External Tangent | AMC 10A, 2018 | Problem 15

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

External Tangent - AMC-10A, 2018- Problem 15

Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$, as shown in the diagram. The distance $A B$ can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n ?$

$21$
$29$
$58$
$69$
$93$

Key Concepts

Geometry

Triangle

Pythagoras

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Source of the problem

AMC-10A, 2018 Problem-15

Check the answer here, but try the problem first

$69$

Try with Hints

First Hint

Given that two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$. we have to find out the length $AB$.

Now join $A$ & $B$ and the points $Y$ & $Z$. If we can show that $\triangle XYZ \sim \triangle XAB$ then we can find out the length of $AB$.

Now can you finish the problem?

Second Hint

now the length of $YZ=5+5=10$ (as the length of the radius of smaller circle is $5$) and $XY=XA-AY=13-5=8$. Now $YZ|| AB$.therefore we can say that $\triangle XYZ \sim \triangle XAB$. therefore we can write $\frac{X Y}{X A}=\frac{Y Z}{A B}$

Now Can you finish the Problem?

Third Hint

From the relation we can say that $\frac{X Y}{X A}=\frac{Y Z}{A B}$

$\Rightarrow \frac{8}{13}=\frac{10}{AB}$

$\Rightarrow AB=\frac{13\times 10}{8}$

$\Rightarrow AB=\frac{65}{4}$ which is equal to $\frac{m}{n}$

Therefore $m+n=65+4=69$

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Dice Problem | AMC 10A, 2014| Problem No 17

Try this beautiful Problem on Probability based on Dice from AMC 10 A, 2014. You may use sequential hints to solve the problem.

Dice Problem - AMC-10A, 2014 - Problem 17

Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

$\frac{1}{6}$
$\frac{13}{72}$
$\frac{7}{36}$
$\frac{5}{24}$
$\frac{2}{9}$

Key Concepts

combinatorics

Dice-problem

Probability

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2014 Problem-17

Check the answer here, but try the problem first

$\frac{5}{24}$

Try with Hints

First Hint

Total number of dice is $3$ and each dice $6$ possibility. therefore there are total $6^{3}=216$ total possible rolls. we have to find out the probability that the values shown on two of the dice sum to the value shown on the remaining die.

Without cosidering any order of the die , the possible pairs are $(1,1,2),(1,2,3),(1,3,4)$,$(1,4,5),(1,5,6),(2,2,4),(2,3,5)$,$(2,4,6),(3,3,6)$

Now can you finish the problem?

Second Hint

Clearly $(1,1,1).(2,2,4),(3,3,6)$ this will happen in $\frac{3 !}{2}=3$ way

$(1,2,3),(1,3,4)$,$(1,4,5),(1,5,6),(2,3,5)$,$(2,4,6),$this will happen in $3 !=6$ ways

Now Can you finish the Problem?

Third Hint

Therefore, total number of ways $3\times3+6\times6=45$ so that sum of the two dice will be the third dice

Therefore the required answer is $\frac{45}{216}$=$\frac{5}{24}$

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?

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

Key Concepts

Algebra

greatest integer

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2018 Problem-14

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

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

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

Right-angled Triangle | AMC 10A, 2018 | Problem No 16

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

Right-angled triangle - AMC-10A, 2018- Problem 16

Right triangle $A B C$ has leg lengths $A B=20$ and $B C=21$. Including $\overline{A B}$ and $\overline{B C}$, how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{A C} ?$

$5$
$8$
$12$
$13$
$15$

Key Concepts

Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2018 Problem-16

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$13$

Try with Hints

First Hint

Given that $\triangle ABC$ is a Right-angle triangle and $AB=20$ and $BC=21$. we have to find out how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$?

Let $P$ be the foot of the altitude from $B$ to $AC$. therefore $BP$ is the shortest legth . $B P=\frac{20 \cdot 21}{29}$ which is between $14$ and $15$.

Now can you finish the problem?

Second Hint

let us assume a line segment $BY$ with $Y$ on $AC$which is starts from $A$ to $P$ . So if we move this line segment the length will be decreases and the values will be look like as $20,.....,15$. similarly if we moving this line segment $Y$ from $P$ to $C$ hits all the integer values from $15, 16, \dots, 21$.

Now Can you finish the Problem?

Third Hint

Therefore numbers of total line segments will be $13$

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 $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

$80$
$81$
$96$
$97$
$625$

Key Concepts

Algebra

greatest integer

Suggested Book | Source | Answer

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 $80$

Length of the crease | AMC 10A, 2018 | Problem No 13

Try this beautiful Problem on Geometry based on Length of the crease from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Length of the crease- AMC-10A, 2018- Problem 13

A paper triangle with sides of lengths $3,4,$ and 5 inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?

$1+\frac{1}{2} \sqrt{2}$
$\sqrt 3$
$\frac{7}{4}$
$\frac{15}{8}$
$2$

Key Concepts

Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2018 Problem-13

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$\frac{15}{8}$

Try with Hints

First Hint

Given that ABC is a right-angle triangle shape paper. Now by the problem the point $A$ move on point $B$ . Therefore a crease will be create i.e $DE$ . noe we have to find out the length of $DE$?

If you notice very carefully then $DE$ is the perpendicular bisector of the line $AB$. Therefore the $\triangle ADE$ is Right-angle triangle. Now the side lengths of $AC$,$AB$,$BC$ are given. so if we can so that the $\triangle ADE$ $\sim$ $\triangle ABC$ then we can find out the side length of $DE$?

Now can you finish the problem?

Second Hint

In $\triangle ABC$ and $\triangle ADE$ we have ...

$\angle A=\angle A$( common angle)

$\angle C=\angle ADE$ (Right angle)

Therefore the remain angle will be equal ....

Therefore we can say that $\triangle ADE$ $\sim$ $\triangle ABC$

Now Can you finish the Problem?

Third Hint

As $\triangle ADE$ $\sim$ $\triangle ABC$ therefore we can write

$\frac{B C}{A C}=\frac{D E}{A D} \Rightarrow \frac{3}{4}=\frac{D E}{\frac{5}{2}} \Rightarrow D E=\frac{15}{8}$

Therefore the length in inches of the crease is $\frac{15}{8}$

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?

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

Key Concepts

Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer

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=$6- \frac{4}{49}$=$\frac{145}{147}$

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 $x^{2}+y^{2}=|x|+|y| ?$

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

Key Concepts

Geometry

Semi circle

graph

Suggested Book | Source | Answer

Source of the problem

AMC-10B, 2016 Problem-21

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$\pi+2$

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 $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$

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 $q-p ?$

Key Concepts

combinatorics

Coin toss

Probability

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2017 Problem-18

Check the answer here, but try the problem first

$4$

Try with Hints

First Hint

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 $q-p=9-5=4$

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 $r+s+t ?$

$7$
$9$
$12$
$15$
$20$

Key Concepts

Geometry

Rectangle

Diagonal

Suggested Book | Source | Answer

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 $r+s+t$=$20$