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AMC 8 Geometry Math Olympiad USA Math Olympiad

Circles and semi-circles| AMC 8, 2010|Problem 23

Try this beautiful problem from Geometry based on Ratio of the area of circle and semi-circles.

Area of circles and semi-circles – AMC-8, 2010 – Problem 23


Semicircles POQ and ROS pass through the center O. What is the ratio of the combined areas of the two semicircles to the area of circle O?

circles and semi-circles
  • $\frac{1}{2}$
  • $\frac{2}{\pi}$
  • $ \frac{3}{2} $

Key Concepts


Geometry

Circle

co-ordinate geometry

Check the Answer


Answer:$\frac{1}{2}$

AMC-8 (2010) Problem 23

Pre College Mathematics

Try with Hints


Find the radius of the circle

Can you now finish the problem ……….

ratio of the areas

Join O and Q

can you finish the problem……..

ratio of the areas of circles and semi-circles

The co-ordinate of Q is (1,1), So OB=1 and BQ=1

By the Pythagorean Theorem, the radius of the larger circle i.e OQ=\(\sqrt{1^2+1^2}\)=\(\sqrt 2\).

Therefore the area of the larger circle be \(\pi (\sqrt 2)^2=2\pi\)

Now for the semicircles, radius OB=OC=1(as co-ordinate of P=(1,1) and S=(1,-1))

So, the area of the two semicircles is  \(2\times\frac{\pi(1)^2}{2}=\pi\)

 Finally, the ratio of the combined areas of the two semicircles to the area of circle O is

\(\frac{\pi}{2\pi}\)=\(\frac{1}{2}\)

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AMC 8 Math Olympiad Number Theory

Page number counting |AMC 8- 2010 -|Problem 21

Try this beautiful problem from Algebra about Page number counting

Page number counting | AMC-8, 2010 |Problem 21


Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read \(\frac{1}{5}\) of the pages plus more, and on the second day she read  \(\frac{1}{4}\) of the remaining pages plus 15 pages. On the third day she read \(\frac{1}{3}\) of the remaining pages plus 18 pages. She then realized that there were only  62 pages left to read, which she read the next day. How many pages are in this book?

  • 320
  • 240
  • 200

Key Concepts


Algebra

Arithmetic

multiplication

Check the Answer


Answer:$240$

AMC-8, 2010 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints


assume that the number of all pages be \(x\)

Can you now finish the problem ……….

count day by day

can you finish the problem……..

Let x be the number of pages in the book

First day ,Hui Read \(\frac{x}{5} + 12\) pages

After first day Remaining pages=\(\{x-(\frac{x}{5}+12)\}\)=\(\frac{4x}{5} -12\)

Second day ,Hui Read \(\frac{1}{4} (\frac{4x}{5} -12) +15=\frac{x}{5} +12\)

After Second day Remaining pages= \((\frac{4x}{5} -12) -(\frac{x}{5} +12)\)=\(\frac{4x}{5} -\frac{x}{5}-24\)=\(\frac{3x}{5} -24\)

Third day,Hui read \(\frac {1}{3} (\frac{3x}{5} -24) +18\) =\((\frac{x}{5} -8+18)\)=\(\frac{x}{5} +10\)

After Third day Remaining pages = \((\frac{3x}{5} -24) -(\frac{x}{5} +10)\) =\(\frac{2x}{5} – 34\)

Now by the condition, \(\frac{2x}{5} – 34 = 62\)

\(\Rightarrow 2x-170=310\)

\(\Rightarrow 2x=480\)

\(\Rightarrow x=240\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of square and circle | AMC 8, 2011|Problem 25

Try this beautiful problem from Geometry based on Ratio of the area of square and circle.

Area of the star and circle – AMC-8, 2011 – Problem 25


A circle with radius 1 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle’s shaded area to the area between the two squares?

area of square and circle
  • $\frac{3}{2}$
  • $\frac{1}{2}$
  • $1$

Key Concepts


Geometry

Circle

Square

Check the Answer


Answer:$\frac{1}{2}$

AMC-8 (2011) Problem 25

Pre College Mathematics

Try with Hints


Join the diagonals of the smaller square (i.e GEHF)

figure simplified 1

Can you now finish the problem ……….

The circle’s shaded area is the area of the smaller square(i.e. GEHF) subtracted from the area of the circle

and The area between the two squares is Area of the square ABCD – Area of the square EFGH

can you finish the problem……..

figure solution 2

Given that the Radius of the circle with centre O is 1.Therefore The area of the circle is \(\pi (1)^2\)=\(\pi\) sq.unit

The diameter of the circle is 2 i.e \(EF=BC=2\) unit

The area of the big square i.e \(ABCD=2^2=4\) sq.unit

\(OE=OH=1\) i.e \(EH=\sqrt{(1^2+1^2)}=\sqrt 2\)

Therefore the area of the smaller square is \((\sqrt 2)^2=2\)

The circle’s shaded area is the area of the smaller square(i.e. GEHF) subtracted from the area of the circle =\(\pi\) – 2

The area between the two squares is Area of the square ABCD – Area of the square EFGH=4-2=2 sq.unit

The ratio of the circle’s shaded area to the area between the two squares is \( \frac{\pi – 2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \frac{1}{2}\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Ratio of the area of Square and Pentagon | AMC 8, 2013

Try this beautiful problem from Geometry: Ratio of the area between Square and Pentagon.

Ratio of the area between Square and Pentagon – AMC-8, 2013 – Problem 24


Squares ABCD ,EFGH and GHIJ are equal in area .Points C and D are the mid points of the side IH and HE ,respectively.what is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares?

ratio of area of square and pentagon

  • $\frac{1}{4}$
  • $\frac{1}{3}$
  • $\frac{3}{8}$

Key Concepts


Geometry

Area of square

Area of Triangle

Check the Answer


Answer:$\frac{1}{3}$

AMC-8(2013) Problem 24

Pre College Mathematics

Try with Hints


extend  IJ until it hits the extension of  AB .

Can you now finish the problem ……….

find the area of the pentagon

can you finish the problem……..

solution figure

First let L=2 (where L is the side length of the squares) for simplicity. We can extend  IJ until it hits the extension of  AB . Call this point  X.

Then clearly length of AX=3 unit & length of XJ = 4 unit .

Therefore area of \(\triangle AXJ= (\frac{1}{2} \times AX \times XJ)=(\frac{1}{2} \times 4 \times 3)=6\) sq.unit

And area of Rectangle BXIC= \(( 1 \times 2)\)=2 sq.unit

Therefore the of the pentagon ABCIJ=6-2=4 sq.unit

The combined area of three given squares be \( (3 \times 2^2)\)=12 sq.unit

Now, the ratio of the shaded area(pentagon) to the combined area of the three squares is \(\frac{4}{12}=\frac{1}{3}\)

.

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of Triangle and Square | AMC 8, 2012 | Problem 25

Try this beautiful problem from AMC-8-2012 (Geometry) based on area of a Triangle and square and congruency.

Area of a Triangle- AMC 8, 2012 – Problem 25


A square with area 4 is inscribed in a square with area 5,  with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?

Area of triangle and square

  • \(\frac{1}{5}\)
  • \(\frac{2}{5}\)
  • \(\frac{1}{2}\)

Key Concepts


Geometry

Square

Triangle

Check the Answer


Answer:\(\frac{1}{2}\)

AMC-8, 2014 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find the area of four triangles

Can you now finish the problem ……….

Four triangles are congruent

can you finish the problem……..

Area of Triangle and Square 2

Total area of the big square i.e ABCD is 5 sq.unit

and total area of the small square i.e EFGH is 4 sq.unit

So Toal area of the \((\triangle AEH + \triangle BEF + \triangle GCF + \triangle DGH)=(5-4)=1\) sq.unit

Now clearly Four triangles\(( i.e \triangle AEH , \triangle BEF , \triangle GCF , \triangle DGH )\) are congruent.

Total area of the four congruent triangles formed by the squares is 5-4=1 sq.unit

So area of the one triangle is \(\frac{1}{4}\) sq.unit

Now “a” be the height and “b” be the base of one triangle

The area of one triangle be \((\frac{1}{2} \times base \times height )\)=\(\frac{1}{4}\)

i.e \((\frac{1}{2} \times b \times a)\)= \(\frac{1}{4}\)

i.e \(ab\)= \(\frac{1}{2}\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of star and circle | AMC-8, 2012|problem 24

Try this beautiful problem from Geometry: Ratio of the area of the star figure to the area of the original circle

Area of the star and circle – AMC-8, 2012 – Problem 24


A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?

Area of star and circle
  • $\frac{1}{\pi}$
  • $\frac{4-\pi}{\pi}$
  • $\frac{\pi – 1}{\pi}$

Key Concepts


Geometry

Circle

Arc

Check the Answer


Answer:$\frac{4-\pi}{\pi}$

AMC-8 (2012) Problem 24

Pre College Mathematics

Try with Hints


Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.

Can you now finish the problem ……….

find the area of the star figure

can you finish the problem……..

Star and circle

Draw a square around the star figure. Then the length of one side of the square be 4(as the diameter of the circle is 4)

Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.

circle

The area of the above circle is \(\pi (2)^2 =4\pi\)

circle and square

and the area of the outer square is \((4)^2=16\)

star

Thus, the area of the star figure is \(16-4\pi\)

Therefore \(\frac{(the \quad area \quad of \quad the \quad star \quad figure)}{(the \quad area \quad of \quad the \quad original \quad circle )}=\frac{16-4\pi}{4\pi}\)

= \(\frac{4-\pi}{\pi}\)

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AMC 8 Beautiful Mathematics videos Geometry Math Olympiad

Circumference of a Semicircle | AMC 8, 2014 | Problem 25

Try this beautiful problem from AMC-8-2014 (Geometry) based on Circumference of a Semicircle

Circumference of a Semicircle- AMC 8, 2014 – Problem 25


On A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5miles per hour, how many hours will it take to cover the one-mile stretch?

circumference of a semicircle

  • \(\frac{\pi}{11}\)
  • \(\frac{\pi}{10}\)
  • \(\frac{\pi}{5}\)

Key Concepts


Geometry

Semicircle

Distance

Check the Answer


Answer:\(\frac{\pi}{10}\)

AMC-8, 2014 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find the circumference of a semi-circle

Can you now finish the problem ……….

If Robert rides in a straight line, it will take him \(\frac {1}{5}\) hours

can you finish the problem……..

circumference of a semicircle

If Robert rides in a straight line, it will take him \(\frac {1}{5}\) hours. When riding in semicircles, let the radius of the semicircle r, then the circumference of a semicircle is \({\pi r}\). The ratio of the circumference of the semicircle to its diameter is \(\frac {\pi}{2}\). so the time Robert takes is  \(\frac{1}{5} \times \frac{\pi}{2}\). which is equal to \(\frac{\pi}{10}\)

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Algebra AMC 8 Math Olympiad

Divisibility | AMC 8, 2014 |Problem 21

Try this beautiful problem from Algebra based on multiplication and divisibility of two given numbers.

Multiplication and Divisibility- AMC 8, 2014


The  7-digit numbers 74A52B1 ana 326AB4C are each multiples of 3.which of the following could be the value of c ?

  • 1
  • 2
  • 3

Key Concepts


Algebra

Division algorithm

Integer

Check the Answer


Answer:1

AMC-8, 2014 problem 21

Challenges and Thrills of Pre College Mathematics

Try with Hints


Use the rules of Divisibility ……..

Can you now finish the problem ……….

If both numbers are divisible by 3 then the sum of their digits has to be divisible by 3……

can you finish the problem……..

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8… and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10… and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7… and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1. so the answer is 1

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AMC 8 Geometry Math Olympiad

Rolling ball Problem | Semicircle |AMC 8- 2013 -|Problem 25

Try this beautiful problem from Geometry based on a rolling ball on a semicircular track.

A Ball rolling Problem from AMC-8, 2013


A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3  semicircular arcs whose radii are \(R_1=100\) inches ,\(R_2=60\) inches ,and \(R_3=80\) inches respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from  A to B?

Path of the rolling ball- Problem

  • \( 235 \pi\)
  • \( 238\pi\)
  • \( 240 \pi\)

Key Concepts


Geometry

circumference of a semicircle

Circle

Check the Answer


Answer:\( 238 \pi\)

AMC-8, 2013 problem 25

Pre College Mathematics

Try with Hints


Find the circumference of semicircle….

Can you now finish the problem ……….

Find the total distance by the ball….

can you finish the problem……..

rolling ball problem

The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses \(2\pi \times \frac{2}{2}=2\pi\)  inches each, and it gains \(2\pi\) inches on B .

So, the departure from the length of the track means that the answer is

\(\frac{200+120+160}{2} \times \pi\) + (-2-2+2) \(\times \pi\)=240\(\pi\) -2\(\pi\)=238\(\pi\)

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AMC 8 Geometry Math Olympiad

Hexagon and Triangle |AMC 8- 2015 -|Problem 21

Try this beautiful problem from Geometry based on hexagon and Triangle.

Area of Triangle | AMC-8, 2015 |Problem 21


In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK $ is equilateral and FE=BC. What is the area of $\triangle KBC$?

hexagon and triangle

  • 9
  • 12
  • 32

Key Concepts


Geometry

Triangle

hexagon

Check the Answer


Answer:$12$

AMC-8, 2015 problem 21

Pre College Mathematics

Try with Hints


Clearly FE=BC

Can you now finish the problem ……….

$\triangle KBC$ is a Right Triangle

can you finish the problem……..

hexagon and triangle

Clearly ,since FE is a side of square with area 32

Therefore FE=$\sqrt 32$=$4\sqrt2$

Now since FE=BC,We have BC=$4\sqrt2$

Now JB is a side of a square with area 18

so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$

Lastly $\triangle KBC$ is a right triangle ,we see that

$\angle JBA + \angle ABC +\angle CBK +\angle KBJ$ =$360^\circ$

i.e$ 90^\circ + 120^\circ +\angle CBK + 60^\circ=360^\circ$

i.e $\angle CBK=90^\circ $

So $\triangle KBC $ is a right triangle with legs $3\sqrt 2$ and $4\sqrt2$

Now its area is $3\sqrt2 \times 4\sqrt 2 \times \frac {1}{2}$=$\frac{24}{2}$=12

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