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

Revision as of 23:51, 23 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 $OG=\frac13 OC=4.$

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 $\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.$

~MRENTHUSIASM

2018 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2018 AMC 12B Problems/Problem 8"

Revision as of 23:51, 23 October 2021

Problem

Solution

See Also

@@ Line 1: / Line 1: @@
 ==Problem ==
-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?
+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) } 25 \qquad \textbf{(B) } 38  \qquad \textbf{(C) } 50  \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75  </math>
 ==Solution==
-Draw the Median connecting C to the center O of the circle. Note that the centroid is <math>\frac{1}{3}</math> of the distance from O to C.
+For each <math>\triangle ABC,</math> note that the length of one median is <math>OC=12.</math> Let <math>G</math> be the centroid of <math>\triangle ABC.</math> It follows that <math>OG=\frac13 OC=4.</math>
-Thus, as C traces a circle of radius 12, the Centroid will trace a circle of radius <math>\frac{12}{3}=4</math>.
-The area of this circle is <math>\pi\cdot4^2=16\pi \approx 50</math>.
+As shown below, <math>\triangle ABC_1</math> and <math>\triangle ABC_2</math> are two shapes of <math>\triangle ABC</math> with centroids <math>G_1</math> and <math>G_2,</math> 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 <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>
+~MRENTHUSIASM
 ==See Also==
 {{AMC12 box|year=2018|ab=B|num-a=9|num-b=7}}
 {{MAA Notice}}
+[[Category:Intermediate Geometry Problems]]