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## Circle Problem | AMC 10A, 2006 | Problem 23

Try this beautiful problem from Geometry: Circle

## Circle Problem – AMC-10A, 2006- Problem 23

Circles with centers $A$ and $B$ have radii 3 and 8 , respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at $E,$ and $A E=5 .$ What is $C D ?$

,

i

• $13$
• $\frac{44}{3}$
• $\sqrt{221}$
• $\sqrt{255}$
• $\frac{55}{3}$

### Key Concepts

Geometry

Circle

Tangents

Answer: $\frac{44}{3}$

AMC-10 (2006) Problem 23

Pre College Mathematics

## Try with Hints

Given that Circles with centers $A$ and $B$ have radii 3 and 8 and $A E=5 .$.we have to find out $CD$.So join $BC$ and $AD$.then clearly $\triangle BCE$ and $\triangle ADE$ are Right-Triangle(as $CD$ is the common tangent ).Now $\triangle BCE$ and $\triangle ADE$ are similar.Can you proof $\triangle BCE$ and $\triangle ADE$?

Can you now finish the problem ……….

$\angle A E D$ and $\angle B E C$ are vertical angles so they are congruent, as are angles $\angle A D E$ and $\angle B C E$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle A C E \sim \triangle B D E$.

By the Pythagorean Theorem, line segment $DE=4$

Therefore from the similarity we can say that $\frac{D E}{A D}=\frac{C E}{B C} \Rightarrow \frac{4}{3}=\frac{C E}{8}$ .

Therefore $C E=\frac{32}{3}$

can you finish the problem……..

Therefore $CD=CE+DE=4+\frac{32}{3}=\frac{44}{3}$

Categories

## Centroid Problem: Ratio of the areas of two Triangles

Try this beautiful problem from Geometry based on Centroid.

## Centroid Problem: Ratio of the areas of two Triangles

$\triangle ABC$ has centroid $G$.$\triangle ABG$,$triangle BCG$, and $\triangle CAG$ have centroids $G_1$, $G_2$, $G_3$ respectively. The value of $\frac{[G1G2G3]}{[ABC]}$ can BE represented by $\frac{p}{q}$ for positive integers $p$ and $q$.

Find $p+q$ where$[ABCD]$ denotes the area of ABCD.

• $14$
• $10$
• $7$

### Key Concepts

Geometry

Triangle

centroid

Answer: $10$

Question Papers

Pre College Mathematics

## Try with Hints

$\triangle ABC$ has centroid $G$.$\triangle ABG$,$triangle BCG$, and $\triangle CAG$ have centroids $G1$,$G2$,$G3$ respectively.we have to find out value of $\frac{[G1G2G3]}{[ABC]}$ i.e area of $\frac{[G1G2G3]}{[ABC]}$

Let D, E, F be the midpoints of BC, CA, AB respectively.
Area of $\frac{[DEF]}{[ABC]}$=$\frac{1}{4}$

we know that any median is divided at the centroid $2:1$. Now can you find out $GG_1,GG_2,GG_3$ ?

Can you now finish the problem ……….

we know that any median is divided at the centroid $2:1$
Now  $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G_1F$
Similarly,$GG_2 = 2G_2D$ and$GG_3 = 2G_3E$
Thus, From  homothetic transformation  $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$
Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$

can you finish the problem……..

Therefore we say that $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]}$= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$

So $p+q$=$9+1$=$10$

Categories

## Triangle Problem | AMC 10B, 2013 | Problem 16

Try this beautiful problem from Geometry from AMC-10B, 2013, Problem-16, based on triangle

## Triangle | AMC-10B, 2013 | Problem 16

In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$?

• $12$
• $13.5$
• $15.5$

### Key Concepts

Geometry

Triangle

Pythagorean

Answer:$13.5$

AMC-10B, 2013 problem 16

Pre College Mathematics

## Try with Hints

We have to find out the area of AEDC which is divided in four triangles i.e$\triangle APC$,$\triangle APE$, $\triangle PED$, $\triangle CPD$

Now if we find out area of four triangle then we can find out the required area AEDC. Now in the question they supply the information only one triangle i.e $\triangle PED$ such that $PE=1.5$, $PD=2$, and $DE=2.5$ .If we look very carefully the given lengths of the $\triangle PED$ then $( 1.5 )^2 +(2)^2=(2.5)^2$ $\Rightarrow$ $(PE)^2 +(PD)^2=(DE)^2$ i,e $\triangle PED$ is a right angle triangle and $\angle DEP =90^{\circ}$ .so we can easily find out the area of the $\triangle PED$ using formula $\frac{1}{2} \times base \times height$ . To find out the area of other three triangles, we must need the lengths of the sides.can you find out the length of the sides of other three triangles…

Can you now finish the problem ……….

To find out the lengths of the sides of other three triangles:

Given that AD and CD are the medians of the given triangle and they intersects at the point P. .we know the fact that the centroid ($P$) divides each median in a $2:1$ ratio .Therefore AP:PD =2:1 & CP:PE=2:1. Now $PE=1.5$ and $PD=2$ .Therefore AP=4 and CP=3.

And also the angles i.e ($\angle APC ,\angle CPD, \angle APE$) are all right angles as $\angle DPE= 90^{\circ}$

can you finish the problem……..

Area of four triangles :

Area of the $\triangle APC$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times AP \times PC$ = $\frac{1}{2} \times 4 \times 3$ =6

Area of the $\triangle DPC$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times CP \times PD$ = $\frac{1}{2} \times 3 \times 2$ =3

Area of the $\triangle PDE$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times PD \times DE$ = $\frac{1}{2} \times 2 \times 1.5$=1.5

Area of the $\triangle APE$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times AP \times PE$ = $\frac{1}{2} \times 4 \times 1.5 =6$

Total area of ACDE=area of ($\triangle APC$+$\triangle APE$+ $\triangle PED$+ $\triangle CPD$)=$(6+3+1.5+6)=17.5$ sq.unit

Categories

## Area of Trapezium | AMC-10A, 2018 | Problem 24

Try this beautiful problem from Geometry based on the Area of the Trapezium

## Area of the Trapezium – AMC-10A, 2018- Problem 24

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

• $79$
• $75$
• $82$

### Key Concepts

Geometry

Triangle

Trapezium

Answer: $75$

AMC-10A (2018) Problem 24

Pre College Mathematics

## Try with Hints

We have to find out the area of BGFD.Given that AG is the angle bisector of $\angle BAC$ ,$D$ and $E$ are the mid points of $AB$ and $AC$. so we may say that $DE ||BC$ by mid point theorm…

So clearly BGFD is a Trapezium.now area of the trapezium=$\frac{1}{2} (BG+DF) \times height betwween DF and BG$

can you find out the value of $BG,DF$ and height between them….?

Can you now finish the problem ……….

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.

Therefore area of $\triangle ABC$=$\frac{ah}{2}$=$120$………………..(1)

From the angle bisector theorem, we have that$\frac{50}{x} = \frac{10}{y}$ i.e $\frac{x}{y}=5$

Let $BC$=$a$ then $BG$=$\frac{5a}{6}$ and $DF$=$\frac{1}{2 } \times BG$ i.e $\frac{5a}{12}$

now can you find out the area of Trapezium ?

can you finish the problem……..

Therefore area of the Trapezium=$\frac{1}{2} (BG+DF) \times FG$=$\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}$=$\frac{ah}{2} \times \frac{15}{24}$=$120 \times \frac{15}{24}$=$75$ $(from ……..(1))$

Categories

## Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

## Finding side of Triangle | PRMO | Problem 15

Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

• $28$
• $26$
• $30$

### Key Concepts

Geometry

Triangle

Pythagoras

Answer:$26$

PRMO-2014, Problem 15

Pre College Mathematics

## Try with Hints

Given that $\angle XOY=90^{\circ}$ .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now $\triangle XON$ & $\triangle MOY$ are Right angle Triangle. Use Pythagoras theorem …….

Can you now finish the problem ……….

Let $XM=MO=p$ and $ON=NY=q$.Now using Pythagoras theorm on $\triangle XON$ & $\triangle MOY$ we have…

$OX^2 +ON^2=XN^2$ $\Rightarrow 4p^2 +q^2=19^2$ $\Rightarrow 4p^2 +q^2=361$………..(1) and $OM^2 +OY^2=MY^2$ $\Rightarrow p^2 +4q^2=22^2$ $\Rightarrow p^2 +4q^2=484$……(2)

Now Adding (1)+(2)=$(4p^2 +q^2=361)$+$(p^2 +4q^2=484$ $\Rightarrow 5(p^2+q^2)=845$ $\Rightarrow (p^2+q^2)=169$ $\Rightarrow 4(p^2+q^2)=676$ $\Rightarrow (OX)^2+(OY)^2=(26)^2$ $\Rightarrow (XY)^2=(26)^2$ $\Rightarrow XY=26$.

Categories

## The area of trapezoid | AMC 8, 2003 | Problem 21

Try this beautiful problem from Geometry: The area of trapezoid

## The area of trapezoid – AMC-8, 2003- Problem 21

The area of trapezoid ABCD is 164 $cm^2$. The altitude is  8 cm, AB is 10 cm, and CD is 17 cm. What is BC in centimeters?

,

i

• $8$
• $10$
• $15$

### Key Concepts

Geometry

trapezoid

Triangle

Answer: $10$

AMC-8 (2003) Problem 21

Pre College Mathematics

## Try with Hints

Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Can you now finish the problem ……….

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

can you finish the problem……..

Given that the area of the trapezoid is 164 sq.unit

Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

Using Pythagorean rules on the triangle ABD,we have…

$(AD)^2 + (BD)^2 =(AB)^2$

$\Rightarrow (AD)^2 + (8)^2 =(10)^2$

$\Rightarrow (AD)^2 =(10)^2 – (8)^2$

$\Rightarrow (AD)^2 = 36$

$\Rightarrow (AD) =6$

Using Pythagorean rules on the triangle CED,we have…

$(CE)^2 + (DE)^2 =(DC)^2$

$\Rightarrow (CE)^2 + (8)^2 =(17)^2$

$\Rightarrow (CE)^2 =(17)^2 – (8)^2$

$\Rightarrow (CE)^2 = 225$

$\Rightarrow (CE) =15$

Let BC= DE=x

Therefore area of the trapezoid=$\frac{1}{2} \times (AD+BC) \times 8$=164

$\Rightarrow \frac{1}{2} \times (6+x+15) \times 8$ =164

$\Rightarrow x=10$

Therefore BC=10 cm

Categories

## Largest area Problem | AMC 8, 2003 | Problem 22

Try this beautiful problem from Geometry based Largest area.

## Largest area – AMC-8, 2003 – Problem 22

The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

• $A$
• $B$
• $C$

### Key Concepts

Geometry

Circle

Square

Answer:$C$

AMC-8 (2003) Problem 22

Pre College Mathematics

## Try with Hints

To find out the largest area at first we have to find out the radius of the circles . all the circles are inscribed ito the squares .now there is a relation between the radius and the side length of the squares….

Can you now finish the problem ……….

area of circle =$\pi r^2$

can you finish the problem……..

In A:

Total area of the square =$2^2=4$

Now the radius of the inscribed be 1(as the diameter of circle = side length of the side =2)

Area of the inscribed circle is $\pi (1)^2=\pi$

Therefore the shaded area =$4- \pi$

In B:

Total area of the square =$2^2=4$

There are 4 circle and radius of one circle be $\frac{1}{2}$

Total area pf 4 circles be $4 \times \pi \times (\frac{1}{2})^2=\pi$

Therefore the shaded area =$4- \pi$

In C:

Total area of the square =$2^2=4$

Now the length of the diameter = length of the diagonal of the square=2

Therefore radius of the circle=$\pi$ and lengthe of the side of the square=$\sqrt 2$

Thertefore area of the shaded region=Area of the square-Area of the circle=$\pi (1)^2-(\sqrt 2)^2$=$\pi – 2$

Categories

## Triangle and Quadrilateral | AMC-10A, 2005 | Problem 25

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral from AMC 10A, 2005.

## Ratios of the areas of Triangle and Quadrilateral – AMC-10A, 2005- Problem 25

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?

• $\frac{19}{56}$
• $\frac{19}{66}$
• $\frac{17}{56}$
• $\frac{11}{56}$
• $\frac{19}{37}$

### Key Concepts

Geometry

Triangle

Answer: $\frac{19}{56}$

AMC-10A (2005) Problem 25

Pre College Mathematics

## Try with Hints

Given that $AB = 25$, $BC = 39$, and $AC=42$.we have to find out Ratios of the areas of Triangle$\triangle ADE$ and the quadrilateral $CBED$.So if we can find out the area the $\triangle ADE$ and area of the $\triangle ABC$ ,and subtract $\triangle ADE$ from $\triangle ABC$ then we will get area of the region $CBDE$.Can you find out the area of $CBDE$?

Can you find out the required area…..?

Now $\frac{\triangle ADE}{\triangle ABC}=\frac{AD}{AB}.\frac{AE}{AC}=\frac{19}{25}.\frac{14}{42}=\frac{19}{75}$

Therefore area of $BCED$=area of $\triangle ABC$-area of $\triangle ADE$.Now can you find out Ratios of the areas of Triangle and the quadrilateral?

can you finish the problem……..

$\frac{[A D E]}{[B C E D]}=\frac{[A D E]}{[A B C]-[A D E]}$
=$\frac{1}{[A B C] /[A D E]-1}$
=$\frac{1}{75 / 19-1}$

=$\frac{19}{56}$

Categories

## Lengths of Rectangle Problem | AMC-10A, 2009 | Problem 14

Try this beautiful problem from geometry based on Lengths of Rectangle Problem.

## Lengths of Rectangle Problem – AMC-10A, 2009- Problem 14

Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?

• $3$
• $\sqrt 10$
• $2+\sqrt 2$
• $2\sqrt 3$
• $4$

### Key Concepts

Geometry

Square

Rectangle

Answer: $3$

AMC-10A (2009) Problem 14

Pre College Mathematics

## Try with Hints

Given that The area of the outer square is $4$ times that of the inner square.therefore we can say that Therefore the side of the outer square is $\sqrt 4 = 2$ times that of the inner square.Can you find out length of the longer side of each rectangle?

Can you now finish the problem ……….

Let the side length of the outer square is $4x$ then the side length of the inner square be $2x$.Hence the side length of the red region is $2x$ .As the rectangles are congruent ,therefore side length of green shaded region and the side length of blue shaded region will be x

Therefore the length of the longer side of each rectangle be $3x$ and length of the shoter side will be $x$

Therefore the ratio of the length of the longer side of each rectangle to the length of its shorter side will be $\frac{3x}{x}=3$

Categories

## Reflection Problem | AIME I, 1988 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988, Question 14, based on Reflection.

## Reflection Problem – AIME I, 1988

Let C be the graph of xy=1 and denote by C’ the reflection of C in the line y=2x. let the equation of C’ be written in the form $12x^{2}+bxy +cy^{2}+d=0$, find the product bc.

• is 107
• is 84
• is 840
• cannot be determined from the given information

### Key Concepts

Geometry

Equation

Algebra

AIME I, 1988, Question 14

Coordinate Geometry by Loney

## Try with Hints

Let P(x,y) on C such that P'(x’,y’) on C’ where both points lie on the line perpendicular to y=2x

slope of PP’=$\frac{-1}{2}$, then $\frac{y’-y}{x’-x}$=$\frac{-1}{2}$

or, x’+2y’=x+2y

also midpoint of PP’, $(\frac{x+x’}{2},\frac{y+y’}{2})$ lies on y=2x

or, $\frac{y+y’}{2}=x+x’$

or, 2x’-y’=y-2x

solving these two equations, x=$\frac{-3x’+4y’}{5}$ and $y=\frac{4x’+3y’}{5}$

putting these points into the equation C $\frac{(-3x’+4y’)(4x’+3y’)}{25}$=1

which when expanded becomes

$12x’^{2}-7x’y’-12y’^{2}+25=0$

or, bc=(-7)(-12)=84.