Categories

## 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 $\overline{A C} ?$

,

• $5$
• $8$
• $12$
• $13$
• $15$

Geometry

Triangle

Pythagoras

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-16

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

$13$

## 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 $13$

Categories

## 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?

,

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

Geometry

Triangle

Pythagoras

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-13

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

$\frac{15}{8}$

## 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 $\frac{15}{8}$

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

## 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

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra based on Set of Fractions from AMC 10 A, 2015. You may use sequential hints to solve the problem.

## Set of Fractions – 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 $10 \% ?$

,

• $0$
• $1$
• $2$
• $3$
• $infinitely many$

Algebra

fraction

factorization

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2015 Problem-15

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

$1$

## Try with Hints

#### First Hint

According to the questation we can write $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

$\Rightarrow xy +11x-10y=0$

$\Rightarrow (x-10)(y-11)=-110$

Now can you finish the problem?

#### Second Hint

Here $x$ and $y$ must positive, so $x>0$ and $y>0$, so $x-10>-10$ and $y+11>11$

Now we have to find out the factors of $110$ and find out the possible pairs to fulfill the condition….

Now Can you finish the Problem?

#### Third Hint

uses the factors of $110$ , we can get the factor pairs: $(-1,110),(-2,55),$ and $(-5,22)$
But we can’t stop here because $x$ and $y$ must be relatively prime.
$(-1,110 )$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn’t work.
$(-2,55 )$ gives $x=8$ and $y=44$. This doesn’t work.
$(-5,22)$ gives $x=5$ and $y=11$. This does work.

Therefore the one solution exist

Categories

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Geometry based on Positive Integers and Quadrilateral from AMC 10 A, 2015. You may use sequential hints to solve the problem.

## Positive Integers and Quadrilateral – AMC-10A, 2015- Problem 24

For some positive integers $p$, there is a quadrilateral $A B C D$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C, A B=2$, and $C D=A D$. How many different values of $p<2015$ are possible?

,

• $30$
• $31$
• $61$
• $62$
• $63$

Geometry

Rectangle

Integer

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2015 Problem-24

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

$31$

## Try with Hints

#### First Hint

Given that $ABCD$ is a quadrilateral whose perimeter $p$, right angles at $B$ and $C, A B=2$, and $C D=A D$. we have to find out How many different values of $p<2015$ are possible.

Now draw a perpendicular $AE$ on $CD$ . Let us assume that $BC=AE=x$ Then $CE=2$ and $DE=y-2$

Now can you finish the problem?

#### Second Hint

Now from the $\triangle ADE$ we can write $x^{2}+(y-2)^{2}=y^{2}$

$\Rightarrow x^{2}-4 y+4=0$

$\Rightarrow x^2=4(y-1)$, Thus, $y$ is one more than a perfect square.

Therefore the perimeter will be $p=2+x+2 y=2 y+2 \sqrt{y-1}+2$

Now according to the problem $p<2015$

So, $p=2+x+2 y=2 y+2 \sqrt{y-1}+2 <2015$

Now Can you finish the Problem?

#### Third Hint

Now $y=31^{2}+1=962$ and $y=32^{2}+1=1025$

Here $y=31^{2}+1=962$ is valid but $y=32^{2}+1=1025$ is not. On the lower side, $y=1$ does not work (because $x>0$ ), but $y=1^{2}+1$ does work. Hence, there are $31$ valid $y$ (all $y$ such that $y=n^{2}+1$ for $1 \leq n \leq 31$ )

Therefore the correct answer is $31$

Categories

## Rectangular Piece of Paper | AMC 10A, 2014| Problem No 22

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

## Rectangular Piece of PaperÂ  – AMC-10A, 2014- Problem 23

A rectangular piece of paper whose length is $\sqrt{3}$ times the width has area $A$. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area $B$. What is the ratio $B: A ?$

,

• $2:3$
• $1:2$
• $1:3$
• $3:2$
• $2:5$

Geometry

Rectangle

Ratio

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2014 Problem-23

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

$2:3$

## Try with Hints

#### First Hint

we have to find out the  The ratio of the area of the folded paper to that of the original paper.

At first we have to find out the midpoint of the dotted line. Now draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. Now it will form a triangle . we have to find out the area of the triangle ….

Now can you finish the problem?

#### Second Hint

Let us assume the width of the paper is $1$  and the length is $\sqrt 3$.Now the side length of the triangle be $\frac{2 \sqrt{3}}{3}$, $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$, and $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$

Now for the area of the of the Paper,

It is an equilateral triangle with height $\frac{\sqrt{3}}{3} \cdot \sqrt{3}=1$ and the area will be $\frac{\frac{2 \sqrt{3}}{3} \cdot 1}{2}=\frac{\sqrt{3}}{3}$

Therefore the area of the paper will be $1 . \sqrt 3=\sqrt 3$

Now Can you finish the Problem?

#### Third Hint

Now the area of the folded paper is $\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2 \sqrt{3}}{3}$

Therefore the ratio of the area of the folded paper to that of the original paper is $2:3$