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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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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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AMC 8 Geometry Math Olympiad

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?

area of a square- AMC 8
  • 9
  • 15
  • 17

Key Concepts


Geometry

Area

Square

Check the Answer


Answer:15

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 ……….

square that fits into the area

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

can you finish the problem……..

square that fits into the area

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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AMC 8 Geometry Math Olympiad

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?

Semicircle in an isosceles triangle

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

Key Concepts


Geometry

Area

pythagoras

Check the Answer


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

semicircle inscribed in an isosceles triangle

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……..

radius of a semicircle in an isosceles triangle

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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AMC 8 Geometry Math Olympiad

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

Check the Answer


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……..

Area of triangle- figure

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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AMC 8 Geometry Math Olympiad USA Math Olympiad

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 $R^2$ ?

area of cube's cross section

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

Key Concepts


Geometry

Area

Pythagorean theorem

Check the Answer


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……..

area of cube's cross section

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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AMC 8 Geometry Math Olympiad

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?

radius of a semi circle

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

Key Concepts


Geometry

congruency

similarity

Check the Answer


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……..

Radius of a semi circle

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 =$\frac{10}{3}$

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