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

m (→‎Solution)
m (→‎Solution)
Line 35: Line 35:
 
draw(A--B^^A--C1--B^^A--C2--B);
 
draw(A--B^^A--C1--B^^A--C2--B);
 
draw(O--C1^^O--C2);
 
draw(O--C1^^O--C2);
−
dot(M1,red+linewidth(1),UnFill);
+
dot(M1,red+linewidth(0.9),UnFill);
−
dot(M2,red+linewidth(1),UnFill);
+
dot(M2,red+linewidth(0.9),UnFill);
 
</asy>
 
</asy>
 
Therefore, point <math>G</math> traces out a circle (missing two points) with the center <math>O</math> and the radius <math>\overline{OG},</math> as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is <math>\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.</math>
 
Therefore, point <math>G</math> traces out a circle (missing two points) with the center <math>O</math> and the radius <math>\overline{OG},</math> as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is <math>\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.</math>

Revision as of 21:51, 6 October 2021

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?

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

Solution

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 $OG=\frac13 OC=4.$

Two shapes of $\triangle ABC,$ namely $\triangle ABC_1$ and $\triangle ABC_2$ with their respective centroids $G_1$ and $G_2,$ are shown below: [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(0.9),UnFill); dot(M2,red+linewidth(0.9),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 $\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.$

~MRENTHUSIASM

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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