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## 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

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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## 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

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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## 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

least value

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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## 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

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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## 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

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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## 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

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=$21$

Categories

## 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

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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## 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

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 $78$

Categories

## 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

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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## 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

Answer: $1$

AMC-10A (2015) Problem 15

Pre College Mathematics

## Try with Hints

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 ……….

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……..

Therefore the Possible answer will be $1$