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AMC 10 Geometry Math Olympiad USA Math Olympiad

Surface area of Cube Problem | AMC-10A, 2007 | Problem 21

Try this beautiful problem from Geometry based on Surface area of a cube.

Surface area of cube – AMC-10A, 2007- Problem 21


A sphere is inscribed in a cube that has a surface area of \(24\) square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?

  • \(3\)
  • \(4\)
  • \(8\)
  • \(4\)

Key Concepts


Geometry

Cube

square

Check the Answer


Answer: \(8\)

AMC-10A (2007) Problem 21

Pre College Mathematics

Try with Hints


Surface area of Cube - problem

We have to find out the surface area of the inner cube.but to find out the surface area of inner cube the side length is require.but we don’t know the side length of the inner cube.but if you see the above diagram you must notice that  two opposite vertices of the cube are on opposite faces of the larger cube.Therefore two opposite vertices of the cube are on opposite faces of the larger cube.Thus the diagonal of the smaller cube is the side length of the outer square…..

Can you now finish the problem ……….

surface area of inner cube

The area of each face of the outer cube is \(\frac {24}{6} = 4\).Therefore  the edge length of the outer cube is \(2\). so the diagonal of the inner cube is \(2\).let the side length of inner cube is \(x\)

Therefore diagonal =\(\sqrt {x^2 +({\sqrt 2}x)^2}=2\) \(\Rightarrow x=\frac{2}{\sqrt 3}\)

can you finish the problem……..

Therefore surface area of the inner cube is \(6x^2\)=\(6 \times (\frac{2}{\sqrt 3})^2\)=\(8\)

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Geometry Math Olympiad USA Math Olympiad

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.

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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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AMC 10 Math Olympiad USA Math Olympiad

Greatest Common Divisor | AMC-10A, 2018 | Problem 22

Try this beautiful problem from Algebra based on Greatest Common Divisor from AMC 10A, 2018, Problem 22.

Greatest Common Divisor – AMC-10A, 2018- Problem 22


Let \(a, b, c,\) and \(d\) be positive integers such that \(\gcd(a, b)=24\), \(\gcd(b, c)=36\), \(\gcd(c, d)=54\), and \(70<\gcd(d, a)<100\). Which of the following must be a divisor of \(a\)?

  • \(5\)
  • \(7\)
  • \(13\)

Key Concepts


Number theory

Gcd

Divisior

Check the Answer


Answer: \(13\)

AMC-10A (2018) Problem 22

Pre College Mathematics

Try with Hints


TO find the divisor of \(a\) at first we have to find the value of \(a\).can you find the value of \(a\)?

Given that \(\gcd(a, b)=24\), \(\gcd(b, c)=36\), \(\gcd(c, d)=54\), and \(70<\gcd(d, a)<100\)

so we can say \(a=24 \times\) some integer and \(b=24 \times\) some another integer (according to gcd rules)

similarly for the others c & d…..

now if we can find out the value of \(\gcd(d, a)\) then we may use the condition \(70<\gcd(d, a)<100\)

Can you now finish the problem ……….

so we may say that \(gcd(a, b)\) is \(2^3 * 3\) and the \(gcd\) of \((c, d)\) is \(2 * 3^3\). However, the \(gcd\) of \((b, c) = 2^2 * 3^2\) (meaning both are divisible by 36). Therefore, \(a\) is only divisible by \(3^1\) (and no higher power of 3), while \(d\) is divisible by only \(2^1\) (and no higher power of 2).

can you finish the problem……..

so we can say that \(gcd\) of \((a, d)\) can be expressed in the form \(2 \times 3 \times \) some positve integer and now \(k\) is a number not divisible by \(2\) or \(3\). so from the given numbes it will be \(13\) because \(2 \times 3 \times k\) must lie \(70<\gcd (d, a)<100\). so the required ans is \(13\)

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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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Algebra AMC 10 Math Olympiad Theory of Equations USA Math Olympiad

Quadratic Equation Problem | AMC-10A, 2005 | Problem 10

Try this beautiful problem from Algebra based on Quadratic Equation….

Quadratic equation – AMC-10A, 2005- Problem 10


There are two values of $a$ for which the equation $4 x^{2}+a x+8 x+9=0$ has only one solution for $x$. What is the sum of those values of $a$ ?

  • \(5\)
  • \(20\)
  • \(-16\)
  • \(25\)
  • \(36\)

Key Concepts


algebra

Quadratic equation

Equal roots

Check the Answer


Answer: \(-16\)

AMC-10A (2005) Problem 10

Pre College Mathematics

Try with Hints


The given equation is $4 x^{2}+a x+8 x+9=0$

\(\Rightarrow 4 x^{2}+x(a+8)+9=0\)

comparing the above equation with \(Ax^2-Bx+C=0\) we will get \(A=4\),\(B=(a+8)\),\(C=9\)

Now for equal roots of a quadratic equation \(B^2-4Ac=0\)

Can you now finish the problem ……….

Now \(B^2-4Ac=0\) becomes

\((a+8)^2-4\times 9 \times 4=0\)

\(\Rightarrow (a+8)^2=144\)

\(\Rightarrow (a+8)=\pm 12\)

\(\Rightarrow a=+4 \) & \(-20\)

Therefore The sum of the values of \(a=-20+4=-16\)

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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

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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