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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of Isosceles Triangle | AMC 8, 2005 | Problem 23

Try this beautiful problem from AMC-8, 2005, Problem-23 based on the area of an isosceles triangle.

Area of the Triangle- AMC- 8, 2005 – Problem 23


Isosceles right triangle ABC encloses a semicircle of area \(2\pi\) . The circle has its center O on hypotenuse AB and is tangent to sides AC and BC . What is the area of triangle ABC ?

Area of an isoceles triangle
  • $6$
  • $8$
  • $10$

Key Concepts


Geometry

Triangle

Semi-circle

Check the Answer


Answer:$8$

AMC-8 (2005) Problem 23

Pre College Mathematics

Try with Hints


area of triangle - solution

Join Oand D

Can you now finish the problem ……….

This is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments

can you finish the problem……..

area of triangle - solution

Given that AB and AC are the two sides of the Isosceles right triangle ABC and encloses a semicircle of area \(2\pi\), center of the semicircle is O.

Let OD=r be the radius of the semi-circle

then area of semi-circle be \(\frac{\pi r^2}{2}\)

Now \(\frac{\pi r^2}{2}\) = \(2\pi\)

\(\Rightarrow r^2=4\)

\(\Rightarrow r=2\)

this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments

They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, 4

The area of the triangle is \(\frac{1}{2} \times 4 \times 4\)=8

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AMC 8 Math Olympiad Number Theory USA Math Olympiad

Prime numbers | AMC 8, 2006| Problem 25

Try this beautiful problem from Algebra based on Prime numbers.

Algebra based on Number theory – AMC-8, 2009 – Problem 23


Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?

  • $14$
  • $12$
  • $16$

Key Concepts


Algebra

Number theory

card number

Check the Answer


Answer:$14$

AMC-8 (2006) Problem 25

Pre College Mathematics

Try with Hints


Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained.

Can you now finish the problem ……….

Obtain this even number would be to add another even number to 44

Can you finish the problem……..

Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38. Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23

Since the sum of the hidden primes is 2+17+23=42, the average of the primes is \(\frac{42}{3}=14\)

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AMC 8 Math Olympiad Number Theory

Largest and smallest numbers | AMC 8, 2006 | Problem 22

Try this beautiful problem from Algebra about Largest and smallest numbers from AMC-8, 2006.

Largest and smallest numbers | AMC-8, 2006|Problem 22


Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?

Largest and smallest numbers

  • 34
  • 12
  • 26

Key Concepts


Number theory

Number counting

integer

Check the Answer


Answer:$26$

AMC-8, 2006 problem 22

Challenges and Thrills in Pre College Mathematics

Try with Hints


the lower cells contain A,B and C,

Can you now finish the problem ……….

the second row will contain A+B and B+C and the top cell will contain A+2B+C.

can you finish the problem……..

Largest and smallest numbers

If the lower cells contain A,B and C, then the second row will contain A+B and B+C and the top cell will contain A+2B+C. To obtain the smallest sum, place 1 in the center cell and 2 and 3 in the outer ones. The top number will be 7 . For the largest sum, place 9 in the center cell and 7 and 8 in the outer ones. This top number will be 33. The difference is 33-7=26

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of pinwheel | AMC 8, 2007 | Problem 23

Try this beautiful problem from Geometry based on area of pinwheel.

Pinwheel – AMC 8, 2007


What is the area of the shaded pinwheel shown in the \(5 \times 5\) grid?

Area of pinwheel
  • $8$
  • $6$
  • $9$

Key Concepts


Geometry

Triangle

Square

Check the Answer


Answer:$6$

AMC-8 (2007) Problem 23

Pre College Mathematics

Try with Hints


Find the area of white space

Area of pinwheel- solution

Can you now finish the problem ……….

The area of the pinwheel =Total area of the square – The white space

can you finish the problem……..

Area of pinwheel- solution

Now The area of the square around the pinwheel is \(5 \times 5\)=25 sq.unit

Clearly in the above picture, there are four white triangles i.e (\(\triangle EOF, \triangle KOL, \triangle HOG, \triangle JOI)\) and four white small square(at the corner)

clearly HG=3 unit & OP=2.5 unit

Therefore the area of four Triangle = \(4 \times \frac{1}{2} \times HG \times OP= 4 \times \frac{1}{2} \times 3 \times\) 2.5=15 sq.unit

And also the area of  four white corner squares =\((4 \times 1)\)=4 sq.unit

The area of the pinwheel =Total area of the square – The white space =(25-15-4)=6 sq.unit

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AMC 8 Combinatorics Math Olympiad Probability

Probability | AMC 8, 2010 | Problem no. 24

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

Probability | AMC-8, 2007 |Problem 24


A bag contains four pieces of paper, each labeled with one of the digits 1,2,3 or 4.  with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

  • \(\frac{3}{4}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{4}\)

Key Concepts


probability

combination

Number counting

Check the Answer


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

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

Try with Hints


there are two ways that The combination of digits that give multiples of 3

Can you now finish the problem ……….

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4)

can you finish the problem……..

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4) . The number of ways to choose three digits out of four is 4. Therefore, the probability is \(\frac{1}{2}\)

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AMC 8 Geometry Math Olympiad

Area of Circle Problem | AMC 8, 2008 | Problem 25

Try this beautiful problem from Geometry based on the Area of a Circle.

Area of Circle | AMC-8, 2008 | Problem 25


Margie’s winning art design is shown. The smallest circle has radius  2 inches, with each successive circle’s radius increasing by 2 inches. Approximately what percent of the design is black?

Area of circle problem
  • $44$
  • $42$
  • $45$

Key Concepts


Geometry

Area

Circle

Check the Answer


Answer:$42$

AMC-8, 2008 problem 25

Pre College Mathematics

Try with Hints


Area of the square is \(\pi (r)^2\),where \(r\)=radius of the circle

Can you now finish the problem ……….

Find the total area of the black region……..

can you finish the problem……..

Area of circle problem

Given that The smallest circle has radius  2 inches, with each successive circle’s radius increasing by 2 inches .

The radius of the 1st circle is 2, So the area is \(\pi(2)^2\)=4\(\pi\) sq.unit

The radius of the 2nd circle is 4, So the area is \(\pi(4)^2\)=16\(\pi\) sq.unit

The radius of the 3rd circle is 6 So the area is \(\pi(6)^2\)=36\(\pi\) sq.unit

The radius of the 4th circle is 8, So the area is \(\pi(8)^2\)=64\(\pi\) sq.unit

The radius of the 5th circle is 10, So the area is \(\pi(10)^2\)=100\(\pi\) sq.unit

The radius of the 6th circle is 12, So the area is \(\pi(12)^2\)=144\(\pi\) sq.unit

Therefore The entire circle’s area is 144\(\pi\)

The area of the black regions is \((100\pi-64\pi)+(36\pi-16\pi)+4\pi=60\pi \)sq.unit

The percentage of the design that is black is  \((\frac{60\pi}{144\pi} \times 100)\%=(\frac{5}{12} \times 100) \% \approx 42\%\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Total surface area of a cube | AMC-8, 2009 | Problem 25

Try this beautiful problem from Geometry: Total surface area of a cube

Total surface area of a cube – AMC-8, 2009- Problem 25


A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is \(\frac{1}{2}\) foot from the top face. The second cut is  \(\frac{1}{3}\) foot below the first cut, and the third cut is \(\frac{1}{17}\) foot below the second cut. From the top to the bottom the pieces are labeled A,B,C and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?

 a cube
stair
  • $10$
  • $11$
  • $12$

Key Concepts


Geometry

Cube

surface area

Check the Answer


Answer:$11$

AMC-8 (2009) Problem 25

Pre College Mathematics

Try with Hints


Calculate the surface area side by side

Can you now finish the problem ……….

The total height be 1

can you finish the problem……..

total surface of a cube
stair

Clearly The tops of A,B,C, and D in the figure such that 1+1+1+1=4 as do the bottoms .

Thus the total surface area is 8.

 Now, one of the sides has area one, since it combines all of the heights of A,B,C and D which is 1

The other side also satisfies this. Thus the total area now is  10.

Now From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like A, with surface area half. From the back it is the same thing. Thus, the total is \(10+\frac{1}{2}+\frac{1}{2}\)=11

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Algebra AMC 8 Math Olympiad USA Math Olympiad

Quadratic equation Problem | AMC 8, 2009 | Problem 23

Try this beautiful problem from Algebra based on Quadratic equation.

Algebra based on Quadratic equation – AMC-8, 2009 – Problem 23


On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought  400 jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

  • $34$
  • $28$
  • $25$

Key Concepts


Algebra

quadratic equation

Factorization

Check the Answer


Answer:$28$

AMC-8 (2009) Problem 23

Pre College Mathematics

Try with Hints


Let the number of girls be x

so the number of boys be x+2

Can you now finish the problem ……….

She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

can you finish the problem……..

Let the number of girls be x

so the number of boys be x+2

she gave each girl x jellybeans and each boy x+2 jellybeans,

Therefore she gave total number of  jelly beans to girls be \(x^2\)

Therefore she gave total number of  jelly beans to boys be \((x+2)^2\)

 She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

\(\Rightarrow x^2 + x^2 +4x+4=394\)

\(\Rightarrow 2x^2 +4x-390=0\)

\(\Rightarrow x^2 +2x -195=0\)

\((x+15)(x-13)=0\)

i.e x=-15 , 13 (we neglect negetive as number of students can not be negetive )

Therefore x=13 i.e number os girls be 13 and number of boys be 13+2=15

total number of students ne 13+15=28

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of the triangle and square | AMC 8, 2008 | Problem 23

Try this beautiful problem from AMC-8, 2008 based on the Ratio of the area of the triangle and square.

Area of the star and circle – AMC- 8, 2008 – Problem 23


In square ABCE, AF=2FE and CD=2DE .what is the ratio of the area of \(\triangle BFD\) to the area of square ABCE?

Area of the triangle and square
  • $\frac{7}{20}$
  • $\frac{5}{18}$
  • $\frac{11}{20}$

Key Concepts


Geometry

Triangle

Square

Check the Answer


Answer:$\frac{5}{18}$

AMC-8 (2008) Problem 23

Pre College Mathematics

Try with Hints


Area of the square =\((side)^2\)

Area of triangle =\(\frac{1}{2} \times base \times height\)

Can you now finish the problem ……….

Area of the triangle and square

The area of the \(\triangle BFD\)=(The area of the square ABCE- The area of \(\triangle ABF\) -The area of\( \triangle BCD\) -Area of the\( \triangle EFD) \)

can you finish the problem……..

Square and triangle

Let us assume that the side length of the given square is 6 unit

Then clearly AB=BC=CE=EA=6 unit & AF=4 unit,EF=2 unit, CD=4 unit

Total area of the square is \(6^2\)=36 sq.unit

Area of the \(\triangle ABF=\frac{1}{2}\times AB \times AF= \frac{1}{2}\times 6 \times 4= 12\) sq.unit

Area of the \(\triangle BCD=\frac{1}{2}\times BC \times CD= \frac{1}{2}\times 6 \times 4= 12\) sq.unit

Area of the \(\triangle EFD=\frac{1}{2}\times EF \times ED= \frac{1}{2}\times 2 \times 2= 2\) sq.unit

The area of the \(\triangle BFD\)=(The area of the square ABCE- The area of \(\triangle ABF\) -The area of\( \triangle BCD\) -Area of the\( \triangle EFD)=(36-12-12-2)=10 \)sq.unit

the ratio of the area of \(\triangle BFD\) to the area of square ABCE=\(\frac{10}{36}=\frac{5}{18}\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of trapezoid | AMC 8, 2011|Problem 20

Try this beautiful problem from Geometry: The area of trapezoid.

The area of the trapezoid – AMC-8, 2011 – Problem 20


Quadrilateral  ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid?

area of trapezoid
  • $700$
  • $750$
  • $800$

Key Concepts


Geometry

Trapezoid

Area of Triangle

Check the Answer


Answer:$750$

AMC-8(2011) Problem 20

Pre College Mathematics

Try with Hints


Draw altitudes from the top points A and B to CD at X and Y points

Can you now finish the problem ……….

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times\) (height between AB and CD)

can you finish the problem……..

a trapezoid

Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle .

Using The Pythagorean theorem on \(\triangle ADX and \triangle BYC\) ,

\((DX)^2+(AX)^2=(AD)^2\)

\(\Rightarrow (a)^2+(12)^2=(15)^2\)

\(\Rightarrow a=\sqrt{(15)^2-(12)^2}=\sqrt {81} =9\)

and

\((BY)^2+(YC)^2=(BC)^2\)

\(\Rightarrow (12)^2+(b)^2=(20)^2\)

\(\Rightarrow b=\sqrt{(20)^2-(12)^2}=\sqrt {256} =16\)

Now ABYX is a Rectangle so \(XY=AB=50\)

\(CD=DX+XY+YC=a+XY+b=9+50+16=75\)

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times (height between AB and CD)=\frac {1}{2} \times (AB+CD) \times 12=750\)

.

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