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Geometry Math Olympiad PRMO USA Math Olympiad

Area of the Trapezium | PRMO-2017 | Question 30

Try this beautiful Problem from Geometry based on Area of the Trapezium from PRMO 2017.

Area of the Trapezium – PRMO 2017, Problem 30


Consider the areas of the four triangles obtained by drawing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576 , determine the square root of the maximum possible area of the trapezium to the nearest integer.

  • $9$
  • $40$
  • $13$
  • $20$

Key Concepts


Geometry

Triangle

Trapezium

Check the Answer


Answer:$13$

PRMO-2017, Problem 30

Pre College Mathematics

Try with Hints


Let $x, y, z, w$ be areas of the four triangles as shown in figure.
then area of \(\triangle ADB\)= Area of \(\triangle ACB\)
$\Rightarrow x+y=x+w \Rightarrow y=w$

Also \(\frac{AE}{EC}\)=\(\frac{area of \triangle ADE}{area of \triangle DEF}\)=\(\frac{area of \triangle AEB}{area of \triangle BEC}\)
$\Rightarrow \frac{y}{z}=\frac{x}{w}=\frac{x}{y} \Rightarrow y^{2}=z x$
$\Rightarrow z, y, x$ are in G.P.

Can you now finish the problem ……….

Let $y=z r$ and $x=z r^{2},$ where $r \geq 1$ To make area of trapezium ABCD maximum, we take $z y=z^{2} r=576$
and $y w=z^{2} r^{2}=1296$
As $( z \leq y \leq x)$
Therefore $ \frac{z^{2} r^{2}}{z^{2} r}=\frac{1296}{576} \Rightarrow r=\frac{9}{4} \Rightarrow z=16$

Can you finish the problem……..

Therefore area of trapezium $\mathrm{ABCD}$
$=x+y+z+w=z r^{2}+2 z r+z$
$=z(1+r)^{2}=16\left(1+\frac{9}{4}\right)^{2}=13^{2}$
Therefore Answer is $13 .$

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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


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

AMC-10A (2016) Problem 21

Pre College Mathematics

Try with Hints


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 ……….

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

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

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

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

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

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

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

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

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

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Algebra AMC 10 Math Olympiad USA Math Olympiad

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

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


Answer: \((2018)\)

AMC-10A (2019) Problem 19

Pre College Mathematics

Try with Hints


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 ……….

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……..

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


Answer: $2$

AMC-10A (2015) Problem 12

Pre College Mathematics

Try with Hints


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 ……….

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……..

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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Algebra Math Olympiad PRMO USA Math Olympiad

Positive Integer | PRMO-2017 | Question 1

Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

Positive Integer – PRMO 2017, Question 1


How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by $3 ?$

  • $9$
  • $7$
  • $28$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$28$

PRMO-2017, Problem 1

Pre College Mathematics

Try with Hints


Let $n$ be the positive integer less than 1000 and $s$ be the sum of its digits, then $3 \mid n$ and $7 \mid s$
$3|n \Rightarrow 3| s$
therefore$21| s$

Can you now finish the problem ……….

Also $n<1000 \Rightarrow s \leq 27$
therefore $\mathrm{s}=21$
Clearly, n must be a 3 digit number Let $x_{1}, x_{2}, x_{3}$ be the digits, then $x_{1}+x_{2}+x_{3}=21$
where $1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9$
$\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18$
$\Rightarrow x_{1} \geq 3$

Can you finish the problem……..

For $x_{1}=3,4, \ldots ., 9,$ the equation (1) has $1,2,3, \ldots ., 7$ solutions
therefore total possible solution of equation (1)

=$1+2+\ldots+7=\frac{7 \times 8}{2}=28$

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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


Answer: \(21\)

AMC-10A (2015) Problem 7

Pre College Mathematics

Try with Hints


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 ……….

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

\(\Rightarrow n=21\)

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


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

AMC-10A (2015) Problem 9

Pre College Mathematics

Try with Hints


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 ……….

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……..

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

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


Answer: \(78\)

AMC-10A (2010) Problem 21

Pre College Mathematics

Try with Hints


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……..

\(2010\) 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……..

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 8 Math Olympiad USA Math Olympiad

Median of numbers | AMC-10A, 2020 | Problem 11

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020.

Median of numbers – AMC-10A, 2020- Problem 11


What is the median of the following list of $4040$ numbers$?$

\(1,2,3,…….2020,1^2,2^2,3^2………..{2020}^2\)

  • $1989.5$
  • $ 1976.5$
  • $1972.5$

Key Concepts


Median

Algebra

square numbers

Check the Answer


Answer: $1976.5$

AMC-10A (2020) Problem 11

Pre College Mathematics

Try with Hints


To find the median we need to know how many terms are there and the position of the numbers .here two types of numbers, first nonsquare i.e (1,2,3……2020) and squares numbers i.e \((1^2,2^2,3^2……2020^2)\).so We want to know the \(2020\)th term and the \(2021\)st term to get the median.

Can you now finish the problem ……….

Now less than 2020 the square number is \({44}^2\)=1936 and if we take \({45}^2\)=2025 which is greater than 2020.therefore we take the term that \(1,2,3…2020\) trms + 44 terms=\(2064\) terms.

can you finish the problem……..

since $44^{2}$ is $44+45=89$ less than $45^{2}=2025$ and 84 less than 2020 we will only need to consider the perfect square terms going down from the 2064 th term, 2020, after going down $84$ terms. Since the $2020$th and $2021$st terms are only $44$ and $43$ terms away from the $2064$th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$. Averaging the two,=\(1976.5\)

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Algebra Math Olympiad PRMO USA Math Olympiad

Pen & Note Books Problem| PRMO-2017 | Question 8

Try this beautiful Pen & Note Books Problem from Algebra, from PRMO 2017.

Pen & Note Books – PRMO 2017, Question 8


A pen costs $Rs 11$ and a notebook costs $Rs. 13 .$ Find the number of ways in which a person can spend exactly Rs.1000 to buy pens and notebooks.

  • $9$
  • $7$
  • $11$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$7$

PRMO-2017, Problem 8

Pre College Mathematics

Try with Hints


Given A pen costs Rs.\(11\) and a note book costs Rs.\(13\)

$11 x+13 y=1000$………………….(1)

Can you now finish the problem ……….

Now $11 x+13 y=1000$
$\Rightarrow 11 x=1000-13 y=(1001-11 y)-(2 y+1)$

=$11(91-y)-(2 y+1)$
$\Rightarrow 11 | 2 y+1$
Let $2 y+1=11(2 k-1), k \in I^{+}$
$\Rightarrow y=11 \mathrm{k}-6$
therefore $11 x=11(97-11 k)-11(2 k-1)$
$\Rightarrow x=98-13 k$
But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$

Can you finish the problem……..

But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$
therefore for each $\mathrm{k} \in{1,2, \ldots ., 7},$ we get a unique pair $(\mathrm{x}, \mathrm{y})=(98-13 \mathrm{k}, 11 \mathrm{k}-6)$ satisfying equation
Hence 7 ways are possible.

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