Categories

## Chosing Program | AMC 10A, 2013 | Problem 7

Try this beautiful problem from Combinatorics: Chosing Program

## Chosing Program – AMC-10A, 2013- Problem 7

A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

,

• $6$
• $8$
• $9$
• $12$
• $16$

### Key Concepts

Combinatorics

Answer: $9$

AMC-10A (2013) Problem 7

Pre College Mathematics

## Try with Hints

There are six programms: English, Algebra, Geometry, History, Art, and Latin. Since the student must choose a program of four course with the condition that there must contain English and at least one mathematics course. Therefore one course( i.e English) are already fixed and we have to find out the other subjects combinations…….

Can you now finish the problem ……….

There are Two cases :
Case 1: The student chooses both algebra and geometry.
This means that 3 courses have already been chosen. We have 3 more options for the last course, so there are 3 possibilities here.
case 2: The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by 2 , and then add to the previous.

Let us choose the mathematics course is algebra. so we can choose 2 of History, Art, and Latin, which is simply $3 \choose 2$=$3$. If it is geometry, we have another 3 options, so we have a total of 6 options if only one mathematics course is chosen.

can you finish the problem……..

Therefore the require ways are $6+3=9$

Categories

## Area of Triangle | AMC 10A, 2006 | Problem 21

Try this beautiful problem from Geometry: Area of a triangle

## Triangle – AMC-10A, 2006- Problem 21

A circle of radius 1 is tangent to a circle of radius 2 . The sides of $\triangle A B C$ are tangent to the circles as shown, and the sides $\overline{A B}$ and $\overline{A C}$ are congruent. What is the area of $\triangle A B C ?$

,

i

• $15 \sqrt{2}$
• $\frac{35}{2}$
• $\frac{64}{3}$
• $16 \sqrt{2}$
• $24$

### Key Concepts

Geometry

Circle

Triangle

Answer: $16 \sqrt{2}$

AMC-10A (2006) Problem 21

Pre College Mathematics

## Try with Hints

Given that there are two circle of radius 1 is tangent to a circle of radius 2.we have to find out the area of the $\triangle ABC$.Now draw a perpendicular line $AF$ on $BC$.Clearly it will pass through two centers $O_1$ and $O_2$. and $\overline{A B}$ and $\overline{A C}$ are congruent i.e $\triangle ABC$ is an Isosceles triangle. Therefore $BF=FC$

So if we can find out $AF$ and $BC$ then we can find out the area of the $\triangle ABC$.can you find out $AF$ and $BC$?

Can you now finish the problem ……….

Now clearly $\triangle A D O_{1} \sim \triangle A E O_{2} \sim \triangle A F C$ ( as $O_1D$ and $O_2E$ are perpendicular on $AC$ , R-H-S law )

From Similarity we can say that , $\frac{A O_{1}}{A O_{2}}=\frac{D O_{1}}{E O_{2}} \Rightarrow \frac{A O_{1}}{A O_{1}+3}=\frac{1}{2} \Longrightarrow A O_{1}=3$

By the Pythagorean Theorem we have that $A D=\sqrt{3^{2}-1^{2}}=\sqrt{8}$

Again from $\triangle A D O_{1} \sim \triangle A F C$
$\frac{A D}{A F}=\frac{D O_{1}}{C F} \Longrightarrow \frac{2 \sqrt{2}}{8}=\frac{1}{C F} \Rightarrow C F=2 \sqrt{2}$

can you finish the problem……..

The area of the triangle is $\frac{1}{2} \cdot A F \cdot B C=\frac{1}{2} \cdot A F \cdot(2 \cdot C F)=A F \cdot C F=8(2 \sqrt{2})$=$16\sqrt2$

Categories

## Problem on Cube | AMC 10A, 2008 | Problem 21

Try this beautiful problem from Geometry: Problem on Cube.

## Problem on Cube – AMC-10A, 2008- Problem 21

A cube with side length 1 is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $A B C D ?$

,

i

• $\frac{\sqrt{6}}{2}$
• $\frac{5}{4}$
• $\sqrt{2}$
• $\frac{5}{8}$
• $\frac{3}{4}$

### Key Concepts

Geometry

Square

Pythagoras

Answer: $\frac{\sqrt{6}}{2}$

AMC-10A (2008) Problem 21

Pre College Mathematics

## Try with Hints

The above diagram is a cube and given that side length $1$ and $B$ and $D$ are the mid points .we have to find out area of the $ABCD$.Now since $A B=A D=C B=C D=\sqrt{\frac{1}{2}^{2}}+1^{2},$ it follows that $A B C D$ is a rhombus. can you find out area of the rhombus?

Can you now finish the problem ……….

The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^{2}+1^{2}}=\sqrt{2} \cdot A C$ is a space diagonal, with length $\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$

can you finish the problem……..

Therefore area $A=\frac{1}{2} \times \sqrt{2} \times \sqrt{3}=\frac{\sqrt{6}}{2}$

Categories

## Television Problem | AMC 10A, 2008 | Problem 14

Try this beautiful Television Problem from AMC – 10A, 2008.

## Television Problem – AMC-10A, 2008- Problem 14

Older television screens have an aspect ratio of 4: 3 . That is, the ratio of the width to the height is 4: 3 . The aspect ratio of many movies is not $4: 3,$ so they are sometimes shown on a television screen by “letterboxing” – darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of 2: 1 and is shown on an older television screen with a 27 -inch diagonal. What is the height, in inches, of each darkened strip?

,

i

• $2$
• $2.25$
• $2.5$
• $2.7$
• $3$

### Key Concepts

Geometry

Square

Pythagoras

Answer: $2.7$

AMC-10A (2008) Problem 14

Pre College Mathematics

## Try with Hints

The above diagram is a diagram of Television set whose aspect ratio of $4: 3$.Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$-inch diagonal. Then we have to find the height, in inches, of each darkened strip.

we assume that the width and height of the screen be $4x$ and $3x$ respectively, and let the width and height of the movie be $2y$ and $y$ respectively. If we can find out the value of $x$ and $y$ then  the height of each strip can be calculate eassily

Can you now finish the problem ……….

By the Pythagorean Theorem, the diagonal is $\sqrt{(3 x)^{2}+(4 x)^{2}}=5 x=27 .$ So $x=\frac{27}{5}$

Now the movie and the screen have the same width, $2 y=4 x \Rightarrow y=2 x$

can you finish the problem……..

Thus, the height of each strip is $\frac{3 x-y}{2}=\frac{3 x-2 x}{2}=\frac{x}{2}=\frac{27}{10}=2.7$

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

## 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

## 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

## Number system | AMC-10A, 2007 | Problem 22

Try this beautiful problem from Number system based on digit problem

## Number system – AMC-10A, 2007- Problem 22

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$?

• $31$
• $37$
• $43$

### Key Concepts

Number system

multiplication

Answer: $37$

AMC-10A (2007) Problem 22

Pre College Mathematics

## Try with Hints

The given condition is “A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term,And also another codition that the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term” so we may assume four integers that be $(xyz,yzm,zmp,qxy)$ i.e$(100x+10y+z,100y+10z+m,100z+10m+p,100q+10x+y)$

Now the sum of the digits be$(110x+111y+111z+11m+p+100q)$

can you finish the problem……..

But “the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term”……so we may say that in last integer $qxy$…$q=m$ & $p=x$.Therefore the sum becomes $(110x+111y+111z+11q+x+100q)$=$111(x+y+z+m)$ i.e $111 K$ (say)

can you finish the problem……..

N ow in $111K$= $3.37.K$………So in the given answers the largest prime number is 37

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

## Quadratic equation Problem | AMC-10A, 2003 | Problem 5

Try this beautiful problem from Algebra based on quadratic equation.

## Quadratic equation – AMC-10A, 2003- Problem 5

Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$?

• $2$
• $0$
• $\frac{7}{2}$

### Key Concepts

algebra

Roots

Answer: $0$

AMC-10A (2003) Problem 5

Pre College Mathematics

## Try with Hints

To find out the value of $(d-1)(e-1)$,at first we have to find out the value of $d$ and $e$.Given that $d$ and $e$ are the solutions of the equations $2x^{2}+3x-5=0$ that means $d$ and $e$ are the roots of the given equation.so if we find out the values of roots from the given equation then we will get $d$ and $e$.Can you find out the roots?

Can you now finish the problem ……….

To find out the roots :

The given equation is $2x^{2}+3x-5=0$ $\Rightarrow (2x+5)(x-1)=0$ $\Rightarrow x=1 or \frac{-5}{2}$

Therefore the values of $d$ and $e$ are $1$ and $\frac{-5}{2}$ respectively

can you finish the problem……..

Therefore $(d-1)(e-1)$=$(1-1)(\frac{-5}{2} -1)$=$0$