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| − | ==Problem ==
| + | #REDIRECT [[2018_AMC_10B_Problems/Problem_12]] |
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| − | 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?
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| − | <math>\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75 </math>
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| − | ==Solution 1==
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| − | 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>
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| − | 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:
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| − | <asy>
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| − | /* Made by MRENTHUSIASM */
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| − | size(200);
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| − | pair O, A, B, C1, C2, G1, G2, M1, M2;
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| − | O = (0,0);
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| − | A = (-12,0);
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| − | B = (12,0);
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| − | C1 = (36/5,48/5);
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| − | C2 = (-96/17,-180/17);
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| − | G1 = O + 1/3 * C1;
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| − | G2 = O + 1/3 * C2;
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| − | M1 = (4,0);
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| − | M2 = (-4,0);
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| − | draw(Circle(O,12));
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| − | draw(Circle(O,4),red);
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| − | dot("$O$", O, (3/5,-4/5), linewidth(4.5));
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| − | dot("$A$", A, W, linewidth(4.5));
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| − | dot("$B$", B, E, linewidth(4.5));
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| − | dot("$C_1$", C1, dir(C1), linewidth(4.5));
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| − | dot("$C_2$", C2, dir(C2), linewidth(4.5));
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| − | dot("$G_1$", G1, 1.5*E, linewidth(4.5));
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| − | dot("$G_2$", G2, 1.5*W, linewidth(4.5));
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| − | draw(A--B^^A--C1--B^^A--C2--B);
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| − | draw(O--C1^^O--C2);
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| − | dot(M1,red+linewidth(0.8),UnFill);
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| − | dot(M2,red+linewidth(0.8),UnFill);
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| − | </asy>
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| − | 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>
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| − | ~MRENTHUSIASM
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| − | ==Solution 2==
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| − | We assign coordinates. Let <math>A = (-12,0)</math>, <math>B = (12,0)</math>, and <math>C = (x,y)</math> lie on the circle <math>x^2 +y^2 = 12^2</math>. Then, the centroid of <math>\triangle ABC</math> is <math>G = \left(\frac{-12 + 12 + x}{3}, \frac{0 + 0 + y}{3}\right) = \left(\frac x3,\frac y3\right)</math>. Thus, <math>G</math> traces out a circle with a radius <math>\frac13</math> of the radius of the circle that point <math>C</math> travels on. Thus, <math>G</math> traces out a circle of radius <math>\frac{12}{3} = 4</math>, which has area <math>16\pi\approx \boxed{\textbf{(C) } 50}</math>.
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| − | ==Solution 3==
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| − | 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 <math>C</math> 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 <math>\angle C = \frac{180^\circ}{2} = 90^\circ</math>. Now we know that all triangles <math>ABC</math> 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 <math>45</math>-<math>45</math>-<math>90</math> triangle in conjunction with the properties of a centroid, we can quickly see that the length of the centroid is <math>4</math> now we can plug it in to the area formula where we get <math>16\pi\approx\boxed{\textbf{(C) } 50}</math>.
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| − | ==See Also==
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| − | {{AMC12 box|year=2018|ab=B|num-a=9|num-b=7}}
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| − | {{MAA Notice}}
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| − | [[Category:Intermediate Geometry Problems]] | |
