2018 AMC 10B Problems/Problem 12

The following problem is from both the 2018 AMC 12B #8 and 2018 AMC 10B #12, so both problems redirect to this page.


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?

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

Solution 1

For each $\triangle ABC,$ note that the length of one median is $OC=12.$ Let $G$ be the centroid of $\triangle ABC.$ It follows that $OG=\frac13 OC=4.$

As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ respectively: [asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C1, C2, G1, G2, M1, M2; O = (0,0); A = (-12,0); B = (12,0); C1 = (36/5,48/5); C2 = (-96/17,-180/17); G1 = O + 1/3 * C1; G2 = O + 1/3 * C2; M1 = (4,0); M2 = (-4,0);  draw(Circle(O,12)); draw(Circle(O,4),red);  dot("$O$", O, (3/5,-4/5), linewidth(4.5)); dot("$A$", A, W, linewidth(4.5)); dot("$B$", B, E, linewidth(4.5)); dot("$C_1$", C1, dir(C1), linewidth(4.5)); dot("$C_2$", C2, dir(C2), linewidth(4.5)); dot("$G_1$", G1, 1.5*E, linewidth(4.5)); dot("$G_2$", G2, 1.5*W, linewidth(4.5)); draw(A--B^^A--C1--B^^A--C2--B); draw(O--C1^^O--C2); dot(M1,red+linewidth(0.8),UnFill); dot(M2,red+linewidth(0.8),UnFill); [/asy] Therefore, point $G$ traces out a circle (missing two points) with the center $O$ and the radius $\overline{OG},$ as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is $\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.$


Solution 2

We assign coordinates. Let $A = (-12,0)$, $B = (12,0)$, and $C = (x,y)$ lie on the circle $x^2 +y^2 = 12^2$. Then, the centroid of $\triangle ABC$ is $G = \left(\frac{-12 + 12 + x}{3}, \frac{0 + 0 + y}{3}\right) = \left(\frac x3,\frac y3\right)$. Thus, $G$ traces out a circle with a radius $\frac13$ of the radius of the circle that point $C$ travels on. Thus, $G$ traces out a circle of radius $\frac{12}{3} = 4$, which has area $16\pi\approx \boxed{\textbf{(C) } 50}$.

Solution 3

First we can draw a few conclusions from the given information. Firstly we can see clearly that the distance from the centroid to the center of the circle will remain the same no matter $C$ is on the circle. Also we can see that because the two legs of every triangles will always originate on the diameter, using inscribed angle rules, we know that $\angle C = \frac{180^\circ}{2} = 90^\circ$. Now we know that all triangles $ABC$ will be a right triangle. We also know that the closed curve will simply be a circle with radius equal to the centroid of each triangle. We can now pick any arbitrary triangle, calculate its centroid, and the plug it in to the area formula. Using a $45^\circ$-$45^\circ$-$90^\circ$ triangle in conjunction with the properties of a centroid, we can quickly see that the length of the centroid is $4$ now we can plug it in to the area formula where we get $16\pi\approx\boxed{\textbf{(C) } 50}$.



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See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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
2018 AMC 12B (ProblemsAnswer KeyResources)
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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