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

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==Problem ==
 
==Problem ==
  
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Line segment <math>\overline{AB}</math> is a diameter of a circle with <math>AB = 24</math>. Point <math>C</math>, not equal to <math>A</math> or <math>B</math>, lies on the circle. As point <math>C</math> moves around the circle, the centroid (center of mass) of (insert triangle symbol)<math>ABC</math> traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
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Line segment <math>\overline{AB}</math> is a diameter of a circle with <math>AB = 24</math>. Point <math>C</math>, not equal to <math>A</math> or <math>B</math>, lies on the circle. As point <math>C</math> moves around the circle, the centroid (center of mass) of <math>\triangle ABC</math> traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
  
 
<math>\textbf{(A)} \indent 25 \qquad \textbf{(B)} \indent 32  \qquad \textbf{(C)} \indent 50  \qquad \textbf{(D)} \indent 63 \qquad \textbf{(E)} \indent 75  </math>
 
<math>\textbf{(A)} \indent 25 \qquad \textbf{(B)} \indent 32  \qquad \textbf{(C)} \indent 50  \qquad \textbf{(D)} \indent 63 \qquad \textbf{(E)} \indent 75  </math>

Revision as of 20:35, 19 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 $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

$\textbf{(A)} \indent 25 \qquad \textbf{(B)} \indent 32 \qquad \textbf{(C)} \indent 50 \qquad \textbf{(D)} \indent 63 \qquad \textbf{(E)} \indent 75$

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 $\pi\cdot4^2=16\pi \approx \boxed{50}$.

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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