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AMC 8 Math Olympiad USA Math Olympiad

Problem on Circumscribed Circle | AMC-10A, 2003 | Problem 17

Try this beautiful problem from Geometry based on Circumscribed Circle

Problem on Circumscribed Circle – AMC-10A, 2003- Problem 17


The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

  • \(\frac{5\sqrt3}{\pi}\)
  • \(\frac{3\sqrt3}{\pi}\)
  • \(\frac{3\sqrt3}{2\pi}\)

Key Concepts


Geometry

Triangle

Circle

Check the Answer


Answer: \(\frac{3\sqrt3}{\pi}\)

AMC-10A (2003) Problem 17

Pre College Mathematics

Try with Hints


Circumscribed circle figure

Let ABC is a equilateral triangle which is inscribed in a circle. with center \(O\). and also given that perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle.so for find out the peremeter of Triangle we assume that the side length of the triangle be \(x\) and the radius of the circle be \(r\). then the side of an inscribed equilateral triangle is \(r\sqrt{3}\)=\(x\)

Can you now finish the problem ……….

circumscribed circle

The perimeter of the triangle is=\(3x\)=\(3r\sqrt{3}\) and Area of the circle=\(\pi r^2\)

Now The perimeter of the triangle=The Area of the circle

Therefore , \(3x\)=\(3r\sqrt{3}\)=\(\pi r^2\)

can you finish the problem……..

Now \(3x\)=\(3r\sqrt{3}\)=\(\pi r^2\) \(\Rightarrow {\pi r}=3\sqrt 3\) \(\Rightarrow r=\frac{3\sqrt3}{\pi}\)

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AMC 10 Math Olympiad USA Math Olympiad

Area of Region in a Circle | AMC-10A, 2011 | Problem 18

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

Area of Region in a Circle – AMC-10A, 2011- Problem 18


Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$. What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ?

  • \(\pi\)
  • \(\frac{3\pi}{2}\)
  • \(2\)
  • \(6\)
  • \(\frac{5\pi}{2}\)

Key Concepts


Geometry

Circle

Rectangle

Check the Answer


Answer: \(2\)

AMC-10A (2011) Problem 18

Pre College Mathematics

Try with Hints


We have to find out the area of the shaded region .Given that three circles with radius \(1\) and Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$.so if we draw a rectangle as shown in given below then we can find out the required region by the area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$

can you finish the problem……..

Now area of the rectangle is \(2\times 1=2\)

Area of the half circle with center (gray shaded region)=\(\frac{\pi (1)^2}{2}\)

The area of the two sectors created by $A$ and $B$(blue region)=\(\frac{2\pi(1)^2}{4}\)

can you finish the problem……..

Therefore, the required region (gray region)=area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$=\(\frac{\pi (1)^2}{2}\)+\(2\times 1=2\)-\(\frac{2\pi(1)^2}{4}\)=\(2\)

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AMC 10 Math Olympiad USA Math Olympiad

Sectors in Circle | AMC-10A, 2012 | Problem 10

Try this beautiful problem from Geometry based on Sectors in Circle.

Sectors in Circle – AMC-10A, 2012- Problem 10


Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

  • \(6\)
  • \(12\)
  • \(14\)
  • \(8\)
  • \(16\)

Key Concepts


Geometry

Circle

AP

Check the Answer


Answer: \(8\)

AMC-10A (2012) Problem 10

Pre College Mathematics

Try with Hints


We have to find out  the degree measure of the smallest possible sector angle.Let $x$ be the smallest sector angle and $r$ be the difference between consecutive sector angles,

Therefore the angles are $x, x+r, a+2r, \cdots. x+11r$. Now we know that sum of the angles of all sectors of a circle is \(360^{\circ}\).Can you find out the values of \(x\) and \(r\)?

can you finish the problem……..

Therefore using the AP formula we will get ,

\(\frac{x+x+11r}{2} . 12=360\)

\(\Rightarrow x=\frac{60-11r}{2}\)

can you finish the problem……..

Since all sector angles are integers so $r$ must be a multiple of 2. Now an even integers for $r$ starting from 2 to minimize $x.$ We find this value to be 4 and the minimum value of $x$ to be \(\frac{60-11(4)}{2}=8\)

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AMC 8 Math Olympiad USA Math Olympiad

Area of The Region | AMC-8, 2017 | Problem 25

Try this beautiful problem from Geometry based on the area of Region

Find the area – AMC-8, 2017- Problem 25


In the figure shown, US and UT and are line segments each of length 2, and\(\angle TUS=60^{\circ}\). Arcs TR and SR  and are each one-sixth of a circle with radius 2. What is the area of the figure shown?

Area of the region

  • (4 – \(\frac{4\pi}{3})\)
  • (\(4\sqrt3\) – \(\frac{4\pi}{3})\)
  • (\(4\sqrt3\) – \(4\pi\)

Key Concepts


Geometry

Triangle

Circle

Check the Answer


Answer: (\(4\sqrt3\) – \(\frac{4\pi}{3})\)

AMC-8 (2017) Problem 25

Pre College Mathematics

Try with Hints


shaded region

We have to find out the shaded region. The above diagram is not a standard geometrical figure (such as triangle, square or circle, etc.). So we can not find out the area of the shaded region by any standard formula.

shaded portion

Now if we extend US and UT and joined XY( shown in the above figure ) then it becomes a Triangle shape i.e \(\triangle UXY\) and region XSR and region TRY are circular shapes. Now you have to find out the area of these geometrical figures…

Can you now finish the problem ……….

shaded portion

Given that US and UT and are line segments each of length 2 and Arcs TR and SR  and are each one-sixth of a circle with radius 2. Therefore UX=2+2=4,UY=2+2=4 and SX=2+2=4.Therefore \(\triangle UXY\) is an equilateral triangle with side length 4 and area of equilateral triangle =\(\frac{\sqrt 3}{4} (side)^2\) and the region XSR and region TRY are each one-sixth of a circle with radius 2.nor area of the circle =\(\pi (radius)^2\)

can you finish the problem……..

The area of the \(\triangle UXY=\frac{\sqrt 3}{4} (4)^2=4\sqrt3\) sq.unit

Now area of (region SXR + region TRY)=\(2 \times \frac{\pi (2)^2}{6}=\frac{4\pi}{3}\)

Therefore the area of the region USRT=Area of \(\triangle UXY\)- (region SXR + region TRY)=(\(4\sqrt3\) – \(\frac{4\pi}{3})\)

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AMC 8 Math Olympiad USA Math Olympiad

Area of the figure | AMC-8, 2014 | Problem 20

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

Area of the figure – AMC-8, 2014- Problem 20


Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle of radius $1$ is centered at $A$, a circle of radius $2$ is centered at $B$, and a circle of radius $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?

Area of the figure
  • $3.5$
  • $ 4.0$
  • $4.5$

Key Concepts


Geometry

Rectangle

Circle

Check the Answer


Answer: $4.0$

AMC-8 (2014) Problem 20

Pre College Mathematics

Try with Hints


Area of the red shaded portion

To Find out the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure), we have to find out the area of the Rectangle -the area of three quarter circle inside the circle(i.e green shaded region)

Can you now finish the problem ……….

To find out the area of the rectangle, AD=5 and CD=3 are given. to find out the area of the three-quarter circles, the radii are 1,2 & 3 respectively.

Now area of Rectangle=\(AD \times CD\) and area of  quarter circles =\(\frac{\pi r^2}{4}\),where \(r\)=Radius of the circle

can you finish the problem……..

Area of the red shaded portion

Area of the rectangle=\( 5 \times 3\)=15 sq.unit

Area of the quarter circle with the center C=\(\frac{\pi (3)^2}{4}\)=\(\frac{9 \pi}{4}\) sq.unit

Area of the quarter circle with the center B= \(\frac{\pi (2)^2}{4}\) =\(\frac{4 \pi}{4}\)=\(\pi\) sq.unit

Area of the quarter circle with the center C= \(\frac{\pi (1)^2}{4}\) =\(\frac{\pi}{4}\) sq.unit

Therefore the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure) =(15- \(\frac{9 \pi}{4}\)- \(\pi\) – \(\frac{\pi}{4}\) )=15-\(\frac{7\pi}{2}\)=15-11=4 sq.unit (Taking \(\pi =\frac{22}{7}\))

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AMC 8 Math Olympiad

Intersection of two Squares | AMC 8, 2004 | Problem 25

Try this beautiful problem from Geometry based on Intersection of two Squares.

When 2 Squares intersect | AMC-8, 2004 | Problem 25


Two \(4\times 4\) squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

Intersection of two Squares
  • \(28-2\pi\)
  • \(25-2\pi\)
  • \(30-2\pi\)

Key Concepts


Geometry

square

Circle

Check the Answer


Answer: \(28-2\pi\)

AMC-8, 2004 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 ……….

Clearly, if 2 squares intersect, it would be a smaller square with half the side length, 2.

can you finish the problem……..

Intersection of two Squares

Clearly the intersection of 2 squares would be a smaller square with half the side length, 2.

The area of this region =Total area of larger two squares – the area of the intersection, the smaller square i.e \(4^2+4^2 -2^2=28 \)

Now The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle

Using the Pythagorean th. diameter of the circle be \(\sqrt{2^2 +2^2}=2\sqrt 2\)

Radius=\(\sqrt 2\)

area of the square=\(\pi (\sqrt2)^2\)=\(2\pi\)

Area of the shaded region= 28-2\(\pi\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of Rectangle Problem | AMC 8, 2004 | Problem 24

Try this beautiful problem from Geometry from AMC-8, 2004 ,Problem-24, based on area of Rectangle.

Rectangle | AMC-8, 2004 | Problem 24


In the figure ABCD is a rectangle and EFGH  is a parallelogram. Using the measurements given in the figure, what is the length  d  of the segment that is perpendicular to  HE and FG?

Area of Rectangle Problem
  • $7.1$
  • $7.6$
  • $7.8$

Key Concepts


Geometry

Rectangle

Parallelogram

Check the Answer


Answer:$7.6$

AMC-8, 2004 problem 24

Pre College Mathematics

Try with Hints


Find Area of the Rectangle and area of the Triangles i.e \((\triangle AHE ,\triangle EBF , \triangle FCG , \triangle DHG) \)

Can you now finish the problem ……….

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of\((\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) \)

can you finish the problem……..

Area of Rectangle Problem

Area of the Rectangle =\(CD \times AD \)=\(10 \times 8\)=80 sq.unit

Area of the \(\triangle AHE\) =\(\frac{1}{2} \times AH \times AE \)= \(\frac{1}{2} \times 4 \times 3\) =6 sq.unit

Area of the \(\triangle EBF\) =\(\frac{1}{2} \times EB \times BE \)= \(\frac{1}{2} \times 6 \times 5\) =15 sq.unit

Area of the \(\triangle FCG\) =\(\frac{1}{2} \times GC \times FC\)= \(\frac{1}{2} \times 4\times 3\) =6 sq.unit

Area of the \(\triangle DHG\) =\(\frac{1}{2} \times DG \times DH \)= \(\frac{1}{2} \times 6 \times 5\) =15 sq.unit

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of\((\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) \)=\(80-(6+15+6+15)=80-42=38\)

As ABCD is a Rectangle ,\(\triangle GCF\) is a Right-angle triangle,

Therefore GF=\(\sqrt{4^2 + 3^2}\)=5 sq.unit

Now Area of the parallelogram EFGH=\( GF \times d\)=38

\(\Rightarrow 5 \times d\)=38

\(\Rightarrow d=7.6\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Radius of the Circle | AMC-8, 2005 | Problem 25

Try this beautiful problem from Geometry: Radius of a circle

Radius of a circle – AMC-8, 2005- Problem 25


A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

radius of the circle
  • $\frac{5}{\sqrt \pi}$
  • $ \frac{2}{\sqrt \pi} $
  • $\sqrt \pi$

Key Concepts


Geometry

Cube

square

Check the Answer


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

AMC-8 (2005) Problem 25

Pre College Mathematics

Try with Hints


The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Can you now finish the problem ……….

Shaded region of the figure

Region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square

The area of the circle -x=Area of the square – x

can you finish the problem……..

Shaded region of the figure

Given that The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Let the region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square .

Let r be the radius of the circle

Therefore, The area of the circle -x=Area of the square – x

so, \(\pi r^2 – x=4-x\)

\(\Rightarrow \pi r^2=4\)

\(\Rightarrow r^2 = \frac{4}{\pi}\)

\(\Rightarrow r=\frac{2}{\sqrt \pi}\)

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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

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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