Difference between revisions of "2018 AMC 12B Problems/Problem 8"
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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>. | 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 <math>\pi\cdot4^2=16\pi \approx 50</math>. | + | The area of this circle is <math>\pi\cdot4^2=16\pi \approx \boxed{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 17:01, 16 February 2018
Problem
Line Segment is a diameter of a circle with . Point , not equal to or , lies on the circle. As point moves around the circle, the centroid (center of mass) of (insert triangle symbol) 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 of the distance from O to C. Thus, as C traces a circle of radius 12, the Centroid will trace a circle of radius .
The area of this circle is .
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
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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