a if a and b are integers such that \(x^{2}-x-1\) is a factor of \(ax^{17}+bx^{16}+1\).

  • is 107
  • is 987
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Sets

Check the Answer


Answer: is 987.

AIME I, 1988, Question 13

Elementary Number Theory by David Burton

Try with Hints


Let F(x)=\(ax^{17}+bx^{16}+1\)

Let P(x) be polynomial such that

\(P(x)(x^{2}-x-1)=F(x)\)

constant term of P(x) =(-1)

now \((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\) where \(c_{i}\)=coefficient

comparing the coefficients of x we get the terms

since F(x) has no x term, then \(c_{15}\)=1

getting \(c_{14}\)

\((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\)

=terms +\(0x^{2}\) +terms

or, \(c_{14}=-2\)

proceeding in the same way \(c_{13}=3\), \(c_{12}=-5\), \(c_{11}=8\) gives a pattern of Fibonacci sequence

or, coefficients of P(x) are Fibonacci sequence with alternating signs

or, a=\(c_1=F_{16}\) where \(F_{16}\) is 16th Fibonacci number

or, a=987.

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Function Problem | AIME I, 1988 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on function.

Function Problem – AIME I, 1988


For any positive integer k, let \(f_1(k)\) denote the square of the sum of the digits of k. For \(n \geq 2\), let \(f_n(k)=f_1(f_{n-1}(k))\), find \(f_{1988}(11)\).

  • is 107
  • is 169
  • is 634
  • cannot be determined from the given information

Key Concepts


Functions

Equations

Algebra

Check the Answer


Answer: is 169.

AIME I, 1988, Question 2

Functional Equation by Venkatchala

Try with Hints


\(f_1(11)=4\)

or, \(f_2(11)=f_1(4)=16\)

or, \(f_3(11)=f_1(16)=49\)

or, \(f_4(11)=f_1(49)=169\)

or, \(f_5(11)=f_1(169)=256\)

or, \(f_6(11)=f_1(256)=169\)

or, \(f_7(11)=f_1(169)=256\)

This goes on between two numbers with this pattern, here 1988 is even,

or, \(f_1988(11)=f_4(11)=169\).

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Reflection Problem | AIME I, 1988 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988, Question 14, based on Reflection.

Reflection Problem – AIME I, 1988


Let C be the graph of xy=1 and denote by C’ the reflection of C in the line y=2x. let the equation of C’ be written in the form \(12x^{2}+bxy +cy^{2}+d=0\), find the product bc.

  • is 107
  • is 84
  • is 840
  • cannot be determined from the given information

Key Concepts


Geometry

Equation

Algebra

Check the Answer


Answer: is 84.

AIME I, 1988, Question 14

Coordinate Geometry by Loney

Try with Hints


Let P(x,y) on C such that P'(x’,y’) on C’ where both points lie on the line perpendicular to y=2x

slope of PP’=\(\frac{-1}{2}\), then \(\frac{y’-y}{x’-x}\)=\(\frac{-1}{2}\)

or, x’+2y’=x+2y

also midpoint of PP’, \((\frac{x+x’}{2},\frac{y+y’}{2})\) lies on y=2x

or, \(\frac{y+y’}{2}=x+x’\)

or, 2x’-y’=y-2x

solving these two equations, x=\(\frac{-3x’+4y’}{5}\) and \(y=\frac{4x’+3y’}{5}\)

putting these points into the equation C \(\frac{(-3x’+4y’)(4x’+3y’)}{25}\)=1

which when expanded becomes

\(12x’^{2}-7x’y’-12y’^{2}+25=0\)

or, bc=(-7)(-12)=84.

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AIME I Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Problem on Complex plane | AIME I, 1988| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Complex Plane.

Problem on Complex Plane – AIME I, 1988


Let w_1,w_2,….,w_n be complex numbers. A line L in the complex plane is called a mean line for the points w_1,w_2,….w_n if L contains points (complex numbers) z_1,z_2, …..z_n such that \(\sum_{k=1}^{n}(z_{k}-w_{k})=0\) for the numbers \(w_1=32+170i, w_2=-7+64i, w_3=-9+200i, w_4=1+27i\) and \(w_5=-14+43i\), there is a unique mean line with y-intercept 3. Find the slope of this mean line.

  • is 107
  • is 163
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


Answer: is 163.

AIME I, 1988, Question 11

Elementary Algebra by Hall and Knight

Try with Hints


\(\sum_{k=1}^{5}w_k=3+504i\)

and \(\sum_{k-1}^{5}z_k=3+504i\)

taking the numbers in the form a+bi

\(\sum_{k=1}^{5}a_k=3\) and \(\sum_{k=1}^{5}b_k=504\)

or, y=mx+3 where \(b_k=ma_k+3\) adding all 5 equations given for each k

or, 504=3m+15

or, m=163.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Problem on Real Numbers | AIME I, 1990| Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Real Numbers.

Problem on Real Numbers – AIME I, 1990


Find \(ax^{5}+by^{5}\) if real numbers a,b,x,y satisfy the equations

ax+by=3

\(ax^{2}+by^{2}=7\)

\(ax^{3}+by^{3}=16\)

\(ax^{4}+by^{4}=42\)

  • is 107
  • is 20
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


Answer: is 20.

AIME I, 1990, Question 15

Elementary Algebra by Hall and Knight

Try with Hints


Let S=x+y, P=xy

\((ax^{n}+by^{n})(x+y)\)

\(=(ax^{n+1}+by^{n+1})+(xy)(ax^{n-1}+by^{n-1})\)

or,\( (ax^{2}+by^{2})(x+y)=(ax^{3}+by^{3})+(xy)(ax+by)\) which is first equation

or,\( (ax^{3}+by^{3})(x+y)=(ax^{4}+by^{4})+(xy)(ax^{2}+by^{2})\) which is second equation

or, 7S=16+3P

16S=42+7P

or, S=-14, P=-38

or, \((ax^{4}+by^{4})(x+y)=(ax^{5}+by^{5})+(xy)(ax^{3}+by^{3})\)

or, \(42S=(ax^{5}+by^{5})+P(16)\)

or, \(42(-14)=(ax^{5}+by^{5})+(-38)(16)\)

or, \(ax^{5}+by^{5}=20\).

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Digits and Integers | AIME I, 1990 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Digits and Integers.

Digits and Integers – AIME I, 1990


Let T={\(9^{k}\): k is an integer, \(0 \leq k \leq 4000\)} given that \(9^{4000}\) has 3817 digits and that its first (leftmost) digit is 9, how many elements of T have 9 as their leftmost digit?

  • is 107
  • is 184
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Sets

Check the Answer


Answer: is 184.

AIME I, 1990, Question 13

Elementary Number Theory by David Burton

Try with Hints


here \(9^{4000}\) has 3816 digits more than 9,

or, 4000-3816=184

or, 184 numbers have 9 as their leftmost digits.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Consecutive positive Integers | AIME I, 1990| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on consecutive positive integers.

Consecutive positive integer – AIME I, 1990


Someone observed that 6!=(8)(9)(10). Find the largest positive integer n for which n! can be expressed as the product of n-3 consecutive positive integers.

  • is 107
  • is 23
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Inequality

Algebra

Check the Answer


Answer: is 23.

AIME I, 1990, Question 11

Elementary Number Theory by David Burton

Try with Hints


The product of (n-3) consecutive integers=\(\frac{(n-3+a)!}{a!}\) for a is an integer

\(n!=\frac{(n-3+a)!}{a!}\) for \(a \geq 3\) \((n-3+a)! \geq n!\)

or, \(n!=\frac{n!(n+1)(n+2)….(n-3+a)}{a!}\)

for a=4, n+1=4! or, n=23 which is greatest here

n=23.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Convex polyhedron Problem | AIME I, 1988 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on convex polyhedron.

Convex polyhedron Problem – AIME I, 1988


A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

  • is 107
  • is 840
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Edges

Algebra

Check the Answer


Answer: is 840.

AIME I, 1988, Question 10

Geometry Revisited by Coxeter

Try with Hints


\({48 \choose 2}\)=1128

Every vertex lies on exactly one vertex of a square/hexagon/octagon

V=(12)(4)=(8)(6)=(6)(8)=48

each vertex is formed by the trisection of three edges and every edge is counted twice, once at each of its endpoints, the number of edges E=\(\frac{3V}{2}\)=72

each of the segment on face of polyhedron is diagonal of that face, so each square gives \(\frac{n(n-3)}{2}=2\) diagonals, each hexagon=9,each octagon=20. The number of diagonals is \((2)(12)+(9)(8)+(20)(6)\)=216

or, number of space diagonals =1128-72-216=840.

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