Difference between revisions of "2018 AMC 12B Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | Draw the Median connecting C to the center O of the circle. Note that the centroid is 1/ | + | 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 | + | 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. | + | 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 16:57, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
