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Difference between revisions of "2018 AMC 10B Problems/Problem 12"
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<math>\textbf{(A)} \text{ 25} \qquad \textbf{(B)} \text{ 38} \qquad \textbf{(C)} \text{ 50} \qquad \textbf{(D)} \text{ 63} \qquad \textbf{(E)} \text{ 75}</math>
 
<math>\textbf{(A)} \text{ 25} \qquad \textbf{(B)} \text{ 38} \qquad \textbf{(C)} \text{ 50} \qquad \textbf{(D)} \text{ 63} \qquad \textbf{(E)} \text{ 75}</math>
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==Solution==
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Let <math>A=(-12,0),B=(12,0)</math>. Therefore, <math>C</math> lies on the circle with equation <math>x^2+y^2=144</math>. Let it have coordinates <math>(x,y)</math>. Since we know the centroid of a triangle with vertices  with coordinates of <math>(x_1,y_1),(x_2,y_2),(x_3,y_3)</math> is <math>\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)</math>, the centroid of <math>\triangle ABC</math> is <math>\left(\frac{x}{3},\frac{y}{3}\right)</math>. Because <math>x^2+y^2=144</math>, we know that <math>\left(\frac{x}{3}\right)^2+\left(\frac{y}{3}\right)^2=16</math>, so the curve is a circle centered at the origin. Therefore, its area is <math>16\pi\approx \boxed{\text{(C) }50}</math>.
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-tdeng
  
 
==See Also==
 
==See Also==

Revision as of 15:53, 16 February 2018

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?

$\textbf{(A)} \text{ 25} \qquad \textbf{(B)} \text{ 38} \qquad \textbf{(C)} \text{ 50} \qquad \textbf{(D)} \text{ 63} \qquad \textbf{(E)} \text{ 75}$

Solution

Let $A=(-12,0),B=(12,0)$. Therefore, $C$ lies on the circle with equation $x^2+y^2=144$. Let it have coordinates $(x,y)$. Since we know the centroid of a triangle with vertices with coordinates of $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$, the centroid of $\triangle ABC$ is $\left(\frac{x}{3},\frac{y}{3}\right)$. Because $x^2+y^2=144$, we know that $\left(\frac{x}{3}\right)^2+\left(\frac{y}{3}\right)^2=16$, so the curve is a circle centered at the origin. Therefore, its area is $16\pi\approx \boxed{\text{(C) }50}$. -tdeng

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 10 Problems and Solutions

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