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AMC 10 Geometry Math Olympiad USA Math Olympiad

Problem on Circle and Triangle | AMC 10A, 2016 | Problem 21

Try this beautiful problem from Geometry: Problem on Circle and Triangle

Problem on Circle and Triangle – AMC-10A, 2016- Question 21


Circles with centers $P, Q$ and $R,$ having radii 1,2 and 3 , respectively, lie on the same side of line $l$ and are tangent to $l$ at $P^{\prime}, Q^{\prime}$ and $R^{\prime}$ respectively, with $Q^{\prime}$ between $P^{\prime}$ and $R^{\prime}$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $P Q R ?$

,

 

  • $0$
  • $\sqrt{6} / 3$
  • $1$
  • $\sqrt{6}-\sqrt{2}$
  • $\sqrt{6} / 2$

Key Concepts


Geometry

Circle

Triangle

Check the Answer


But try the problem first…

Answer: $\sqrt{6}-\sqrt{2}$

Source
Suggested Reading

AMC-10A (2016) Problem 21

Pre College Mathematics

Try with Hints


First hint

We have to find out area of the Triangle PQR. But PQR is not a Standard Triangle that we can find out eassily. Join $PP^{\prime}$, $QQ^{\prime}$, $RR^{\prime}$. Now we can find out PQR such that $\left[P^{\prime} P Q R R^{\prime}\right]$ in two different ways: $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]$ and $[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$, so $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

Can you now finish the problem ……….

Second Hint

$P^{\prime} Q^{\prime}=\sqrt{P Q^{2}-\left(Q Q^{\prime}-P P^{\prime}\right)^{2}}=\sqrt{9-1}=\sqrt{8}=2 \sqrt{2}$

$Q^{\prime} R^{\prime}=\sqrt{Q R^{2}-\left(R R^{\prime}-Q Q^{\prime}\right)^{2}}=\sqrt{5^{2}-1^{2}}=\sqrt{24}=2 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]=\frac{P^{\prime} P+Q^{\prime} Q}{2} * 2 \sqrt{2}=\frac{1+2}{2} * 2 \sqrt{2}=3 \sqrt{2}$

$\left[Q^{\prime} Q R R^{\prime}\right]=5 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]$ = $P^{\prime} R^{\prime}=P^{\prime} Q^{\prime}+Q^{\prime} R^{\prime}=2 \sqrt{2}+2 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]=4 \sqrt{2}+4 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

$3 \sqrt{2}+5 \sqrt{6}=4 \sqrt{2}+4 \sqrt{6}+[P Q R]$

$[P Q R]=\sqrt{6}-\sqrt{2}$

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Algebra AMC 10 Math Olympiad USA Math Olympiad

Least Possible Value Problem | AMC-10A, 2019 | Quesstion19

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    Try this beautiful problem from Algebra based on Least Possible Value.

    Least Possible Value – AMC-10A, 2019- Problem 19


    What is the least possible value of \(((x+1)(x+2)(x+3)(x+4)+2019)\)

    where (x) is a real number?

    • \((2024)\)
    • \((2018)\)
    • \((2020)\)

    Key Concepts


    Algebra

    quadratic equation

    least value

    Check the Answer


    But try the problem first…

    Answer: \((2018)\)

    Source
    Suggested Reading

    AMC-10A (2019) Problem 19

    Pre College Mathematics

    Try with Hints


    First hint

    To find out the least positive value of \((x+1)(x+2)(x+3)(x+4)+2019\), at first we have to expand the expression .\(((x+1)(x+2)(x+3)(x+4)+2019)\) \(\Rightarrow (x+1)(x+4)(x+2)(x+3)+2019=(x^2+5x+4)(x^2+5x+6)+2019)\)

    Let us take \(((x^2+5x+5=m))\)

    then the above expression becomes \(((m-1)(m+1)+2019)\) \(\Rightarrow m^2-1+2019\) \(\Rightarrow m^2+2018\)

    Can you now finish the problem ……….

    Second Hint

    Clearly in \((m^2+2018)…….(m^2)\) is positive ( squares of any number is non-negative) and least value is 0

    can you finish the problem……..

    Final Step

    Therefore minimum value of \(m^2+2108\) is \(2018\) since \(m^2 \geq 0\) for all m belongs to real .

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    AMC 10 Geometry Math Olympiad USA Math Olympiad

    Graph Coordinates | AMC 10A, 2015 | Question 12

    Try this beautiful Problem on Graph Coordinates from coordinate geometry from AMC 10A, 2015.

    Graph Coordinates – AMC-10A, 2015- Problem 12


    Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1 .$ What is $|a-b| ?$

    ,

    • $0$
    • $1$
    • $2$
    • $3$
    • \(4\)

    Key Concepts


    Co-ordinate geometry

    graph

    Distance Formula

    Check the Answer


    But try the problem first…

    Answer: $2$

    Source
    Suggested Reading

    AMC-10A (2015) Problem 12

    Pre College Mathematics

    Try with Hints


    First hint

    The given points are $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ which are satisfying the equation $y^{2}+x^{4}=2 x^{2} y+1$.

    So we can write $y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1$

    Can you now finish the problem ……….

    Second Hint

    Therefore

    $y^{2}+\pi^{2}=2 \pi y+1$
    $y^{2}-2 \pi y+\pi^{2}=1$
    $(y-\pi)^{2}=1$
    $y-\pi=\pm 1$
    $y=\pi+1$
    $y=\pi-1$

    can you finish the problem……..

    Third Hint:

    There are only two solutions to the equation, so one of them is the value of $a$ and the other is $b$. our requirement is $|a-b|$ so between a and b which is greater is not importent…………

    So, $|(\pi+1)-(\pi-1)|=2$

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    AMC 10 Geometry Math Olympiad USA Math Olympiad

    Arithmetic sequence | AMC 10A, 2015 | Problem 7

    Try this beautiful problem from Algebra: Arithmetic sequence from AMC 10A, 2015, Problem.

    Arithmetic sequence – AMC-10A, 2015- Problem 7


    How many terms are in the arithmetic sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$?

    • \(20\)
    • \(21\)
    • \(24\)
    • \(60\)
    • \(61\)

    Key Concepts


    Algebra

    Arithmetic sequence

    Check the Answer


    But try the problem first…

    Answer: \(21\)

    Source
    Suggested Reading

    AMC-10A (2015) Problem 7

    Pre College Mathematics

    Try with Hints


    First hint

    The given terms are $13$, $16$, $19$, $\dotsc$, $70$, $73$. We have to find out the numbers of terms…..

    If you look very carefully then the distance between two digits is \(3\).Therefore this is an Arithmetic Progression where the first term is \(13\) and common difference is \(3\)

    Can you now finish the problem ……….

    Second Hint

    $a+(n-1) d \Longrightarrow 13+(n-1) 3=73$

    \(\Rightarrow n=21\)

    The number of terms=\(21\)

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    AMC 10 Geometry Math Olympiad USA Math Olympiad

    Problem based on Cylinder | AMC 10A, 2015 | Question 9

    Try this beautiful problem from Mensuration: Problem based on Cylinder from AMC 10A, 2015.

    Cylinder – AMC-10A, 2015- Problem 9


    Two right circular cylinders have the same volume. The radius of the second cylinder is $10 \%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?

    • (A) The second height is $10 \%$ less than the first.
    • (B) The first height is $10 \%$ more than the second.
    • (C) The second height is $21 \%$ less than the first.
    • (D) The first height is $21 \%$ more than the second.
    • (E) The second height is $80 \%$ of the first.

    Key Concepts


    Mensuration

    Cylinder

    Check the Answer


    But try the problem first…

    Answer: (D) The first height is $21 \%$ more than the second.

    Source
    Suggested Reading

    AMC-10A (2015) Problem 9

    Pre College Mathematics

    Try with Hints


    First hint

    Let the radius of the first cylinder be $r_{1}$ and the radius of the second cylinder be $r_{2}$. Also, let the height of the first cylinder be $h_{1}$ and the height of the second cylinder be $h_{2}$.

    Can you now finish the problem ……….

    Second Hint

    According to the problem,

    $r_{2}=\frac{11 r_{1}}{10}$
    $\pi r_{1}^{2} h_{1}=\pi r_{2}^{2} h_{2}$

    can you finish the problem……..

    Third Hint:

    $r_{1}^{2} h_{1}=\frac{121 r_{1}^{2}}{100} h_{2} \Rightarrow h_{1}=\frac{121 h_{2}}{100}$

    Therefore the Possible answer will be (D) The first height is $21 \%$ more than the second.

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    AMC 10 Math Olympiad USA Math Olympiad

    Cubic Equation | AMC-10A, 2010 | Problem 21

    Try this beautiful problem based on Cubic Equation from AMC 10A, 2010.

    Cubic Equation – AMC-10A, 2010- Problem 21


    The polynomial $x^{3}-a x^{2}+b x-2010$ has three positive integer roots. What is the smallest possible value of $a ?$

    • \(31\)
    • \(78\)
    • \(43\)

    Key Concepts


    Algebra

    Cubic Equation

    Roots

    Check the Answer


    But try the problem first…

    Answer: \(78\)

    Source
    Suggested Reading

    AMC-10A (2010) Problem 21

    Pre College Mathematics

    Try with Hints


    First hint

    The given equation is $x^{3}-a x^{2}+b x-2010$

    Comparing the equation with \(Ax^3+Bx^2+Cx+D=0\) we get \(A=1,B=-a,C=b,D=0\)

    Let us assume that \(x_1,x_2,x_3\) are the roots of the above equation then using vieta’s formula we can say that \(x_1.x_2.x_3=2010\)

    Therefore if we find out the factors of \(2010\) then we can find out our requirement…..

    can you finish the problem……..

    Second Hint

    \(2010\) factors into $2 \cdot 3 \cdot 5 \cdot 67 .$ But, since there are only three roots to the polynomial,two of the four prime factors must be multiplied so that we are left with three roots and we have to find out the smallest positive values of \(a\)

    can you finish the problem……..

    Final Step

    To minimize $a, 2$ and 3 should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is \(78\)

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    AMC 10 Geometry Math Olympiad USA Math Olympiad

    Problem on Fraction | AMC 10A, 2015 | Question 15

    Try this beautiful Problem on Fraction from Algebra from AMC 10A, 2015.

    Fraction – AMC-10A, 2015- Problem 15


    Consider the set of all fractions $\frac{x}{y},$ where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by 1 , the value of the fraction is increased by $10 \%$ ?

    ,

    • $0$
    • $1$
    • $2$
    • $3$
    • \(4\)

    Key Concepts


    algebra

    Fraction

    Check the Answer


    But try the problem first…

    Answer: $1$

    Source
    Suggested Reading

    AMC-10A (2015) Problem 15

    Pre College Mathematics

    Try with Hints


    First hint

    Given that $\frac{x}{y},$ is a fraction where $x$ and $y$ are relatively prime positive integers. We have to find out the numbers of fraction if both numerator and denominator are increased by 1.

    According to the question we have $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

    Can you now finish the problem ……….

    Second Hint

    Now from the equation we can say that $x+1>\frac{11}{10} \cdot x$ so $x$ is at most 9
    By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11 x}{10-x}$. since $x$ and $y$ are integer, this only has solutions for $x \in{5,8,9} .$ However, only the first yields a $y$ that is relative prime to $x$

    can you finish the problem……..

    Third Hint:

    Therefore the Possible answer will be \(1\)

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    AMC 10 Geometry Math Olympiad USA Math Olympiad

    Side of Square | AMC 10A, 2013 | Problem 3

    Try this beautiful problem from Geometry: Side of Square.

    Sides of Square – AMC-10A, 2013- Problem 3


    Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?

    ,

     i

    Side of Square - Problem
    • $4$
    • $5$
    • $6$
    • $7$
    • \(8\)

    Key Concepts


    Geometry

    Square

    Triangle

    Check the Answer


    But try the problem first…

    Answer: $8$

    Source
    Suggested Reading

    AMC-10A (2013) Problem 3

    Pre College Mathematics

    Try with Hints


    First hint

    Side of Square

    Given that Square $ABCD$ has side length $10$ and area of $\triangle ABE$ is $40$.we have to find out length of \(BE\) where \(E\) is the point on \(BC\). we know area of the \(\triangle ABE=\frac{1}{2} AB.BE=40\)

    Can you find out the side length of \(BE\)?

    Can you now finish the problem ……….

    Second Hint

    Side of Square

    \(\triangle ABE=\frac{1}{2} AB.BE=40\)

    \(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

    \(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

    \(\Rightarrow BE=8\)

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    AMC 10 Geometry Math Olympiad USA Math Olympiad

    Counting Days | AMC 10A, 2013 | Problem 17

    Try this beautiful problem from Algebra: Counting Days

    Counting Days – AMC-10A, 2013- Problem 17


    Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365 -day period will exactly two friends visit her?

    ,

    • $48$
    • $54$
    • $60$
    • $65$
    • \(72\)

    Key Concepts


    Algebra

    LCM

    Check the Answer


    But try the problem first…

    Answer: $54$

    Source
    Suggested Reading

    AMC-10A (2013) Problem 17

    Pre College Mathematics

    Try with Hints


    First hint

    Given that Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day.

    According to the questation , Let us assume that $A=3 x B=4 x C=5 x$
    Now, we want the days in which exactly two of these people meet up
    The three pairs are $(A, B),(B, C),(A, C)$
    Can yoiu find out the LCM of each pair……….

    Can you now finish the problem ……….

    Second Hint

    \(LCM(A, B)=12 x, LCM(B, C)=20 x, LCM(A, C)=15 x\)
    Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.
    Hence, $LCM(A, B, C)=60 x$

    can you finish the problem……..

    Third Hint:

    Now, we add all of the days up(including overcount).
    We get $30+18+24=72 .$ Now, because $60(6)=360,$ we have to subtract 6 days from every pair. Hence, our answer is $72-18=54$

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    AMC 10 Math Olympiad USA Math Olympiad

    Order Pair | AMC-10B, 2012 | Problem 10

    Try this beautiful problem from Algebra: Order Pair

    Order Pair – AMC-10B, 2012- Problem 10


    How many ordered pairs of positive integers (M,N) satisfy the equation $\frac{M}{6}=\frac{6}{N}$

    • \(31\)
    • \(78\)
    • \(43\)

    Key Concepts


    Algebra

    Order Pair

    Multiplication

    Check the Answer


    But try the problem first…

    Answer: \(78\)

    Source
    Suggested Reading

    AMC-10A (2010) Problem 21

    Pre College Mathematics

    Try with Hints


    First hint

    Given that $\frac{M}{6}=\frac{6}{N}$ \(\Rightarrow MN=36\).Next we have to find out the the Possibilities to getting \(a \times b=36\)

    can you finish the problem……..

    Second Hint

    Now the possibilities are ….

    $1 \times 36=36$
    $2 \times 18=36$
    $3 \times 12=36$
    $4 \times 9=36$
    $6 \times 6=36$

    We can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order.

    can you finish the problem……..

    Final Step

    Therefore the total Possible order pairs that satisfy the equation $\frac{M}{6}=\frac{6}{N}$=\(9\)

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