Difference between revisions of "2018 AMC 12B Problems/Problem 8"

(Solution)
(Solution)
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==Solution==
 
==Solution==
Draw the Median connecting C to the center O of the circle. Note that the centroid is 1/3 of the distance from O to C.
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Draw the Median connecting C to the center O of the circle. Note that the centroid is <math>\frac{1}{3}</math> of the distance from O to C.
Thus, as C traces a circle of radius 12, the Centroid will trace a circle of radius 12/3=4.
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Thus, as C traces a circle of radius 12, the Centroid will trace a circle of radius <math>\frac{12}{3}=4</math>.
  
The area of this circle is pi*4^2=16pi ~ 50.
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The area of this circle is <math>pi*4^2=16pi ~ 50</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=B|num-a=9|num-b=7}}
 
{{AMC12 box|year=2018|ab=B|num-a=9|num-b=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:57, 16 February 2018

Problem

Line Segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of (insert triangle symbol)$ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

Solution

Draw the Median connecting C to the center O of the circle. Note that the centroid is $\frac{1}{3}$ of the distance from O to C. Thus, as C traces a circle of radius 12, the Centroid will trace a circle of radius $\frac{12}{3}=4$.

The area of this circle is $pi*4^2=16pi ~ 50$.

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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