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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Cones and circle | AIME I, 2008 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Cones and circle.

Cones and circle – AIME I, 2008


A right circular cone has base radius r and height h the cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cones base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making 17 complete rotations. The value of \(\frac{h}{r}\) can be written in the form \(m{n}^\frac{1}{2}\) where m and n are positive integers and n in not divisible by the square of any prime, find m+n.

  • is 107
  • is 14
  • is 840
  • cannot be determined from the given information

Key Concepts


Cones

Circles

Algebra

Check the Answer


Answer: is 14.

AIME I, 2008, Question 5

Geometry Vol I to IV by Hall and Stevens

Try with Hints


The path is circle with radius =\(({r}^{2}+{h}^{2})^\frac{1}{2}\) then length of path=\(2\frac{22}{7}({r}^{2}+{h}^{2})^\frac{1}{2}\)

length of path=17 times circumference of base then \(({r}^{2}+{h}^{2})^\frac{1}{2}\)=17r then \({h}^{2}=288{r}^{2}\)

then \(\frac{h}{r}=12{(2)}^\frac{1}{2}\) then 12+2=14.

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AMC 8 USA Math Olympiad

Perimeter of a circle : AMC 8 2013 Problem 25

What is the area and perimeter of a circle?


A circle is a curve which maintains same distance from a fixed point called center.

The perimeter of a circle is the length of the curve and area of a circle is portion of a plane bounded by the curve.

Try the problem


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?

$\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi$

AMC 8 2013 Problem 25

Geometry : Perimeter of a circle

7 out of 10

Mathematical Circles.

Knowledge Graph


Perimeter of a circle- knowledge graph

Use some hints


First I want to give you the formula required.

You can clearly notice that we have to find the perimeters of all of the semicircles

The perimeter of a circle of radius $r$ unit can be obtained by the formula $2\pi r$. Then can you find perimeter of the semicircles ?!!!

So using the formula, the perimeters of

Semicircle 1 =$\frac{2\pi\times 100}{2}$ inches.

Semicircle 2 =$\frac{2\pi\times 60}{2}$ inches.

Semicircle 3 =$\frac{2\pi\times 80}{2}$ inches.

So the total path covered by the ball is

$\pi(100+60+80)=240\pi$ inches.

Is it the final answer??? Or have we ignored something ?

OK !!! please notice that they have asked for the distance covered by the center of the ball.

And the ball is of radius \(2\) inches.

So for the \(1^{st}\) and \(3^{rd}\) semicircle : The center will roll along a semicircular path of radius \(R_1-2\) and \(R_3-2\).

See this image :

And for the \(2^{nd}\) semicircle : The center will roll along a semicircular path of radius \(R_2+2\).

See the image below :

So the length of the path covered by the center of the ball is

\([\pi(100-2)+\pi(60+2)+\pi(80-2)] \quad \text{inches} \\=\pi(98+62+78) \quad \text{inches}\\=238\pi \quad \text{inches}\).

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AMC 8 USA Math Olympiad

2D Geometry – Areas related to circle AMC 8 2017 Problem 25

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What are we learning ?

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Competency in Focus: 2D Geometry (Areas related to circle)

This problem from American Mathematics Contest 8 (AMC 8, 2017) is based on calculation of areas related to circle. It is Question no. 25 of the AMC 8 2017 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/amc8_2017_25.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″]In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $TR$ and $SR$ are each one-sixth of a circle with radius 2. What is the area of the region shown? $\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$[/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=”on”]American Mathematical Contest 2017, AMC 8 Problem 25[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″ inline_fonts=”Abhaya Libre”]

Finding the area of a triangle and sector of a circle. (Area related to circles)

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]5/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.2.2″ open=”off”]Pre college 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″]C0nstruction : Let $X$ and $Y$ are the centres of the scetors $ST$ and $TR$ Now Let us join $SX$ and $TY$ What do you think? Will the points $U,S,\textbf{ and}\quad X$ be in a straightline?[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]$U,S,\textbf{ and}\quad X$ will be in a straight line because $\angle STU =60^{\circ}$ And angle of a  circle is $360$  i.e., $\angle SXR = \angle TYR = 60^{\circ}$ [Since sector($SXR$)=$\frac{1}{6}circle$] Then $UXY$ will make an equilateral triangle.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]So after construction the figure will look like this : Therefore, The required area = Area of $\triangle UXY$ – $2 \times$ Area of the sector $SXR$.  [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Area of equilateral triangle $\triangle UXY= 4\sqrt{3}$ And the are of sector $SXR= \frac{2\pi}{3}$ ANS : $4\sqrt{3}-\frac{4\pi}{3}$[/et_pb_tab][et_pb_tab title=”Formulas Used ” _builder_version=”4.2.2″]Area of an equilateral triangle =$\frac{a^2\sqrt{3}}{4}$ [where $a$ is a sied of the triangle] Area of a sector of a circle of angle $\theta$ = $\frac{\theta}{360}\pi r^2$ [where $r$ is the radius of the circle][/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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