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## Area of a square | AMC 8- 2015| Problem 25

Try this beautiful problem from AMC-8-2015 (Geometry) based on area of square.

## Area of a square – AMC 8, 2015 – Problem 25

One-inch Squares are cut from the corners of this 5 inch square.what is the area in square inches of the largest square that can be fitted into the remaining space?

• 9
• 15
• 17

### Key Concepts

Geometry

Area

Square

AMC-8, 2015 problem 25

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Find the Length of HG……

Can you now finish the problem ……….

Draw the big square in the remaining space of the big sqare and find it’s area …….

can you finish the problem……..

We want to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base  3 and height 1 .  so the combined area of the four triangles is  $4 \times \frac {3}{2}$=6.

The area of the smaller square is  9+6=15.

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## LCM – AMC 8, 2016 – Problem 20

The least common multiple of a and b is 12 .and the lest common multiple of b and c is 15.what is the least possible value of the least common multiple of a and c?

• 30
• 60
• 20

### Key Concepts

Algebra

Division algorithm

Integer

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Find greatest common factors

Can you now finish the problem ……….

Find Least common multiple….

can you finish the problem……..

we wish to find possible values of a,b and c .By finding the greatest common factor 12 and 15, algebrically ,it’s some multiple of b and from looking at the numbers ,we are sure that it is 3.Moving on to a and c ,in order to minimize them,we wish to find the least such that the LCM of a and 3 is 12,$\to 4$.similarly with 3 and c,we obtain 5.the LCM of 4 and 5 is 20 .

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## Radius of a Semicircle | AMC 8, 2016 | Problem 25

Try this beautiful problem from Geometry based on Radius of a semicircle inscribed in an isosceles triangle.

## Radius of a Semi circle – AMC-8, 2016 – Problem 25

A semicircle is inscribed in an isoscles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle as shown .what is the radius of the semicircle?

• $\frac{110}{19}$
• $\frac{120}{17}$
• $\frac{9}{5}$

### Key Concepts

Geometry

Area

pythagoras

Answer:$\frac{120}{17}$

AMC-8, 2016 problem 25

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Draw a perpendicular from the point C on base AB

Can you now finish the problem ……….

D be the midpoint of the AB(since $\triangle ABC$ is an isoscles Triangle)

Find AC and area

can you finish the problem……..

Area of the $\triangle ABC= \frac{1}{2} \times AB \times CD$

= $\frac{1}{2} \times 16 \times 15$

=120 sq.unit

Using the pythagoras th. $AC^2= AD^2+CD^2$

i.e $AC^2=(8)^2+(15)^2$

i.e $AC=17$

Let$ED = x$ be the radius of the semicircle

Therefore Area of $\triangle CAD = \frac{1}{2} \times AC \times ED$=$\frac {1}{2} area of \triangle ABC$

i.e $\frac{1}{2} \times AC \times ED$=60

i.e $\frac{1}{2} \times 17 \times x$ =60

i.e $x=\frac {120}{7}$

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## Area of a Triangle -AMC 8, 2018 – Problem 20

Try this beautiful problem from Geometry based on Area of a Triangle Using similarity

## Area of Triangle – AMC-8, 2018 – Problem 20

In $\triangle ABC$ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.

What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?

• $\frac{2}{3}$
• $\frac{4}{9}$
• $\frac{3}{5}$

### Key Concepts

Geometry

Area

similarity

Answer:$\frac{4}{9}$

AMC-8, 2018 problem 20

Pre College Mathematics

## Try with Hints

$\triangle ADE$ $\sim$ $\triangle ABC$

Can you now finish the problem ……….

$\triangle BEF$ $\sim$ $\triangle ABC$

can you finish the problem……..

Since $\triangle ADE$$\sim$ $\triangle ABC$

$\frac{ \text {area of} \triangle ADE}{ \text {area of} \triangle ABC}$=$\frac{AE^2}{AB^2}$

i.e $\frac{\text{area of} \triangle ADE}{\text{area of} \triangle ABC}$ =$\frac{(1)^2}{(3)^2}$=$\frac{1}{9}$

Again $\triangle BEF$ $\sim$ $\triangle ABC$

Therefore $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$=$\frac{BE^2}{AB^2}$

i.e $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$ =$\frac{(2)^2}{(3)^2}$=$\frac{4}{9}$

Therefore Area of quad. CDEF =$\frac {4}{9}$ of area $\triangle ABC$

i.e The ratio of the area of quad.CDEF to the area of $\triangle ABC$ is $\frac{4}{9}$

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## Area of cube’s cross section |Ratio | AMC 8, 2018 – Problem 24

Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

## Area of cube’s cross section – AMC-8, 2018 – Problem 24

In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?

• $\frac{5}{4}$
• $\frac{3}{2}$
• $\frac{4}{3}$

### Key Concepts

Geometry

Area

Pythagorean theorem

Answer:$\frac{3}{2}$

AMC-8(2018) Problem 24

Pre College Mathematics

## Try with Hints

EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Area of rhombus is half product of its diagonals….

can you finish the problem……..

Let Side length of a cube be x.

then by the pythagorean  theorem$EC=X \sqrt {3}$

$JI =X \sqrt {2}$

Now the area of the rhombus is half product of its diagonals

therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$

This shows that $R= \frac{\sqrt6}{2}$

i.e$R^2=\frac{3}{2}$

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## Radius of a Semi Circle -AMC 8, 2017 – Problem 22

Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

## AMC-8(2017) – Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$

### Key Concepts

Geometry

congruency

similarity

Answer:$\frac{10}{3}$

AMC-8(2017)

Pre College Mathematics

## Try with Hints

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

Can you now finish the problem ……….

Now the $\triangle ODB$and $\triangle OCB$ are congruent

can you finish the problem……..

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree`

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar….

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$

Categories

## Big idea – stabilization

In a counting problem, you may need to stabilize the number of cases. This means, by letting some of the variables change, one may fix the remaining cases.

This is the central idea in the following problem from American Mathematical Contest 8 (AMC 8, 2018, Problem 9).

### Problem

In a sign pyramid a cell gets a “+” if the two cells below it have the same sign, and it gets a “-” if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a “+” at the top of the pyramid?

### Sequential Hints

(How to use this discussion: Do not read the entire solution at one go. First, read more on the Key Idea, then give the problem a try. Next, look into Step 1 and give it another try and so on.)

#### Hint 1

Start from the top! Fix the + sign at the top of the pyramid and try filling in the second row from left. Do you see any pattern?

(What happens if you plugin a + sign in the left most block of second row from top?)

#### Hint 2

If we fix a + sign in the left most block of second row from top, the other box must contain a + sign. (Why?)

On the other hand fixing a – sign in the left most block of second row from top will force the remaining block to contain – sign.

Thus by varying the left most block of the second row, we can fix the entire row.

Will this work for the third row?

#### Hint 3

In fact the same trick will work for third row.

Try this. Fix a plus sign in the first row. Then fix second row as well ( plus, plus or minus, minus).

Then try to plugin some sign in the third row’s first block. And check that the remaining two blocks get automatically fixed. This is precisely known as stabilization.

#### Final Hint

Thus, we should only worry about the first block of second, third and fourth. Fixing them fixes the entire row.

There are 2 choices for the first block of second row, 2 choices for first block of third row and 2 choices for first block of third row. Hence in total we have $2 \times 2 \times 2 = 8$ cases.

Recommended book

Principles and Techniques in Combinatorics Kindle Edition by Chen Chuan-Chong

Try chapter 1.