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AMC 8 USA Math Olympiad

Geometry of circles in AMC 8 2014 problem 25

Try this beautiful problem from AMC 8. It involves geometry of circles. We provide sequential hints so that you can try the problem.

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What are we learning ?

[/et_pb_text][et_pb_text _builder_version=”4.0.9″ 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″]Competency in Focus:Geometry of circles. This problem from American Mathematics contest (AMC 8, 2014) is based on simple counting of semicircles.[/et_pb_text][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″]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/01/AMC8-2014-problem-25.png” align=”center” _builder_version=”4.0.9″][/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″]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.0.9″ 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 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 5 miles per hour, how many hours will it take to cover the one-mile stretch?
Note: 1 mile = 5280feet amc 8 2014 problem no 25[/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=”4.0.9″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0.9″]American Mathematical Contest 2014, AMC 8 Problem 25[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.0.9″ open=”off”]Geometry of circles[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0.9″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics¬†

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Start with hints

[/et_pb_text][et_pb_tabs _builder_version=”4.0.9″ hover_enabled=”0″][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=”4.0.9″]How many lanes the highway consists of ? 2 right! given highway is 40 feet wide .Then width of each lane will be 40/2=20 feet wide .[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0.9″]Look at the diagram .See that the radius of each semicircle will be 20 feet on which Robert must be riding his bike .Again see that each semicrcle covers 40 feet of highway i.e. the diameter of the semicircle .[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0.9″]Calculate the number of semicircles over the whole mile . Number of semicircles=(length of the highway covered in total by Robert)/(length of highway covered by each semicircle)=5280/40 [since 1 mile=5280 feet] =132.[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0.9″ hover_enabled=”0″]Where the semicircles full circles ,their circumference would be 2.\( \pi \) r =2.\( \pi \).20=40¬†\( \pi \) feet (since r=radius=20 feet). Therefore the circumference of semicircles is half that, or 20.\( \pi \) feet. [/et_pb_tab][et_pb_tab title=”Hint 5″ _builder_version=”4.0.9″ hover_enabled=”0″]Therefore over the stretch of hghway, Robert rides a total of 132.20.\( \pi \)=2640. \( \pi \) feet equivalent to \(\frac{ \pi}{2} \)¬† mile( since 1 mile=5280 feet) Given Robert rides at 5 miles per hour.So, time required by Robert =distance travelled/rate=(\( \frac{\pi}{2} \) miles)/(5 miles per hour)= \( \frac{\pi}{10} \)hours.   [/et_pb_tab][/et_pb_tabs][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″ min_height=”12px” custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Connected Program at Cheenta

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Similar Problems

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2 replies on “Geometry of circles in AMC 8 2014 problem 25”

in the question, it was mentioned plainly as “feet wide” and I thought it was one feet. And also, “mile per hour” and I thought it was just a mile.

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