Categories

## 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 $P Q R ?$

,

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

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 ……….

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

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

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

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

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

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

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

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

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

Categories

## 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?

• $(2024)$
• $(2018)$
• $(2020)$

Algebra

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 .

Categories

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

Categories

## 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=$21$

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.

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.

Categories

## 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 $a ?$

• $31$
• $78$
• $43$

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 $78$

Categories

## 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 $10 \%$ ?

,

• $0$
• $1$
• $2$
• $3$
• $4$

algebra

Fraction

## Check the Answer

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 $1$

Categories

## 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

• $4$
• $5$
• $6$
• $7$
• $8$

Geometry

Square

Triangle

## Check the Answer

Answer: $8$

AMC-10A (2013) Problem 3

Pre College Mathematics

## Try with Hints

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 ……….

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

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

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

$\Rightarrow BE=8$

Categories

## 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?

,

• $48$
• $54$
• $60$
• $65$
• $72$

Algebra

LCM

## Check the Answer

Answer: $54$

AMC-10A (2013) Problem 17

Pre College Mathematics

## Try with Hints

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 ……….

$LCM(A, B)=12 x, LCM(B, C)=20 x, LCM(A, C)=15 x$
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……..

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 $72-18=54$

Categories

## 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 $\frac{M}{6}=\frac{6}{N}$

• $31$
• $78$
• $43$

Algebra

Order Pair

Multiplication

## Check the Answer

Answer: $78$

AMC-10A (2010) Problem 21

Pre College Mathematics

## Try with Hints

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……..

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……..

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