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2018 AMC 10B Problems/Problem 12

Revision as of 19:29, 12 February 2019 by Adityaarora (talk | contribs) (→‎Solution)

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(Coordinate Bash)

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

Solution 2 (no coordinates)

We know the centroid of a triangle splits the medians into segments of ratio 2:1, and the median of the triangle that goes to the center of the circle is the radius (length $12$), so the length from the centroid of the triangle to the center of the circle is always $\dfrac{1}{3} \cdot 12 = 4$. The area of a circle with radius $4$ is $16\pi$, or around $\boxed{\textbf{(C)} \text{ 50}}$. -That_Crazy_Book_Nerd

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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