Categories

## Problem on Curve | AMC 10A, 2018 | Problem 21

Try this beautiful Problem on Algebra based on Problem on Curve from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## Curve- AMC 10A, 2018- Problem 21

Which of the following describes the set of values of $a$ for which the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ in the real $x y$ -plane intersect at
exactly 3 points?

• $a=\frac{1}{4}$
• $\frac{1}{4}<a<\frac{1}{2}$
• $a>\frac{1}{4}$
• $a=\frac{1}{2}$
• $a>\frac{1}{2}$

Algebra

greatest integer

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-14

#### Check the answer here, but try the problem first

$a>\frac{1}{2}$

## Try with Hints

#### First Hint

We have to find out the value of $a$

Given that $y=x^{2}-a$ . now if we Substitute this value in $x^{2}+y^{2}=a^{2}$ we will get a quadratic equation of $x$ and $a$. if you solve this equation you will get the value of $a$

Now can you finish the problem?

#### Second Hint

After substituting we will get $x^{2}+\left(x^{2}-a\right)^{2}$=$a^{2} \Longrightarrow x^{2}+x^{4}-2 a x^{2}=0 \Longrightarrow x^{2}\left(x^{2}-(2 a-1)\right)=0$

therefore we can say that either $x^2=0\Rightarrow x=0$ or $x^2-(2a-1)=0$

$\Rightarrow x=\pm \sqrt {2a-1}$. Therefore

Now Can you finish the Problem?

#### Third Hint

Therefore $\sqrt {2a-1} > 0$

$\Rightarrow a>\frac{1}{2}$

Categories

## Finding Greatest Integer | AMC 10A, 2018 | Problem No 14

Try this beautiful Problem on Algebra based on finding greatest integer from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## Finding Greatest Integer – AMC-10A, 2018- Problem 14

What is the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

• $80$
• $81$
• $96$
• $97$
• $625$

Algebra

greatest integer

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-14

#### Check the answer here, but try the problem first

$80$

## Try with Hints

#### First Hint

The given expression is $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

We have to find out the greatest integer which is less than or equal to the given expression .

Let us assaume that $x=3^{96}$ and $y=2^{96}$

Therefore the given expression becoms $\frac{81 x+16 y}{x+y}$

Now can you finish the problem?

#### Second Hint

Now $\frac{81 x+16 y}{x+y}$

=$\frac{16 x+16 y}{x+y}+\frac{65 x}{x+y}$

$=16+\frac{65 x}{x+y}$

Now if we look very carefully we see that $\frac{65 x}{x+y}<\frac{65 x}{x}=65$

Therefore $16+\frac{65 x}{x+y}<16+65=81$

Now Can you finish the Problem?

#### Third Hint

Therefore less than $81$ , the answer will be $80$

Categories

## Right-angled shaped field | AMC 10A, 2018 | Problem No 23

Try this beautiful Problem on Geometry based on Right-angled shaped field from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## Right-angled shaped field – AMC-10A, 2018- Problem 23

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted?

,

• $\frac{25}{27}$
• $\frac{26}{27}$
• $\frac{73}{75}$
• $\frac{145}{147}$
• $\frac{74}{75}$

Geometry

Triangle

Pythagoras

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-23

#### Check the answer here, but try the problem first

$\frac{145}{147}$

## Try with Hints

#### First Hint

Given that ABC is a right-angle Triangle field . Here The corner at $B$ is shaded region which is unplanted. now we have to find out fraction of the field is planted?

Now if we join the triangle with the dotted lines then it will be divided into three triangles as shown below…

Therefore there are three triangles . Now if we can find out the area of three triangles and area of the smaller square then it will be eassy to say….

Now can you finish the problem?

#### Second Hint

Let $x$ be the side length of the sqare then area will be$x^2$

Now area of two thin triangle will be $\frac{x(3-x)}{2}$ and $\frac{x(4-x)}{2}$

area of the other triangle will be $\frac{1}{2}\times 5 \times 2=5$

area of the $\triangle ABC =\frac{1}{2}\times 3 \times 4=6$

Now Can you finish the Problem?

#### Third Hint

Therefore we can say that $x^{2}+\frac{x(3-x)}{2}+\frac{x(4-x)}{2}+5=6$

$\Rightarrow x=\frac{2}{7}$

Therefore area of the small square will be $\frac{4}{49}$

Thererfore our required fraction =Area of the $\triangle ABC$-area of the smaller square=$6- \frac{4}{49}$=$\frac{145}{147}$

Categories

## Area of region | AMC 10B, 2016| Problem No 21

Try this beautiful Geometry Problem based on area of region from AMC 10 B, 2016. You may use sequential hints to solve the problem.

## Area of region– AMC-10B, 2016- Problem 21

What is the area of the region enclosed by the graph of the equation $x^{2}+y^{2}=|x|+|y| ?$

,

• $\pi+\sqrt{2}$
• $\pi+2$
• $\pi+2 \sqrt{2}$
• $2 \pi+\sqrt{2}$
• $2 \pi+2 \sqrt{2}$

Geometry

Semi circle

graph

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10B, 2016 Problem-21

#### Check the answer here, but try the problem first

$\pi+2$

## Try with Hints

#### First Hint

The given equation is $x^{2}+y^{2}=|x|+|y|$. Expanding this equation we get four equation as mod exist here…

$x^2+y^2-x-y=0$…………………..(1)

$x^2+y^2+x+y=0$………………..(2)

$x^2+y^2-x+y=0$…………………(3)

$x^2+y^2+x-y=0$…………………(4)

using this four equation can you draw the figure ?

Now can you finish the problem?

#### Second Hint

now four equations can be written as $x^{2}-x+y^{2}-y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}+y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}-x+y^{2}+y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}-y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$ which represents four circles and they overlapping…..

The center of the four circles are $\left(\frac{1}{2}, \frac{1}{2}\right)$, $\left(\frac{-1}{2}, \frac{-1}{2}\right)$,$\left(\frac{1}{2}, \frac{-1}{2}\right)$,$\left(\frac{-1}{2}, \frac{1}{2}\right)$Now we have to find out the region union of the four circles.

Now can you finish the problem?

#### Third Hint

There are several ways to find the area, but note that if you connect (0,1),(1,0),(-1,0),(0,-1) to its other three respective points in the other three quadrants, you get a square of area 2 , along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$

Categories

## Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability based on Coin toss from AMC 10 A, 2017. You may use sequential hints to solve the problem.

## Coin Toss – AMC-10A, 2017- Problem 18

Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $q-p ?$

,

• $1$
• $2$
• $3$
• $4$
• $5$

combinatorics

Coin toss

Probability

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2017 Problem-18

#### Check the answer here, but try the problem first

$4$

## Try with Hints

#### First Hint

Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins.

Now can you finish the problem?

#### Second Hint

Let $P$ be the probability Amelia wins. Note that $P=$ chance she wins on her first turn $+$ chance she gets to her second turn $\cdot \frac{1}{3}+$ chance she gets to her third turn $\cdot \frac{1}{3} \ldots$ This can be represented by an infinite geometric series,

Therefore the value of $P$ will be $P=\frac{\frac{1}{3}}{1-\frac{2}{3} \cdot \frac{3}{5}}=\frac{\frac{1}{3}}{1-\frac{2}{5}}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3} \cdot \frac{5}{3}=\frac{5}{9}$ which is of the form $\frac{p}{q}$

Now Can you finish the Problem?

#### Third Hint

Therefore $q-p=9-5=4$

Categories

## GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry based on GCF & Rectangle from AMC 10 A, 2010. You may use sequential hints to solve the problem.

## GCF & Rectangle – AMC-10A, 2016- Problem 19

In rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $B E=E F=F C$. Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q$, respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$ where the greatest common factor of $r, s,$ and $t$ is $1 .$ What is $r+s+t ?$

,

• $7$
• $9$
• $12$
• $15$
• $20$

Geometry

Rectangle

Diagonal

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2016 Problem-19

#### Check the answer here, but try the problem first

$20$

## Try with Hints

#### First Hint

Given that , rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q,$ respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$. we have to find out $r+s+t ?$, where greatest common factor of $r,s,t$ is $1$

Now $\triangle A P D \sim \triangle E P B$. From this relation we can find out a relation between $DP$ and $PB$

Now can you finish the problem?

#### Second Hint

Now $\triangle A P D \sim \triangle E P B$$\Rightarrow$ $\frac{D P}{P B}=\frac{A D}{B E}=3$ Therefore $P B=\frac{B D}{4}$.

SimIarly from the $\triangle AQD \sim \triangle BQF$ $\Rightarrow$$\frac{D Q}{Q B}=\frac{3}{2}$

Therefore we can say that $D Q=\frac{3 \cdot B D}{5}$

Now can you finish the problem?

#### Third Hint

Therefore $r: s: t=\frac{1}{4}: \frac{2}{5}-\frac{1}{4}: \frac{3}{5}=5: 3: 12,$ so $r+s+t$=$20$

Categories

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Geometry Problem based on a fly trapped inside cubical box from AMC 10 A, 2010. You may use sequential hints to solve the problem.

## Fly trapped inside cubical boxÂ – AMC-10A, 2010- Problem 20

A fly trapped inside a cubical box with side length 1 meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

,

• $4+4 \sqrt{2}$
• $2+4 \sqrt{2}+2 \sqrt{3}$
• $2+3 \sqrt{2}+3 \sqrt{3}$
• $4 \sqrt{2}+4 \sqrt{3}$
• $3 \sqrt{2}+5 \sqrt{3}$

Geometry

Cube

Diagonal

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2010 Problem-20

#### Check the answer here, but try the problem first

$4 \sqrt{2}+4 \sqrt{3}$

## Try with Hints

#### First Hint

Suppose the fly starts from the point $A$.we have to find out  the maximum possible length. The maximum possible length will be from one corner to another corner such as …..

$A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A$

Now can you find out this maximum path?

Now can you finish the problem?

#### Second Hint

Given that the side length of the cube is $1$. Therefore the diagonal $AC$=$\sqrt 2$ and the diagonal $AG=\sqrt 3$. Now we have to find out the path $AG+GB+BH+HC+CE+ED+DF+FA$.Can you find it ?

Now can you finish the problem?

#### Third Hint

$AG+GB+BH+HC+CE+ED+DF+FA$=$\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2$=($4\sqrt3+4\sqrt2$)

Categories

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry based on Measure of angle from AMC 10 A, 2014. You may use sequential hints to solve the problem.

## Measure of angle  – AMC-10A, 2019- Problem 13

Let $\triangle A B C$ be an isosceles triangle with $B C=A C$ and $\angle A C B=40^{\circ} .$ Construct the circle with diameter $\overline{B C}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{A C}$ and $\overline{A B}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $B C D E .$ What is the degree measure of $\angle B F C ?$

,

• $90$
• $100$
• $105$
• $110$
• $120$

Geometry

Circle

Triangle

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2019 Problem-`13

#### Check the answer here, but try the problem first

$110^{\circ}$

## Try with Hints

#### First Hint

According to the questation we draw the diagram. we have to find out $\angle BFC$

Now $\angle BEC$ = $\angle BDC$ =$90^{\circ}$ (as they are inscribed in a semicircle)

$\angle A C B=40^{\circ} .$ Therefore we can say that $\angle ABC=70^{\circ}$ (as $\triangle A B C$ be an isosceles triangle with $B C=A C$)

Can you find out the value of $\angle B F C ?$

Now can you finish the problem?

#### Second Hint

As $\angle ABC=70^{\circ}$ and $\angle BEC=90^{\circ}$ Therefore $\angle E C B=20^{\circ}$( as sum of the angles of a triangle is$180^{\circ}$

Similarly $\angle D B C=50^{\circ}$

Now Can you finish the Problem?

#### Third Hint

Now $\angle B D C+\angle D C B+\angle D B C=180^{\circ} \Longrightarrow 90^{\circ}+40^{\circ}+\angle D B C=180^{\circ} \Longrightarrow \angle D B C$=$50^{\circ}$

$\angle B E C+\angle E B C+\angle E C B=180^{\circ} \Longrightarrow 90^{\circ}+70^{\circ}+\angle E C B=180^{\circ} \Rightarrow \angle E C B$=$20^{\circ}$

we take triangle $B F C$, and find $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$

Categories

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra based on Recursion from AMC 10 A, 2019. You may use sequential hints to solve the problem.

## Recursion- AMC-10A, 2019- Problem 15

A sequence of numbers is defined recursively by $a_{1}=1, a_{2}=\frac{3}{7},$ and $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

• $2020$
• $4039$
• $6057$
• $6061$
• $8078$

### Key Concepts

Algebra

Recursive formula

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2019 Problem-15

#### Check the answer here, but try the problem first

$8078$

## Try with Hints

#### First Hint

The given expression is $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$ and given that $a_{1}=1, a_{2}=\frac{3}{7}$. we have to find out $a_{2019}$?

at first we may use recursive formula we can find out $a_3$ , $a_4$ with the help of $a_1$, $a_2$. later we can find out $a_n$

Now can you finish the problem?

#### Second Hint

Given that $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

Now $n=3$ then $a_{3}=\frac{a_{(3-2)} \cdot a_{(3-1)}}{2 a_{(3-2)}-a_{(3-1)}}$

$\Rightarrow$ $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

$\Rightarrow$ $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

$\Rightarrow$ $a_{3}=\frac{1*\frac{3}{7}}{2*1-\frac{3}{7}}$

$\Rightarrow$ $a_{3}=\frac{3}{7}$

Similarly if we put $n=4$ we get $a_4=\frac{3}{15}$ (where $a_{1}=1, a_{2}=\frac{3}{7}$,$a_3=\frac{3}{7}$)

Continue this way we $a_{n}=\frac{3}{4 n-1}$

So can you find out the value of $a_{2019}$?

Now Can you finish the Problem?

#### Third Hint

Now $a_{n}=\frac{3}{4 n-1}$

Put $n=2019$

$a_{2019}=\frac{3}{8075}$ which is the form of $\frac{p}{q}$

Therefore $p+q=8078$

Categories

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra based on Roots of Polynomial from AMC 10 A, 2019. You may use sequential hints to solve the problem.

## Algebra- AMC-10A, 2019- Problem 24

Let $p, q,$ and $r$ be the distinct roots of the polynomial $x^{3}-22 x^{2}+80 x-67$. It is given that there exist real numbers $A, B$, and $C$ such that $\frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r}$

for all $s \notin{p, q, r} .$ What is $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} ?$

,

• $243$
• $244$
• $245$
• $246$
• $247$

Algebra

Linear Equation

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2019 Problem-24

#### Check the answer here, but try the problem first

$244$

## Try with Hints

#### First Hint

The given equation is $\frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r}$…………………(1)

If we multiply both sides we will get

Multiplying both sides by $(s-p)(s-q)(s-r)$ we will get
$$1=A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q)$$

Now can you finish the problem?

#### Second Hint

Now Put $S=P$ we will get $\frac{1}{A}=(p-q)(p-r)$…………(2)

Now Put $S=q$ we will get $\frac{1}{B}=(q-p)(q-r)$………..(3)

Now Put $S=r$ we will get $\frac{1}{C}=(r-p)(r-q)$………..(4)

Now Can you finish the Problem?

#### Third Hint

Adding (2) +(3)+(4) we get,$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^{2}+q^{2}+r^{2}-p q-q r-p r$

Now Using Vieta’s Formulas, $p^{2}+q^{2}+r^{2}=(p+q+r)^{2}-2(p q+q r+p r)=324$ and $p q+q r+p r=80$

Therefore the required answer is $324-80$=$244$