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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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AMC 8 USA Math Olympiad

Number Theory Problem | AMC 10B 2019| Problem 19

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What are we learning ?

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||20px||false|false” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]

Competency in Focus: Number Theory

This problem from American Mathematics Contest 10B (AMC 10B, 2019) is based on calculation of number theory. It is Question no. 19 of the AMC 10B 2019 Problem series.

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”10px||10px||false|false” custom_padding=”10px|10px|10px|10px|false|false” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ inline_fonts=”Aclonica”]

First look at the knowledge graph:-

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/02/p19.png” alt=”calculation of mean and median- AMC 8 2013 Problem” title_text=” mean and median- AMC 8 2013 Problem” align=”center” force_fullwidth=”on” _builder_version=”4.2.2″ min_height=”429px” height=”189px” max_height=”198px” custom_padding=”10px|10px|10px|10px|false|false”][/et_pb_image][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ inline_fonts=”Aclonica”]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$ $\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px” hover_enabled=”0″][et_pb_accordion_item title=”Source of the problem” open=”off” _builder_version=”4.2.2″]American Mathematical Contest 2019, AMC 10B Problem 19[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.2.2″ inline_fonts=”Abhaya Libre” open=”off”]

Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.2.2″ hover_enabled=”0″ open=”on”]

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version=”4.0.9″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”48px||48px” custom_padding=”20px|20px|0px|20px||” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ inline_fonts=”Aclonica”]

Start with hints 

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]Any number is divisible by all of its factors. For eaxmple 50 is divisible by \(2,5,10\) and \(25\) out of these their are some prime numbers called Prime factors. [/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]The prime factor of 100,000 are only 2 and 5, the rest of them are not the prime factor, they are composite factor. Also The prime factorization of $100,000$ is $2^5 \cdot 5^5$.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]Any Number which divides 100,000 must be multiple of 2 and (or) 5. So it can be 10=5×2 or \(200=2^{3} 5^{2}\). [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Since prime factorization of $100,000$ is $2^5 \cdot 5^5$. Thus We can find possible value of a,b,c and d being between 0 and 5.[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]

AMC – AIME – USAMO Boot Camp for brilliant students. Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

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AMC 8 USA Math Olympiad

Probability AMC 10B 2019 problem 17

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What are we learning ?

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Competency in Focus: probability

This problem is from American Mathematics Contest 10B (AMC 10B, 2019). It is Question no. 17 of the AMC 10B 2019 Problem series.

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”10px||10px||false|false” custom_padding=”10px|10px|10px|10px|false|false” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ inline_fonts=”Aclonica”]

First look at the knowledge graph:-

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/02/p17.png” alt=”calculation of mean and median- AMC 8 2013 Problem” title_text=” mean and median- AMC 8 2013 Problem” align=”center” force_fullwidth=”on” _builder_version=”4.2.2″ min_height=”429px” height=”189px” max_height=”198px” custom_padding=”10px|10px|10px|10px|false|false”][/et_pb_image][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ inline_fonts=”Aclonica”]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3….$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.2.2″ open=”off”]American Mathematical Contest 2019, AMC 10B Problem 17[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″ inline_fonts=”Abhaya Libre”]

Probability

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”on”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version=”4.0.9″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”48px||48px” custom_padding=”20px|20px|0px|20px||” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ inline_fonts=”Aclonica”]

Start with hints 

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + … = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}$ by the geometric series sum formula.[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]the probability that the two balls will go into bins that have a distance of $2$ from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}$[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]We can see that each time we add a bin between the two balls, the probability halves.[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots$[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]

AMC – AIME – USAMO Boot Camp for brilliant students. Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

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